Working with practicles requires a deep thought of what one wants to investigate. Constant consultations discontinues the toughts and they get influenced by the external factors derived by the discusions.
It would be wise if the researchers think deeply on the issues at hand explore all the opportunities which will enable them to formulate an action plan. After this point then they can seek limited external clarification and support.
Let be real researchers with a Zeal to solve out the problems that are presented only they can we stop crawling.
Mr. G
Tuesday, October 23, 2007
Saturday, October 6, 2007
GRAVITATIONAL FORCE FIELD
Gravitational force and field - Giancoli, sections 5 - 7
1. State Newton's law of universal gravitation.
· By thinking about the force of gravity that causes an apple to fall to the ground and the force necessary to keep the moon in orbit around the earth, Newton was able to come up with his hypothesis. Namely, that a force exists between any two masses.
· The force between any two point masses is given by the formula
· M1 and M2 are the masses of the attracting bodies, r is the separation between the two bodies, and G is a universal constant which has the value G = 6.67 x 10-11 Nm2kg-2.
· The direction of the force is along the line joining the centers of the two masses. This is a force of attraction and the vector form of the equation introduces a minus (-) sign to show that the force is directed towards the attracting object.
· Objects may be considered as point masses if the objects are small in comparison to the distance between them.
· Uniform objects (constant density and spherical symmetry) like the earth may be taken as point masses, with all of the mass of the object located at its center.
2. Define gravitational field strength.
· A mass, M, creates a gravitational field in the space around it. The concept of a "force field" was invented to explain how forces could act over a distance, with no direct contact between the objects involved.
· The idea is that a small mass, m, at some point in space will interact with the field at that point in space created by the larger mass.
· Gravitational field strength is then defined as the ratio of the gravitational force exerted on the mass m to the amount of mass that m possesses.
· In equation form, , where g is the gravitational field strength, FG is the gravitational force exerted on m, and m is the mass of the object in question.
· The gravitational field strength describes not only the amount of force exerted on a unit mass, but also the direction of that force.
3. Derive an expression for the gravitational field as a function of distance from a point mass. This includes the field outside a spherical mass.
· Consider a large point mass, M, and a smaller test mass, m. We know that the large point mass will attract the small test mass toward the center of the large point mass with a force given by F = .
·
We can think of this interaction as the interaction between the two masses, m and M, or as the interaction between m and the gravitational field created by M. Using our definition of gravitational field strength, it is very easy to show that: .
· Since uniform objects may be taken as point masses, this result applies also to uniform objects as long as r is greater than robject. That is, we must be physically outside the object to use this result.
· This result shows us that the farther we get from an object the smaller the gravitational field strength becomes. We can show this in two ways.
4. Derive an expression for gravitational field at the surface of a planet.
· If we move to the surface of a planet, taking the planet to be uniform, then we can think of the planet's mass as being located at the center of the planet. We, on the surface of the planet, are located a distance of one planet radius from the center of the planet, so: , where Mp is the mass of the planet and rp is the radius of the planet.
· We should note at this point, that the gravitational force at the surface of the planet is what we have been calling the weight of the object. That is, FG = mg. The implication of this is that the gravitational field strength g is nothing more than the acceleration due to gravity experienced by an object!
·
Near the surface of the earth, we have difficulty recognizing the spherical shape of the earth (due to its large size). Moreover, as we move away from the surface of the earth we are not moving appreciably farther from the center of the earth. Taken together this means that as long as we stay near the surface of the earth, we can ignore the small changes in r that are experienced. Then, we can say that the gravitational field near the surface of the earth is uniform. This means that it has the same value everywhere AND it points in the same direction everywhere. We take the value of the uniform gravitational field strength of the earth to be 9.8 Nkg-1 and the direction to be downward (toward the surface of the earth).
· Finally, it is interesting to note that for uniform objects, like the earth, the gravitational field strength inside the surface of the planet behaves very differently from the gravitational field strength outside the planet. It can be shown using calculus and Gauss' law that inside the earth, the gravitational field strength is directly related to the distance from the center of the earth. A plot of g against r looks like this:
5. Solve problems involving gravitational forces and fields. Vector addition is required to find the gravitational field strength due to more than one mass.
An example: Stars A and B have the same radius but the mass of A is double that of B. Find the gravitational field at point P, a distance of 3 star radii from the center of each star if the stars are separated by a distance of 4 star radii.
Solution: First, note that since gravitational field is a vector quantity, we need to find both the magnitude and the direction of the field at point P. A picture will help!
To begin, let's look at the magnitude of g for each planet. gA =G(2M)/ (3r)2 and gB = GM/ (3r)2. So, it looks like gA = 2 gB. To simplify things, let's call gB =g. Then, gA = 2g. So here's what we've got:
Both vectors make the same angle with the horizontal line. To find the vector components of our two gravitational fields, we will need to know the sin and cosine of this angle. We can figure these out from the first picture. The side opposite the angle has a length of 2r. The hypotenuse has a length of 3r. So, the sine of the angle is 2r/3r = 2/3. The adjacent side has a length of r (from the pythagorean theorem). So this means the cosine of the angle is r / 3r = / 3. Now we are ready:
The horizontal component of gA = 2g cos q = 2 g / 3, to the left.
The horizontal component of gB = g cos q = g / 3, to the left.
So, the net horizontal component = (2 g / 3) + ( g / 3) = 3 g / 3 = g , to the left.
The vertical component of gA = 2g sin q = 2g(2/3) = 4g/3, upwards.
The vertical component of gB = g sin q = g(2/3) = 2g/3, downwards.
So, the net vertical component = 4g/3 - 2g/3 = 2g/3, upwards.
Then, to find the net magnitude of the gravitational field: gnet =
And, to find the angle: tan q = vert. comp / horiz comp = = 0.298. So, q = tan-1(0.298) = 16.6o above the horizontal.
Gravitational energy and potential - This material is not in your textbook.
6. Define gravitational potential energy and gravitational potential. Students should understand that the work done in moving a mass between two points in a gravitational field is independent of the path taken and that gravitational potential energy is taken to be zero at infinity.
· It turns out that our previous definition of gravitational potential energy (EP = mgh) is only good for a uniform gravitational field. When we move far enough away from the earth for g to be non-uniform we need to look at things in a different way.
· However, the concepts of energy conservation and potential energy as energy stored by doing work against gravity are still good. So, we need to look at work done against gravity.
· The form of the law of universal gravitation suggests that the force exists everywhere in space. However, as distances become very large, the force becomes very small. So, if we could move things infinitely far apart, no force would exist between them and no work would need to be done against gravity to move them around. This sounds like a place where there is no gravitational potential energy. So, we define gravitational potential energy to be zero when objects are infinitely far apart.
· But, we still have a problem -- the force of universal gravitation is not a constant force. So, to calculate the work done to separate two objects from each other, we must be looking at an area under a force vs. displacement curve. This area is not a simple area (like a rectangle or a triangle) and requires the rules of calculus. We will just look at the result: Work = - .
· Now, let's think about what we know from before and what we know now. Before, when we lifted an object away from the surface of the earth, it's potential energy increased and we could make the potential energy be zero wherever we wanted. Now, it would be nice if the change in potential energy as objects get farther apart is still positive (i.e. potential energy increases with distance) and the potential energy is zero when objects are infinitely far apart. How can we make this work?
· The answer is to make the potential energy negative! . Try it -- if we move two objects from a distance of r apart to a distance of 2rapart what happens? The initial potential energy is - GMm/r and the final potential energy is -GMm/2r. The change in potential enegy = final - initial = -GMm/2r - (-GMm/r) = +GMm/2r. The change is still positive, so it works!
· Finally, let's think about the work done one more time. The force we are working against always points towards the center of the stationary mass (the one we aren't moving). Any movement that is perpendicular to this direction requires no work be done (definition of work is movement parallel to the direction of force). So, we can make all kinds of crazy paths and it won't matter as long as the net distance from the center of the stationary object is the same. Another way of saying this is "the work done is independent of the path". It only depends on the starting and ending distances from the center of the stationary object.
7. State the expression for gravitational potential due to a point mass.
· Just like we defined the concept of gravitational field to describe the force between two masses separated by a distance r, we can define the concept of gravitational potential to describe the stored energy in the gravitational field when two objects are separated by a distance r.
· Gravitational Potential (V) is defined as the gravitational potential energy for a certain separation divided the mass of the moving object. In equation form: .
· So, just as force can be thought of as the interaction between a small mass and a gravitational field at a location in space, gravitational potential energy can be thought of as the interaction between a small mass and the gravitational potential at a location in space.
· Gravitational potential can be calculated using:
8. Explain the concept of escape speed.
· To talk about escape speed, we first need to look at the total energy of a small mass that is near a larger, fixed mass. E = EP + EK = . Now, we know that as the small object moves away from the larger object, its kinetic energy will decreases and its potential energy will increase. When the objects are infinitely far apart, the potential energy will be zero, the force between the objects will be zero, and the kinetic energy will maintain the value that it had when the force became zero.
· So, escape speed is then defined as the initial speed necessary to launch an object so that it will reach infinity and stay there. This means that its final speed will be exactly zero, or some value larger than zero. We usually take the escape speed to be the value necessary to exactly arrive at infinity with no kinetic energy remaining.
9. Derive an expression for the escape speed of an object from the surface of a planet.
· Using our definition of escape speed from above, it's not too hard to work this out. We will start out with our small object located at the surface of the planet. Let R be the radius of the planet. Then, the two masses can be thought of as initially separated by a distance R (because all of the mass of the planet is thought to be located at its center).
· So, the initial energy of our small object is .
· And, by definition the final energy of the small object when it reaches infinity will be Ef = 0.
· Conservation of energy allows us to set these two equations equal to one another.
·
· So,
· And, . This is the initial speed necessary to escape the gravitational field of the planet.
10. Solve problems involving gravitational potential energy and gravitational potential. These should include problems on escape speed.
Do these problems in Chapter 5: 23, 25, 27, 29, 31, 33, 62, 65, 66
Plus the following:
Gravitational Field
1. What is the magnitude of the gravitational field 2.00 m from a 5.00 kg mass?
2. At what distance above the surface of the earth is the magnitude of its gravitational field 4.90 m/s2 if the gravitational field at the surface has magnitude 9.80 m/s2?
3. What is the magnitude of the gravitational field of the moon at a height of 200 m above the moon’s surface? (Rmoon = 1.74 x 106 m and Mmoon = 7.35 x 1022 kg)
Gravitational Potential Energy
4. Calculate the escape velocity for an object for an object (a) from the surface of the moon (b) from the surface of saturn. (Msaturn = 5.69 x 1026 kg and Rsaturn = 6.03 x 107 m)
5. An artillery shell with mass m is shot vertically upward from the surface of the earth. If the shell’s initial speed is 8.00 x 103 m/s, to what height above the surface of the earth will it rise? (neglect air resistance) (Mearth = 5.98 x 1024 kg and Rearth = 6.38 x 106 m)
1. State Newton's law of universal gravitation.
· By thinking about the force of gravity that causes an apple to fall to the ground and the force necessary to keep the moon in orbit around the earth, Newton was able to come up with his hypothesis. Namely, that a force exists between any two masses.
· The force between any two point masses is given by the formula
· M1 and M2 are the masses of the attracting bodies, r is the separation between the two bodies, and G is a universal constant which has the value G = 6.67 x 10-11 Nm2kg-2.
· The direction of the force is along the line joining the centers of the two masses. This is a force of attraction and the vector form of the equation introduces a minus (-) sign to show that the force is directed towards the attracting object.
· Objects may be considered as point masses if the objects are small in comparison to the distance between them.
· Uniform objects (constant density and spherical symmetry) like the earth may be taken as point masses, with all of the mass of the object located at its center.
2. Define gravitational field strength.
· A mass, M, creates a gravitational field in the space around it. The concept of a "force field" was invented to explain how forces could act over a distance, with no direct contact between the objects involved.
· The idea is that a small mass, m, at some point in space will interact with the field at that point in space created by the larger mass.
· Gravitational field strength is then defined as the ratio of the gravitational force exerted on the mass m to the amount of mass that m possesses.
· In equation form, , where g is the gravitational field strength, FG is the gravitational force exerted on m, and m is the mass of the object in question.
· The gravitational field strength describes not only the amount of force exerted on a unit mass, but also the direction of that force.
3. Derive an expression for the gravitational field as a function of distance from a point mass. This includes the field outside a spherical mass.
· Consider a large point mass, M, and a smaller test mass, m. We know that the large point mass will attract the small test mass toward the center of the large point mass with a force given by F = .
·
We can think of this interaction as the interaction between the two masses, m and M, or as the interaction between m and the gravitational field created by M. Using our definition of gravitational field strength, it is very easy to show that: .
· Since uniform objects may be taken as point masses, this result applies also to uniform objects as long as r is greater than robject. That is, we must be physically outside the object to use this result.
· This result shows us that the farther we get from an object the smaller the gravitational field strength becomes. We can show this in two ways.
4. Derive an expression for gravitational field at the surface of a planet.
· If we move to the surface of a planet, taking the planet to be uniform, then we can think of the planet's mass as being located at the center of the planet. We, on the surface of the planet, are located a distance of one planet radius from the center of the planet, so: , where Mp is the mass of the planet and rp is the radius of the planet.
· We should note at this point, that the gravitational force at the surface of the planet is what we have been calling the weight of the object. That is, FG = mg. The implication of this is that the gravitational field strength g is nothing more than the acceleration due to gravity experienced by an object!
·
Near the surface of the earth, we have difficulty recognizing the spherical shape of the earth (due to its large size). Moreover, as we move away from the surface of the earth we are not moving appreciably farther from the center of the earth. Taken together this means that as long as we stay near the surface of the earth, we can ignore the small changes in r that are experienced. Then, we can say that the gravitational field near the surface of the earth is uniform. This means that it has the same value everywhere AND it points in the same direction everywhere. We take the value of the uniform gravitational field strength of the earth to be 9.8 Nkg-1 and the direction to be downward (toward the surface of the earth).
· Finally, it is interesting to note that for uniform objects, like the earth, the gravitational field strength inside the surface of the planet behaves very differently from the gravitational field strength outside the planet. It can be shown using calculus and Gauss' law that inside the earth, the gravitational field strength is directly related to the distance from the center of the earth. A plot of g against r looks like this:
5. Solve problems involving gravitational forces and fields. Vector addition is required to find the gravitational field strength due to more than one mass.
An example: Stars A and B have the same radius but the mass of A is double that of B. Find the gravitational field at point P, a distance of 3 star radii from the center of each star if the stars are separated by a distance of 4 star radii.
Solution: First, note that since gravitational field is a vector quantity, we need to find both the magnitude and the direction of the field at point P. A picture will help!
To begin, let's look at the magnitude of g for each planet. gA =G(2M)/ (3r)2 and gB = GM/ (3r)2. So, it looks like gA = 2 gB. To simplify things, let's call gB =g. Then, gA = 2g. So here's what we've got:
Both vectors make the same angle with the horizontal line. To find the vector components of our two gravitational fields, we will need to know the sin and cosine of this angle. We can figure these out from the first picture. The side opposite the angle has a length of 2r. The hypotenuse has a length of 3r. So, the sine of the angle is 2r/3r = 2/3. The adjacent side has a length of r (from the pythagorean theorem). So this means the cosine of the angle is r / 3r = / 3. Now we are ready:
The horizontal component of gA = 2g cos q = 2 g / 3, to the left.
The horizontal component of gB = g cos q = g / 3, to the left.
So, the net horizontal component = (2 g / 3) + ( g / 3) = 3 g / 3 = g , to the left.
The vertical component of gA = 2g sin q = 2g(2/3) = 4g/3, upwards.
The vertical component of gB = g sin q = g(2/3) = 2g/3, downwards.
So, the net vertical component = 4g/3 - 2g/3 = 2g/3, upwards.
Then, to find the net magnitude of the gravitational field: gnet =
And, to find the angle: tan q = vert. comp / horiz comp = = 0.298. So, q = tan-1(0.298) = 16.6o above the horizontal.
Gravitational energy and potential - This material is not in your textbook.
6. Define gravitational potential energy and gravitational potential. Students should understand that the work done in moving a mass between two points in a gravitational field is independent of the path taken and that gravitational potential energy is taken to be zero at infinity.
· It turns out that our previous definition of gravitational potential energy (EP = mgh) is only good for a uniform gravitational field. When we move far enough away from the earth for g to be non-uniform we need to look at things in a different way.
· However, the concepts of energy conservation and potential energy as energy stored by doing work against gravity are still good. So, we need to look at work done against gravity.
· The form of the law of universal gravitation suggests that the force exists everywhere in space. However, as distances become very large, the force becomes very small. So, if we could move things infinitely far apart, no force would exist between them and no work would need to be done against gravity to move them around. This sounds like a place where there is no gravitational potential energy. So, we define gravitational potential energy to be zero when objects are infinitely far apart.
· But, we still have a problem -- the force of universal gravitation is not a constant force. So, to calculate the work done to separate two objects from each other, we must be looking at an area under a force vs. displacement curve. This area is not a simple area (like a rectangle or a triangle) and requires the rules of calculus. We will just look at the result: Work = - .
· Now, let's think about what we know from before and what we know now. Before, when we lifted an object away from the surface of the earth, it's potential energy increased and we could make the potential energy be zero wherever we wanted. Now, it would be nice if the change in potential energy as objects get farther apart is still positive (i.e. potential energy increases with distance) and the potential energy is zero when objects are infinitely far apart. How can we make this work?
· The answer is to make the potential energy negative! . Try it -- if we move two objects from a distance of r apart to a distance of 2rapart what happens? The initial potential energy is - GMm/r and the final potential energy is -GMm/2r. The change in potential enegy = final - initial = -GMm/2r - (-GMm/r) = +GMm/2r. The change is still positive, so it works!
· Finally, let's think about the work done one more time. The force we are working against always points towards the center of the stationary mass (the one we aren't moving). Any movement that is perpendicular to this direction requires no work be done (definition of work is movement parallel to the direction of force). So, we can make all kinds of crazy paths and it won't matter as long as the net distance from the center of the stationary object is the same. Another way of saying this is "the work done is independent of the path". It only depends on the starting and ending distances from the center of the stationary object.
7. State the expression for gravitational potential due to a point mass.
· Just like we defined the concept of gravitational field to describe the force between two masses separated by a distance r, we can define the concept of gravitational potential to describe the stored energy in the gravitational field when two objects are separated by a distance r.
· Gravitational Potential (V) is defined as the gravitational potential energy for a certain separation divided the mass of the moving object. In equation form: .
· So, just as force can be thought of as the interaction between a small mass and a gravitational field at a location in space, gravitational potential energy can be thought of as the interaction between a small mass and the gravitational potential at a location in space.
· Gravitational potential can be calculated using:
8. Explain the concept of escape speed.
· To talk about escape speed, we first need to look at the total energy of a small mass that is near a larger, fixed mass. E = EP + EK = . Now, we know that as the small object moves away from the larger object, its kinetic energy will decreases and its potential energy will increase. When the objects are infinitely far apart, the potential energy will be zero, the force between the objects will be zero, and the kinetic energy will maintain the value that it had when the force became zero.
· So, escape speed is then defined as the initial speed necessary to launch an object so that it will reach infinity and stay there. This means that its final speed will be exactly zero, or some value larger than zero. We usually take the escape speed to be the value necessary to exactly arrive at infinity with no kinetic energy remaining.
9. Derive an expression for the escape speed of an object from the surface of a planet.
· Using our definition of escape speed from above, it's not too hard to work this out. We will start out with our small object located at the surface of the planet. Let R be the radius of the planet. Then, the two masses can be thought of as initially separated by a distance R (because all of the mass of the planet is thought to be located at its center).
· So, the initial energy of our small object is .
· And, by definition the final energy of the small object when it reaches infinity will be Ef = 0.
· Conservation of energy allows us to set these two equations equal to one another.
·
· So,
· And, . This is the initial speed necessary to escape the gravitational field of the planet.
10. Solve problems involving gravitational potential energy and gravitational potential. These should include problems on escape speed.
Do these problems in Chapter 5: 23, 25, 27, 29, 31, 33, 62, 65, 66
Plus the following:
Gravitational Field
1. What is the magnitude of the gravitational field 2.00 m from a 5.00 kg mass?
2. At what distance above the surface of the earth is the magnitude of its gravitational field 4.90 m/s2 if the gravitational field at the surface has magnitude 9.80 m/s2?
3. What is the magnitude of the gravitational field of the moon at a height of 200 m above the moon’s surface? (Rmoon = 1.74 x 106 m and Mmoon = 7.35 x 1022 kg)
Gravitational Potential Energy
4. Calculate the escape velocity for an object for an object (a) from the surface of the moon (b) from the surface of saturn. (Msaturn = 5.69 x 1026 kg and Rsaturn = 6.03 x 107 m)
5. An artillery shell with mass m is shot vertically upward from the surface of the earth. If the shell’s initial speed is 8.00 x 103 m/s, to what height above the surface of the earth will it rise? (neglect air resistance) (Mearth = 5.98 x 1024 kg and Rearth = 6.38 x 106 m)
NEWTONS LAW OF MOTION
Newton's third law - Giancoli, Ch 4, section 5 visit the Physics Classroom for this topic
1. State Newton's third law of motion.
· Newton's third law of motion describes how objects (more than one) interact with one another
· According to Newton, "For every action there is an equal and opposite reaction". This means that when two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.
· The interaction can be through contact -- like in a collision -- or at a distance -- like the earth and the moon. Regardless of the objects involved -- mass makes no difference here -- the two objects will exert equal forces on each other. The bug hits the windshield with the same force that the windshield hits the bug.
· While the forces are equal, the effect of the force (because of mass -- here it does matter) may be very different. Combining the second and third laws allows us to say: mA = Ma
2. Discuss examples of Newton's third law.
· In order to walk you must exert a force on the ground that is directed backwards. The reaction of the ground is to push you forward.
· The horse and cart
· Rockets lifting off -- the rocket pushes the exhaust particles out and the exhaust particles push the rocket
Inertial Mass, Gravitational Mass and Weight - Giancoli, Ch33, p. 948
3. Define inertial mass.
· Inertial mass is the mass that is present in Newton's second law of motion (Fnet = ma). Therefore, the definition of inertial mass is as follows: mi = Fnet / a. In words, inertial mass is the ratio of net force to acceleration.
4. Compare gravitational mass and inertial mass.
· Gravitational mass is the mass of an object that describes how it "feels" the gravitational force exerted by other masses. Near the earth, this mass is the mass responsible for the weight of an object. The definition of gravitational mass is as follows: mg = W / g
· While the concepts of inertial mass and gravitational mass arise from different areas of physics, they are mathematically equivalent. No experiment has been able to distinguish between the two.
· Consider the simple example of a mass in free fall near the surface of the earth. The only force acting on the object is the gravitational force (its weight). According to Newton's second law of motion, we may set Fnet = W. But, W = mgg and Fnet = mia. Therefore, we may state: mgg = mia. Finally, this means that the acceleration of an object in freefall, a, is given by the relation: a = (mg / mI) g. According to Galileo, all objects fall at the same rate of acceleration. This is only true if the ratio mg / mi has a value of 1. In other words, mg = mi.
5. Discuss the concept of weight.
· Weight, by definition, is the net gravitational force exerted on an object. W = mg
· Weight is typically measured using a support scale (like the bathroom scale you have at home). In most cases, the bathroom scale will report the true weight (defined as mg) of an object. This will be true as long as the scale and object are experiencing static or translational equilibrium -- neither is accelerating.
· In non-equilibrium cases, for example a support scale used in an accelerating elevator, the value given by the scale will reflect the motion of the object and will not report the "true weight" (= mg) of the object. The "apparent weight" of an object on a rising elevator will be greater than its true weight. The "apparent weight" of an object on a descending elevator will be less than its true weight.
6. Distinguish between mass and weight.
· Mass is a measure of the amount of matter contained within an object. We have seen that it can be inertial or gravitational in nature (#4 above). Mass is a fundamental property of matter and does not change with location.
· Weight is a specific force involving the concept of gravitational mass. Weight depends greatly on location and will change if the object is moved to a different location.
Momentum - Giancoli, Ch 7, all sections visit the Physics Classroom for this topic
7. Define linear momentum and impulse.
· Linear momentum is defined as the product of mass and velocity. p = mv
· Newton's second law of motion can be written as follows: F = ma = mDv / Dt = m(vf - vo) / Dt = (mvf - mvo) / Dt
Or, F = Dp / Dt. That is, the net force acting describes the rate at which its momentum changes.
· We can define a new quantity, impulse, which describes how momentum changes. From Newton's second law of motion, we see that Dp = F Dt. Impulse is calculated as F Dt. The result of the impulse is to change the momentum by an amount = Dp.
· Large forces acting over small time intervals produce the same impulse as small forces over large time intervals.
8. State the law of conservation of linear momentum.
· We will define a system of objects to contain all objects that will interact with one another.
· External forces are defined as forces that are exerted from outside the system of objects.
· Internal forces are defined s forces that are exerted by the objects within the system of objects.
· In the absence of external forces (an isolated system), the linear momentum of a system of objects is conserved. It can be neither created nor destroyed. It may, during the course of interactions between objects in the system, be transferred from one object to another.
v1
v1f
v2
v2f9. Derive the law of conservation of momentum for an isolated system consisting of two interacting particles.
The two objects in the diagram will interact. The result of the interaction is that each object will acquire a different final velocity. While they are interacting, Newton's third law of motion applies. Newton's second law describes the result for each object.
During the interaction: F1 = - F2 b/c of Newton's third law
We can write: Dp1 / Dt = - Dp2 / Dt b/c of Newton's second law
We can write: Dp1 = - Dp2 b/c Dt is the same for both objects
This means: p1f - p1o = - (p2f - p2o)
Simplify: p1f - p1o = -p2f + p2o
Rearrange: p1f + p2f = p1o + p2o
(p1 + p2)f = (p1 + p2)o
Group: pf = po Total momentum is conserved!
10. Solve problems involving momentum and impulse.
· Regardless of the type of collision involved, linear momentum for an isolated system (no external forces) is conserved.
· If external forces are present, then the momentum will change according to the impulse relation.
· It might sometimes be useful to classify collisions between objects by the way that energy is distributed in the collision. Those collisions for which kinetic energy is conserved are called elastic collisions. If the kinetic energy is not conserved, but is transformed into other types of energy then the collision is called an inelastic collision.
· To apply momentum conservation, calculate the individual momentum of each object (remembering that momentum is a vector quantity) prior to the collision and add them to find the initial momentum of the system. Similarly, find the individual momentum of each object after the collision and add them to find the final momentum of the system. Then, set the initial vector momentum of the system equal to the final vector momentum of the system.
PROBLEMS - Giancoli, Ch 7 #3, 5, 9, 15, 19, 21, 25, 32, 39, 42, 63, 71
1. State Newton's third law of motion.
· Newton's third law of motion describes how objects (more than one) interact with one another
· According to Newton, "For every action there is an equal and opposite reaction". This means that when two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.
· The interaction can be through contact -- like in a collision -- or at a distance -- like the earth and the moon. Regardless of the objects involved -- mass makes no difference here -- the two objects will exert equal forces on each other. The bug hits the windshield with the same force that the windshield hits the bug.
· While the forces are equal, the effect of the force (because of mass -- here it does matter) may be very different. Combining the second and third laws allows us to say: mA = Ma
2. Discuss examples of Newton's third law.
· In order to walk you must exert a force on the ground that is directed backwards. The reaction of the ground is to push you forward.
· The horse and cart
· Rockets lifting off -- the rocket pushes the exhaust particles out and the exhaust particles push the rocket
Inertial Mass, Gravitational Mass and Weight - Giancoli, Ch33, p. 948
3. Define inertial mass.
· Inertial mass is the mass that is present in Newton's second law of motion (Fnet = ma). Therefore, the definition of inertial mass is as follows: mi = Fnet / a. In words, inertial mass is the ratio of net force to acceleration.
4. Compare gravitational mass and inertial mass.
· Gravitational mass is the mass of an object that describes how it "feels" the gravitational force exerted by other masses. Near the earth, this mass is the mass responsible for the weight of an object. The definition of gravitational mass is as follows: mg = W / g
· While the concepts of inertial mass and gravitational mass arise from different areas of physics, they are mathematically equivalent. No experiment has been able to distinguish between the two.
· Consider the simple example of a mass in free fall near the surface of the earth. The only force acting on the object is the gravitational force (its weight). According to Newton's second law of motion, we may set Fnet = W. But, W = mgg and Fnet = mia. Therefore, we may state: mgg = mia. Finally, this means that the acceleration of an object in freefall, a, is given by the relation: a = (mg / mI) g. According to Galileo, all objects fall at the same rate of acceleration. This is only true if the ratio mg / mi has a value of 1. In other words, mg = mi.
5. Discuss the concept of weight.
· Weight, by definition, is the net gravitational force exerted on an object. W = mg
· Weight is typically measured using a support scale (like the bathroom scale you have at home). In most cases, the bathroom scale will report the true weight (defined as mg) of an object. This will be true as long as the scale and object are experiencing static or translational equilibrium -- neither is accelerating.
· In non-equilibrium cases, for example a support scale used in an accelerating elevator, the value given by the scale will reflect the motion of the object and will not report the "true weight" (= mg) of the object. The "apparent weight" of an object on a rising elevator will be greater than its true weight. The "apparent weight" of an object on a descending elevator will be less than its true weight.
6. Distinguish between mass and weight.
· Mass is a measure of the amount of matter contained within an object. We have seen that it can be inertial or gravitational in nature (#4 above). Mass is a fundamental property of matter and does not change with location.
· Weight is a specific force involving the concept of gravitational mass. Weight depends greatly on location and will change if the object is moved to a different location.
Momentum - Giancoli, Ch 7, all sections visit the Physics Classroom for this topic
7. Define linear momentum and impulse.
· Linear momentum is defined as the product of mass and velocity. p = mv
· Newton's second law of motion can be written as follows: F = ma = mDv / Dt = m(vf - vo) / Dt = (mvf - mvo) / Dt
Or, F = Dp / Dt. That is, the net force acting describes the rate at which its momentum changes.
· We can define a new quantity, impulse, which describes how momentum changes. From Newton's second law of motion, we see that Dp = F Dt. Impulse is calculated as F Dt. The result of the impulse is to change the momentum by an amount = Dp.
· Large forces acting over small time intervals produce the same impulse as small forces over large time intervals.
8. State the law of conservation of linear momentum.
· We will define a system of objects to contain all objects that will interact with one another.
· External forces are defined as forces that are exerted from outside the system of objects.
· Internal forces are defined s forces that are exerted by the objects within the system of objects.
· In the absence of external forces (an isolated system), the linear momentum of a system of objects is conserved. It can be neither created nor destroyed. It may, during the course of interactions between objects in the system, be transferred from one object to another.
v1
v1f
v2
v2f9. Derive the law of conservation of momentum for an isolated system consisting of two interacting particles.
The two objects in the diagram will interact. The result of the interaction is that each object will acquire a different final velocity. While they are interacting, Newton's third law of motion applies. Newton's second law describes the result for each object.
During the interaction: F1 = - F2 b/c of Newton's third law
We can write: Dp1 / Dt = - Dp2 / Dt b/c of Newton's second law
We can write: Dp1 = - Dp2 b/c Dt is the same for both objects
This means: p1f - p1o = - (p2f - p2o)
Simplify: p1f - p1o = -p2f + p2o
Rearrange: p1f + p2f = p1o + p2o
(p1 + p2)f = (p1 + p2)o
Group: pf = po Total momentum is conserved!
10. Solve problems involving momentum and impulse.
· Regardless of the type of collision involved, linear momentum for an isolated system (no external forces) is conserved.
· If external forces are present, then the momentum will change according to the impulse relation.
· It might sometimes be useful to classify collisions between objects by the way that energy is distributed in the collision. Those collisions for which kinetic energy is conserved are called elastic collisions. If the kinetic energy is not conserved, but is transformed into other types of energy then the collision is called an inelastic collision.
· To apply momentum conservation, calculate the individual momentum of each object (remembering that momentum is a vector quantity) prior to the collision and add them to find the initial momentum of the system. Similarly, find the individual momentum of each object after the collision and add them to find the final momentum of the system. Then, set the initial vector momentum of the system equal to the final vector momentum of the system.
PROBLEMS - Giancoli, Ch 7 #3, 5, 9, 15, 19, 21, 25, 32, 39, 42, 63, 71
FORCE AND FREE-BODY DIAGRAM
Forces and free-body diagrams - Giancoli, Ch4, section 7 visit the Physics Classroom for this topic
1. Describe force as the cause of deformation or velocity change.
· Force -- a push or a pull -- may do one of two things to an object. First, it can change the state of motion of the object à change its velocity. Second it can change the physical state of the object à sitting on a cushion does not put the cushion in motion, but it does change the shape of the cushion.
2. Identify the forces acting on an object and draw free-body diagrams representing the forces acting.
· Identify all forces acting on a single object and draw a vector diagram showing the forces. Separate diagrams are drawn for each object in the problem.
· Each force should be labeled by name or given a commonly accepted symbol. Vectors should have lengths approximately proportional to their magnitudes. The direction of the vector should also be labeled.
· Common symbols used for forces include the following. W (=mg) for the weight of an object, T (=tension) for strings or cords pulling an object, f (=friction) for the contact friction force exerted by one object on another, and N (=Normal) for the contact support force exerted by one object on another.
3. Resolve forces into components.
· Using the trig ratios and choosing your triangles carefully, find the perpendicular components that make up a force.
· Some hints about choosing triangles: if the direction of motion is known, choose this as one of the axes for finding force components. This also means that the direction perpendicular to the motion will be your other axis. OR, if the direction of the net force is known, choose this as one of the axes for finding force components.
4. Determine the resultant force in different situations.
· Once all forces acting have been broken down into perpendicular components (you get to decide what the two perpendicular directions are), add up the forces acting along each of the two axes that you chose to find the net force along each axis.
· It may be necessary to use the Pythagorean theorem to add the two perpendicular resultant forces together to find the net resultant force. You might also have to draw a new triangle and use tan-1 to find the angle for the direction of the net resultant force.
5. Describe the behavior of a linear spring and solve related problems.
· Springs exert what is called a restoring force on objects. The spring's force is directed opposite to the displacement of the spring. If I pull on a spring to the right, the spring pulls back to the left.
· The more a spring is stretched, the greater the force it exerts on objects.
· The two statements above can be combined in a force equation, known as Hooke's Law. F = -kx. The minus sign indicates the spring's force is opposite to the displacement, x. The fact that F is directly proportional to x shows that the force increases as x increases.
Newton's first law - Giancoli, Ch 4, section 2 visit the Physics Classroom for this topic
6. State Newton's first law of motion.
· Newton's first law of motion describes the effect of a force on an object.
· Every body continues in its state of rest or uniform motion unless it is compelled to change its state by a net external force acting on it.
· In physics the term uniform means "constant". The interpretation of uniform motion means constant motion -- both speed and direction. Uniform motion denotes moving in a straight line with constant speed.
7. Describe examples of Newton's first law.
· Reference demos done in class
Equilibrium - Giancoli, Ch 9, 1-2
8. State the condition for translational equilibrium.
· Translation refers to linear motion. There is also rotational equilibrium which we will encounter later.
· Translational equilibrium means to have a uniform state of motion. Objects may remain at rest or they may remain in uniform motion.
· Based on Newton's first law of motion, the condition for translational equilibrium to occur is that there may be no net external force acting on the object. Another way of saying it is Fnet = 0.
9. Solve problems involving translational equilibrium.
· Problems will state that certain forces (such as friction) are acting and that the object maintains a constant velocity. This is your clue that the object is in translational equilibrium and that Fnet = 0.
Newton's second law - Giancoli, Ch 4, Section 4, 6-7, 9-10 visit the Physics Classroom for this topic
10. State Newton's second law of motion.
· Newton's second law of motion describes how forces change motion.
· The change in motion (change in velocity) or acceleration is directly proportional to the net force applied.
· The change in motion (change in velocity) or acceleration is inversely proportional to the mass of the object.
· In equation form: a = Fnet / m
11. Solve problems involving Newton's second law.
· In practice, because of the vector nature of determining Fnet, the second law is usually written as Fnet = ma.
· This form of the equation gives us a method for applying the second law.
· First, draw a free body diagram for each object present in the problem.
· Second, find any vector components necessary to make all forces (or components) act in two perpendicular directions. Choose one of the directions to be the direction of motion.
· Write an expression for the net force in each of the two perpendicular directions for each object.
· Set each net force = ma for that object in that direction.
· Apply any conditions present -- e.g. objects linked by string both have the same acceleration.
Friction - Giancoli, Ch 4, section 8 visit the Physics Classroom for this topic
12. Describe the nature and properties of frictional forces.
· Friction is a force that occurs because of contact between two objects. Microscopically, even the smoothest surfaces appear quite rough and as two surfaces interact the ridges of one will "catch" on the valleys of the other, producing a loss of motion.
· The amount of friction depends on the size, shape, and mass of the objects involved.
· Materials are rated for the amount of friction they produce using a property called the coefficient of friction (m).
13. Distinguish between static and dynamic (sliding) friction.
· For objects in motion, the force of dynamic (sliding or kinetic) friction acts opposite to the direction of motion.
· The force of dynamic friction is directly related to the normal (support) force between the two surfaces.
· Static friction occurs when objects are in static equilibrium (Fnet = 0 and there is no motion), for example a book sitting on a table. I you push horizontally on the book, it will initially resist the force and remain at rest. This is the result of static friction, which opposes the force you are applying to the book. At some point you will push hard enough to get the book to move and dynamic friction will take over.
· The amount of static friction present will be less than or equal to some maximum value, which is directly related to the normal force between the two surfaces.
14. Define coefficient of friction.
· The coefficient of friction is a proportionality constant and is defined as the ratio of the friction force to the normal force between the two objects. m = f / N. Because there are two types of friction, there are two different coefficients, mk for kinetic friction and ms for static friction.
· For static friction and for kinetic friction
· Because of the nature of the two friction forces -- you almost always have to push harder to get a box to start to slide across the floor than you do to keep it moving -- ms is almost always greater than mk.
15. Solve static and dynamic problems involving friction.
· The key to problems that include friction is to remember that friction always acts opposite to the direction of motion.
· Since friction is parallel to the surface between two objects and its magnitude is related to the normal force (which acts perpendicular to the surface between the two objects), the equation for the magnitude of the friction (f = mN) will link the two equations that you wrote when applying Newton's second law.
Problems: GIANCOLI - Chapter 4 - #1, 3, 11, 15, 19, 21, 25, 27, 29, 37, 39, 43, 51a, 55
1. Describe force as the cause of deformation or velocity change.
· Force -- a push or a pull -- may do one of two things to an object. First, it can change the state of motion of the object à change its velocity. Second it can change the physical state of the object à sitting on a cushion does not put the cushion in motion, but it does change the shape of the cushion.
2. Identify the forces acting on an object and draw free-body diagrams representing the forces acting.
· Identify all forces acting on a single object and draw a vector diagram showing the forces. Separate diagrams are drawn for each object in the problem.
· Each force should be labeled by name or given a commonly accepted symbol. Vectors should have lengths approximately proportional to their magnitudes. The direction of the vector should also be labeled.
· Common symbols used for forces include the following. W (=mg) for the weight of an object, T (=tension) for strings or cords pulling an object, f (=friction) for the contact friction force exerted by one object on another, and N (=Normal) for the contact support force exerted by one object on another.
3. Resolve forces into components.
· Using the trig ratios and choosing your triangles carefully, find the perpendicular components that make up a force.
· Some hints about choosing triangles: if the direction of motion is known, choose this as one of the axes for finding force components. This also means that the direction perpendicular to the motion will be your other axis. OR, if the direction of the net force is known, choose this as one of the axes for finding force components.
4. Determine the resultant force in different situations.
· Once all forces acting have been broken down into perpendicular components (you get to decide what the two perpendicular directions are), add up the forces acting along each of the two axes that you chose to find the net force along each axis.
· It may be necessary to use the Pythagorean theorem to add the two perpendicular resultant forces together to find the net resultant force. You might also have to draw a new triangle and use tan-1 to find the angle for the direction of the net resultant force.
5. Describe the behavior of a linear spring and solve related problems.
· Springs exert what is called a restoring force on objects. The spring's force is directed opposite to the displacement of the spring. If I pull on a spring to the right, the spring pulls back to the left.
· The more a spring is stretched, the greater the force it exerts on objects.
· The two statements above can be combined in a force equation, known as Hooke's Law. F = -kx. The minus sign indicates the spring's force is opposite to the displacement, x. The fact that F is directly proportional to x shows that the force increases as x increases.
Newton's first law - Giancoli, Ch 4, section 2 visit the Physics Classroom for this topic
6. State Newton's first law of motion.
· Newton's first law of motion describes the effect of a force on an object.
· Every body continues in its state of rest or uniform motion unless it is compelled to change its state by a net external force acting on it.
· In physics the term uniform means "constant". The interpretation of uniform motion means constant motion -- both speed and direction. Uniform motion denotes moving in a straight line with constant speed.
7. Describe examples of Newton's first law.
· Reference demos done in class
Equilibrium - Giancoli, Ch 9, 1-2
8. State the condition for translational equilibrium.
· Translation refers to linear motion. There is also rotational equilibrium which we will encounter later.
· Translational equilibrium means to have a uniform state of motion. Objects may remain at rest or they may remain in uniform motion.
· Based on Newton's first law of motion, the condition for translational equilibrium to occur is that there may be no net external force acting on the object. Another way of saying it is Fnet = 0.
9. Solve problems involving translational equilibrium.
· Problems will state that certain forces (such as friction) are acting and that the object maintains a constant velocity. This is your clue that the object is in translational equilibrium and that Fnet = 0.
Newton's second law - Giancoli, Ch 4, Section 4, 6-7, 9-10 visit the Physics Classroom for this topic
10. State Newton's second law of motion.
· Newton's second law of motion describes how forces change motion.
· The change in motion (change in velocity) or acceleration is directly proportional to the net force applied.
· The change in motion (change in velocity) or acceleration is inversely proportional to the mass of the object.
· In equation form: a = Fnet / m
11. Solve problems involving Newton's second law.
· In practice, because of the vector nature of determining Fnet, the second law is usually written as Fnet = ma.
· This form of the equation gives us a method for applying the second law.
· First, draw a free body diagram for each object present in the problem.
· Second, find any vector components necessary to make all forces (or components) act in two perpendicular directions. Choose one of the directions to be the direction of motion.
· Write an expression for the net force in each of the two perpendicular directions for each object.
· Set each net force = ma for that object in that direction.
· Apply any conditions present -- e.g. objects linked by string both have the same acceleration.
Friction - Giancoli, Ch 4, section 8 visit the Physics Classroom for this topic
12. Describe the nature and properties of frictional forces.
· Friction is a force that occurs because of contact between two objects. Microscopically, even the smoothest surfaces appear quite rough and as two surfaces interact the ridges of one will "catch" on the valleys of the other, producing a loss of motion.
· The amount of friction depends on the size, shape, and mass of the objects involved.
· Materials are rated for the amount of friction they produce using a property called the coefficient of friction (m).
13. Distinguish between static and dynamic (sliding) friction.
· For objects in motion, the force of dynamic (sliding or kinetic) friction acts opposite to the direction of motion.
· The force of dynamic friction is directly related to the normal (support) force between the two surfaces.
· Static friction occurs when objects are in static equilibrium (Fnet = 0 and there is no motion), for example a book sitting on a table. I you push horizontally on the book, it will initially resist the force and remain at rest. This is the result of static friction, which opposes the force you are applying to the book. At some point you will push hard enough to get the book to move and dynamic friction will take over.
· The amount of static friction present will be less than or equal to some maximum value, which is directly related to the normal force between the two surfaces.
14. Define coefficient of friction.
· The coefficient of friction is a proportionality constant and is defined as the ratio of the friction force to the normal force between the two objects. m = f / N. Because there are two types of friction, there are two different coefficients, mk for kinetic friction and ms for static friction.
· For static friction and for kinetic friction
· Because of the nature of the two friction forces -- you almost always have to push harder to get a box to start to slide across the floor than you do to keep it moving -- ms is almost always greater than mk.
15. Solve static and dynamic problems involving friction.
· The key to problems that include friction is to remember that friction always acts opposite to the direction of motion.
· Since friction is parallel to the surface between two objects and its magnitude is related to the normal force (which acts perpendicular to the surface between the two objects), the equation for the magnitude of the friction (f = mN) will link the two equations that you wrote when applying Newton's second law.
Problems: GIANCOLI - Chapter 4 - #1, 3, 11, 15, 19, 21, 25, 27, 29, 37, 39, 43, 51a, 55
WORK ENRGY AND POWER
Work - Giancoli, Ch 6, 1-3 visit the Physics Classroom for this topic
1. Define work.
· In physics, work is defined as a force acting upon an object to cause a displacement.
· In general terms, the amount of work performed is calculated as the product of the force and the displacement.
· To see how the product is calculated, consider the example of a rope pulling at an angle, q, above the floor on a box. The result is that the force of the rope, F, moves the box horizontally a distance d.
When looking at the force, F, that acts, only a portion of the force is responsible for the displacement, d. The vertical component of the force (F sin q) has no effect on the motion. It is only the horizontal component, F cos q, that moves the box. So, to find the amount of work done by F, we multiply the horizontal component of F times the magnitude of the displacement, d.
W = Fd cos q
· In words, to find the work done by a force to produce a displacement, multiply the magnitude of the force times the magnitude of the displacement times the cosine of the angle between them.
· The unit of work is the Nm or Joule.
2. Determine the work done by a non-constant force by interpreting a force–displacement graph.
·
Suppose a constant force acting horizontally moves a box across the floor. If we were to make a graph that shows the magnitude of the pulling force as a function of the position of the box, it would look like the graph below.
We know that in this case, the work done by the force is Fd. Looking at the graph, we see that the area of the shaded rectangle is also Fd. This suggests that the work done by a force to produce a displacement, d, is the area of a force-displacement graph.
· Now suppose that the force that is doing the work does not remain constant, but varies with the position of the object, like the second graph above. The work done in this case is still the area of a force-displacement graph, as shown by the shaded region of the graph.
3. Solve problems involving the work done on a body by a force.
· See Giancoli, Ch 6 problems #5, 7a, 9, 13
· PLUS, do the following problem: The force required to stretch an ideal spring varies with the "stretch" of the spring as F = kx, where k is called the spring constant and x represents the amount of stretch from the equilibrium position of the spring. Show that the work necessary to stretch a spring from a position, x1, to a new position, x2, is given by the expression W = 1/2 kx22 - 1/2 kx12. Hint - look at a graph of the force as a function of position of the spring.
Energy and power - Giancoli, Ch 6, 3-10 visit the Physics Classroom for this topic
4. Define kinetic energy.
· Suppose the result of a constant force acting to move an object from rest through a distance, d, is that the object acquires a final speed, v. We know that the work done by the force is given by W = Fd = mad. We also know from kinematics, that the relation v2 = u2 + 2ad = 2ad (because it starts from rest) can be used to describe this motion. If we solve the kinematic relation for the quantity ad = v2/2 and substitute into the expression for the work done, we get W = Fd = mad = mv2/2.
· The quantity, , is called the Kinetic Energy (K) of the object. This is the form of energy related to motion.
· The result of doing work on the object is to change its energy. In this case it gained kinetic energy (= ).
5. Describe the concepts of gravitational potential energy and elastic potential energy.
· Suppose a box of mass m at rest on the ground is lifted to a height h above the ground so that its final speed is zero. How much work is required to lift the box? The force that we have to overcome to lift the box is its weight, mg. The work to lift it is then found using the definition of work, W = Fd = mgh.
· The quantity, mgh, is called the gravitational potential energy (Ug) of the box. This is the form of energy related to a change in position of the box. It has this energy because we moved it, but it is not in motion.
· The force required to stretch a spring varies with the position of the spring as F = kx. This is called Hooke's Law. You showed in the homework problems for work above that the work done by this force to stretch the spring a distance x is given by the expression W = area under graph = .
· The quantity, , is called the elastic potential energy (Uel) of the spring. This is also a form of energy acquired by the spring because of the work done to change its position.
6. State the principle of conservation of energy.
· The quantity U + K (potential + kinetic energy) of an object is called the mechanical energy (E) of the object.
· If no external forces do work on the object, then the mechanical energy (E) of the object remains constant. This is called conservation of mechanical energy.
· If external forces are present (such as friction or some outside push or pull), then the work done by the external force will equal the change in the mechanical energy of the object. W = DE. This is called the work-energy theorem and describes how energy is given to or taken away from objects. Since work results in a change of energy, energy must have the same unit of measure, Joules, as work.
7. List different forms of energy and describe examples of the transformation of energy from one form into another.
· Energy exists in many forms (heat, nuclear, solar, etc.) which can be ultimately reduced to the kinetic and potential energies of constituent particles in objects.
· Conservation of energy states that in the absence of external forces the mechanical energy of an object remains constant. This means that DE = DU + DK = 0. In other words, DU = - DK. If we allow a ball to roll down a hill influenced only by the force of gravity then the potential energy will decrease as the ball moves down the hill. According to energy conservation, this means that the kinetic energy of the ball changes in the opposite way (increases) by the same amount. The potential energy of the ball has been transformed into kinetic energy.
8. Define power.
· When a quantity of work, W, is performed in a time, Dt, the ratio W / Dt is called the power developed.
· Since we have shown that W = DE, we can also define the quantity DE / Dt to be power.
· Power is the rate at which work is done or energy is changed. The unit of power is the Watt (W) = Js-1.
· A special case is when a constant force acts on object. P = W / Dt = (F Ds) / Dt = Fv.
9. Define and apply the concept of efficiency.
· In a perfect world, forces like friction are assumed to be negligible and the concept of efficiency is not necessary. However, in the real world, friction exists. Suppose that a box is pushed up a ramp. The box gains potential energy = mgh. However, because of friction, we have to push with more force than the weight of the box to get it up the ramp. The useful work is the change in potential energy of the box. The actual work is the amount done to slide it up the ramp. In general, the actual work is greater than the useful work.
· The efficiency is defined as the ratio of useful work / actual work.
10. Solve work, energy and power problems.
· See Giancoli, Ch 6 # 19, 23, 29, 35, 40, 41, 63, 74
Projectile Motion - Giancoli, Ch 3, 5-7 visit the Physics Classroom for this topic
1. State the independence of the vertical and horizontal components of motion for a projectile in a uniform field.
· Vertical forces do not affect horizontal motion. This means the horizontal component of projectile motion is motion with constant speed because there are no horizontal forces acting.
· The vertical component of projectile motion is accelerated motion (affected by the acceleration of gravity). At every point in the trajectory of a projectile, the acceleration vector is vertical (downward) and has the value 9.8ms-2.
· The key to analyzing projectile motion is to consider the two perpendicular components separately.
2. Describe the trajectory of projectile motion as parabolic in the absence of friction.
· The combination of accelerated vertical motion and constant horizontal motion produces a trajectory through the air in the shape of a parabola. (This can be proved mathematically, but is not required here.)
· The key to this process is that both motions are linked through time. It takes the same amount of time to complete both motions.
3. Solve problems on projectile motion. Problems may involve projectiles launched horizontally or at any angle above or below horizontal. Applying conservation of energy may provide a simpler solution to some problems than using projectile motion kinematics equations.
· The key to problem solving is to write the equations of motion for the horizontal and vertical components of the motion separately and remember that they are linked through time.
· Conservation of Energy can used to relate speed and height at any two points in the trajectory of an object.
· See Giancoli, Ch 3 # 37, 39, 43, 51
1. Define work.
· In physics, work is defined as a force acting upon an object to cause a displacement.
· In general terms, the amount of work performed is calculated as the product of the force and the displacement.
· To see how the product is calculated, consider the example of a rope pulling at an angle, q, above the floor on a box. The result is that the force of the rope, F, moves the box horizontally a distance d.
When looking at the force, F, that acts, only a portion of the force is responsible for the displacement, d. The vertical component of the force (F sin q) has no effect on the motion. It is only the horizontal component, F cos q, that moves the box. So, to find the amount of work done by F, we multiply the horizontal component of F times the magnitude of the displacement, d.
W = Fd cos q
· In words, to find the work done by a force to produce a displacement, multiply the magnitude of the force times the magnitude of the displacement times the cosine of the angle between them.
· The unit of work is the Nm or Joule.
2. Determine the work done by a non-constant force by interpreting a force–displacement graph.
·
Suppose a constant force acting horizontally moves a box across the floor. If we were to make a graph that shows the magnitude of the pulling force as a function of the position of the box, it would look like the graph below.
We know that in this case, the work done by the force is Fd. Looking at the graph, we see that the area of the shaded rectangle is also Fd. This suggests that the work done by a force to produce a displacement, d, is the area of a force-displacement graph.
· Now suppose that the force that is doing the work does not remain constant, but varies with the position of the object, like the second graph above. The work done in this case is still the area of a force-displacement graph, as shown by the shaded region of the graph.
3. Solve problems involving the work done on a body by a force.
· See Giancoli, Ch 6 problems #5, 7a, 9, 13
· PLUS, do the following problem: The force required to stretch an ideal spring varies with the "stretch" of the spring as F = kx, where k is called the spring constant and x represents the amount of stretch from the equilibrium position of the spring. Show that the work necessary to stretch a spring from a position, x1, to a new position, x2, is given by the expression W = 1/2 kx22 - 1/2 kx12. Hint - look at a graph of the force as a function of position of the spring.
Energy and power - Giancoli, Ch 6, 3-10 visit the Physics Classroom for this topic
4. Define kinetic energy.
· Suppose the result of a constant force acting to move an object from rest through a distance, d, is that the object acquires a final speed, v. We know that the work done by the force is given by W = Fd = mad. We also know from kinematics, that the relation v2 = u2 + 2ad = 2ad (because it starts from rest) can be used to describe this motion. If we solve the kinematic relation for the quantity ad = v2/2 and substitute into the expression for the work done, we get W = Fd = mad = mv2/2.
· The quantity, , is called the Kinetic Energy (K) of the object. This is the form of energy related to motion.
· The result of doing work on the object is to change its energy. In this case it gained kinetic energy (= ).
5. Describe the concepts of gravitational potential energy and elastic potential energy.
· Suppose a box of mass m at rest on the ground is lifted to a height h above the ground so that its final speed is zero. How much work is required to lift the box? The force that we have to overcome to lift the box is its weight, mg. The work to lift it is then found using the definition of work, W = Fd = mgh.
· The quantity, mgh, is called the gravitational potential energy (Ug) of the box. This is the form of energy related to a change in position of the box. It has this energy because we moved it, but it is not in motion.
· The force required to stretch a spring varies with the position of the spring as F = kx. This is called Hooke's Law. You showed in the homework problems for work above that the work done by this force to stretch the spring a distance x is given by the expression W = area under graph = .
· The quantity, , is called the elastic potential energy (Uel) of the spring. This is also a form of energy acquired by the spring because of the work done to change its position.
6. State the principle of conservation of energy.
· The quantity U + K (potential + kinetic energy) of an object is called the mechanical energy (E) of the object.
· If no external forces do work on the object, then the mechanical energy (E) of the object remains constant. This is called conservation of mechanical energy.
· If external forces are present (such as friction or some outside push or pull), then the work done by the external force will equal the change in the mechanical energy of the object. W = DE. This is called the work-energy theorem and describes how energy is given to or taken away from objects. Since work results in a change of energy, energy must have the same unit of measure, Joules, as work.
7. List different forms of energy and describe examples of the transformation of energy from one form into another.
· Energy exists in many forms (heat, nuclear, solar, etc.) which can be ultimately reduced to the kinetic and potential energies of constituent particles in objects.
· Conservation of energy states that in the absence of external forces the mechanical energy of an object remains constant. This means that DE = DU + DK = 0. In other words, DU = - DK. If we allow a ball to roll down a hill influenced only by the force of gravity then the potential energy will decrease as the ball moves down the hill. According to energy conservation, this means that the kinetic energy of the ball changes in the opposite way (increases) by the same amount. The potential energy of the ball has been transformed into kinetic energy.
8. Define power.
· When a quantity of work, W, is performed in a time, Dt, the ratio W / Dt is called the power developed.
· Since we have shown that W = DE, we can also define the quantity DE / Dt to be power.
· Power is the rate at which work is done or energy is changed. The unit of power is the Watt (W) = Js-1.
· A special case is when a constant force acts on object. P = W / Dt = (F Ds) / Dt = Fv.
9. Define and apply the concept of efficiency.
· In a perfect world, forces like friction are assumed to be negligible and the concept of efficiency is not necessary. However, in the real world, friction exists. Suppose that a box is pushed up a ramp. The box gains potential energy = mgh. However, because of friction, we have to push with more force than the weight of the box to get it up the ramp. The useful work is the change in potential energy of the box. The actual work is the amount done to slide it up the ramp. In general, the actual work is greater than the useful work.
· The efficiency is defined as the ratio of useful work / actual work.
10. Solve work, energy and power problems.
· See Giancoli, Ch 6 # 19, 23, 29, 35, 40, 41, 63, 74
Projectile Motion - Giancoli, Ch 3, 5-7 visit the Physics Classroom for this topic
1. State the independence of the vertical and horizontal components of motion for a projectile in a uniform field.
· Vertical forces do not affect horizontal motion. This means the horizontal component of projectile motion is motion with constant speed because there are no horizontal forces acting.
· The vertical component of projectile motion is accelerated motion (affected by the acceleration of gravity). At every point in the trajectory of a projectile, the acceleration vector is vertical (downward) and has the value 9.8ms-2.
· The key to analyzing projectile motion is to consider the two perpendicular components separately.
2. Describe the trajectory of projectile motion as parabolic in the absence of friction.
· The combination of accelerated vertical motion and constant horizontal motion produces a trajectory through the air in the shape of a parabola. (This can be proved mathematically, but is not required here.)
· The key to this process is that both motions are linked through time. It takes the same amount of time to complete both motions.
3. Solve problems on projectile motion. Problems may involve projectiles launched horizontally or at any angle above or below horizontal. Applying conservation of energy may provide a simpler solution to some problems than using projectile motion kinematics equations.
· The key to problem solving is to write the equations of motion for the horizontal and vertical components of the motion separately and remember that they are linked through time.
· Conservation of Energy can used to relate speed and height at any two points in the trajectory of an object.
· See Giancoli, Ch 3 # 37, 39, 43, 51
KINEMATICS
Kinematic concepts - Giancoli, Ch 2: 1-7 visit the Physics Classroom
1. Define displacement, velocity, speed and acceleration.
2. Define and explain the difference between instantaneous and average values of speed, velocity and acceleration.
· Displacement or change in position is the straight line length connecting the starting and ending points of motion, along with the direction of travel à displacement is a vector quantity
· Velocity describes how the position changes over time à velocity is a vector quantity, because it depends on displacement
· Average velocity is calculated as Ds / Dt (where s represents the displacement vector)
· Instantaneous velocity is defined as the velocity at a particular location -- in reality it is the average velocity over an indefinitely small interval of time (Dt approaches the value 0)
· Speed describes how the distance traveled changes over time à speed is a scalar quantity, not a vector quantity. Speed can be average or instantaneous, just like velocity.
· Acceleration describes how the velocity changes over time à acceleration is a vector quantity because it depends on the velocity.
· Average acceleration is calculated as Dv / Dt (where v represents the velocity vector)
· Instantaneous acceleration is defined as the acceleration at a particular location -- in reality it is the average acceleration over an indefinitely small interval of time (Dt approaches the value 0).
3. Describe an object's motion from more than one frame of reference.
· All motion is described from some point of view AND the description of the motion depends greatly on that point of view. The point of view is called a frame of reference since it defines the coordinate system (frame) in which motion is to be described.
· An example: an observer at the side of the road sees two cars traveling N at 20 m/s (car A) and 30 m/s (car B). What is the relative velocity of Car B from the point of view of (or relative to) car A? The motion of car A will change the observed velocity of car B -- Car B is still moving North, but from A's point of view it is only pulling away at 10 ms-1 (30 - 20). What is the velocity of Car A from the point of view of (or relative to) car B? The motion of Car B will change the observed velocity of car A -- Car A appears to moving at a velocity of -10 ms-1 (20 - 30). In other words, Car A appears to going South (-North) at 10 ms-1.
· To calculate the relative velocity of one object relative to a second object, subtract the velocity vector of the second object from the velocity vector of the first object.
Graphical representation of motion - Giancoli, Ch 2: 11 view a Java Applet
4. Draw and analyse distance–time graphs, displacement–time graphs, velocity–time graphs and acceleration–time graphs. Students should be able to sketch and label these graphs for various situations. They should also be able to write descriptions of the motions represented by such graphs.
· Forward (positive) motion is represented by a positive slope on a position-time graph. Backward (negative) motion is represented by a negative slope on a position-time graph.
· Constant velocity is represented by a linear, sloped position-time graph or a horizontal line on a velocity-time graph.
· Forward (positive) velocity is drawn in the first quadrant (above the time axis) on a v-t graph. Backward (negative) velocity is drawn in the fourth quadrant (below the time axis) on a v-t graph.
· Uniformly accelerated motion is represented by a linear, sloped v-t graph. Positive acceleration has a positive slope and negative acceleration has a negative slope.
· Uniformly accelerated motion is represented by a curved (parabola) position-time graph. Positive acceleration is an upward curved parabola. Negative acceleration is a downward curved parabola.
5. Analyze and calculate the slopes of displacement–time graphs and velocity– time graphs, and the areas under velocity–time graphs and acceleration–time graphs. Relate these to the relevant kinematic quantity.
· Slope of a position-time graph is the velocity of the motion. To find instantaneous velocity, find the slope of the line tangent to the curve at a specific location (for curved graphs).
· Slope of a velocity-time graph is the acceleration of the motion. To find the instantaneous acceleration, find the slope of the line tangent to the curve at a specific location (for curved graphs).
· The area between the velocity-time curve and the time axis represents the displacement of the motion.
· The area between the acceleration-time curve and the time axis represents the change in velocity for the motion.
Uniformly accelerated motion - Giancoli, Ch 2: 8-10
6. Determine the velocity and acceleration from simple timing situations.
Students should be able to interpret data from devices such as a light gate, strobe photograph or ticker timer. Analysis may involve graphing the data, taking measurements and applying kinematics concepts.
An example: the dots in the diagram represent successive positions of a car, taken at 1-second intervals, beginning with time = 0 seconds. 1 cm on the diagram represents 10 m. To find the velocity during any interval, you would measure the distance between successive dots, convert the measurement to meters, and divide by the elapsed time. To find the acceleration of the motion, you would need to find the average velocity for two successive intervals and then divide by the same time interval.
7. Derive the equations for uniformly accelerated motion.
· v = u + at this is the equation of a line on a v-t graph with u as the vertical intercept and a as the slope
· Ds = ut + 1/2at2 this is the area under a velocity time graph with u as the vertical intercept and a as the slope
· Ds = (u + v)t / 2 this is still the area under a velocity-time graph (u + v)/2 is the average velocity
· v2 = u2 + 2aDs solve the first equation for t, sub into the second equation and simplify
8. Describe the vertical motion of an object in a uniform gravitational field.
· Freefall motion is the motion of an object falling only under the influence of gravity. Air resistance is assumed negligible .
· Objects fall with uniform acceleration (9.8 ms-2 directed downward)
· All objects, regardless of their mass, fall with the same uniform acceleration.
· The same equations of motion can be used for freefall motion. If the upward direction is taken as positive motion, then a = - 9.8 ms-2 (the minus sign denotes a downward direction for the acceleration).
· The acceleration of an object moving vertically up or down is always 9.8 ms-2 directed downward, regardless of where in the trajectory the object is located.
9. Describe the effects of air resistance on falling objects.
· If air resistance is included in the problem, mathematically it becomes much more complicated. This is because the amount of air resistance changes in relation to the speed of the falling object.
· As objects fall, the amount of air resistance (directed upward) increases until it is exactly balanced by the downward pull of gravity. At this point, there is no net force on the falling object and it will fall with a constant speed from that point on. The constant speed that objects reach is called terminal velocity.
10. Solve problems involving uniformly accelerated motion.
· See Giancoli, Ch 2 Problems Level I: #2, 4, 15a, 19, 21, 35
Level II: #8, 11, 37, 41, 45, 61
Level III: #17a, 29, 51
1. Define displacement, velocity, speed and acceleration.
2. Define and explain the difference between instantaneous and average values of speed, velocity and acceleration.
· Displacement or change in position is the straight line length connecting the starting and ending points of motion, along with the direction of travel à displacement is a vector quantity
· Velocity describes how the position changes over time à velocity is a vector quantity, because it depends on displacement
· Average velocity is calculated as Ds / Dt (where s represents the displacement vector)
· Instantaneous velocity is defined as the velocity at a particular location -- in reality it is the average velocity over an indefinitely small interval of time (Dt approaches the value 0)
· Speed describes how the distance traveled changes over time à speed is a scalar quantity, not a vector quantity. Speed can be average or instantaneous, just like velocity.
· Acceleration describes how the velocity changes over time à acceleration is a vector quantity because it depends on the velocity.
· Average acceleration is calculated as Dv / Dt (where v represents the velocity vector)
· Instantaneous acceleration is defined as the acceleration at a particular location -- in reality it is the average acceleration over an indefinitely small interval of time (Dt approaches the value 0).
3. Describe an object's motion from more than one frame of reference.
· All motion is described from some point of view AND the description of the motion depends greatly on that point of view. The point of view is called a frame of reference since it defines the coordinate system (frame) in which motion is to be described.
· An example: an observer at the side of the road sees two cars traveling N at 20 m/s (car A) and 30 m/s (car B). What is the relative velocity of Car B from the point of view of (or relative to) car A? The motion of car A will change the observed velocity of car B -- Car B is still moving North, but from A's point of view it is only pulling away at 10 ms-1 (30 - 20). What is the velocity of Car A from the point of view of (or relative to) car B? The motion of Car B will change the observed velocity of car A -- Car A appears to moving at a velocity of -10 ms-1 (20 - 30). In other words, Car A appears to going South (-North) at 10 ms-1.
· To calculate the relative velocity of one object relative to a second object, subtract the velocity vector of the second object from the velocity vector of the first object.
Graphical representation of motion - Giancoli, Ch 2: 11 view a Java Applet
4. Draw and analyse distance–time graphs, displacement–time graphs, velocity–time graphs and acceleration–time graphs. Students should be able to sketch and label these graphs for various situations. They should also be able to write descriptions of the motions represented by such graphs.
· Forward (positive) motion is represented by a positive slope on a position-time graph. Backward (negative) motion is represented by a negative slope on a position-time graph.
· Constant velocity is represented by a linear, sloped position-time graph or a horizontal line on a velocity-time graph.
· Forward (positive) velocity is drawn in the first quadrant (above the time axis) on a v-t graph. Backward (negative) velocity is drawn in the fourth quadrant (below the time axis) on a v-t graph.
· Uniformly accelerated motion is represented by a linear, sloped v-t graph. Positive acceleration has a positive slope and negative acceleration has a negative slope.
· Uniformly accelerated motion is represented by a curved (parabola) position-time graph. Positive acceleration is an upward curved parabola. Negative acceleration is a downward curved parabola.
5. Analyze and calculate the slopes of displacement–time graphs and velocity– time graphs, and the areas under velocity–time graphs and acceleration–time graphs. Relate these to the relevant kinematic quantity.
· Slope of a position-time graph is the velocity of the motion. To find instantaneous velocity, find the slope of the line tangent to the curve at a specific location (for curved graphs).
· Slope of a velocity-time graph is the acceleration of the motion. To find the instantaneous acceleration, find the slope of the line tangent to the curve at a specific location (for curved graphs).
· The area between the velocity-time curve and the time axis represents the displacement of the motion.
· The area between the acceleration-time curve and the time axis represents the change in velocity for the motion.
Uniformly accelerated motion - Giancoli, Ch 2: 8-10
6. Determine the velocity and acceleration from simple timing situations.
Students should be able to interpret data from devices such as a light gate, strobe photograph or ticker timer. Analysis may involve graphing the data, taking measurements and applying kinematics concepts.
An example: the dots in the diagram represent successive positions of a car, taken at 1-second intervals, beginning with time = 0 seconds. 1 cm on the diagram represents 10 m. To find the velocity during any interval, you would measure the distance between successive dots, convert the measurement to meters, and divide by the elapsed time. To find the acceleration of the motion, you would need to find the average velocity for two successive intervals and then divide by the same time interval.
7. Derive the equations for uniformly accelerated motion.
· v = u + at this is the equation of a line on a v-t graph with u as the vertical intercept and a as the slope
· Ds = ut + 1/2at2 this is the area under a velocity time graph with u as the vertical intercept and a as the slope
· Ds = (u + v)t / 2 this is still the area under a velocity-time graph (u + v)/2 is the average velocity
· v2 = u2 + 2aDs solve the first equation for t, sub into the second equation and simplify
8. Describe the vertical motion of an object in a uniform gravitational field.
· Freefall motion is the motion of an object falling only under the influence of gravity. Air resistance is assumed negligible .
· Objects fall with uniform acceleration (9.8 ms-2 directed downward)
· All objects, regardless of their mass, fall with the same uniform acceleration.
· The same equations of motion can be used for freefall motion. If the upward direction is taken as positive motion, then a = - 9.8 ms-2 (the minus sign denotes a downward direction for the acceleration).
· The acceleration of an object moving vertically up or down is always 9.8 ms-2 directed downward, regardless of where in the trajectory the object is located.
9. Describe the effects of air resistance on falling objects.
· If air resistance is included in the problem, mathematically it becomes much more complicated. This is because the amount of air resistance changes in relation to the speed of the falling object.
· As objects fall, the amount of air resistance (directed upward) increases until it is exactly balanced by the downward pull of gravity. At this point, there is no net force on the falling object and it will fall with a constant speed from that point on. The constant speed that objects reach is called terminal velocity.
10. Solve problems involving uniformly accelerated motion.
· See Giancoli, Ch 2 Problems Level I: #2, 4, 15a, 19, 21, 35
Level II: #8, 11, 37, 41, 45, 61
Level III: #17a, 29, 51
MECHANICS
2. MECHANICS
2.1. Mechanics - the foundation of physics
The first and most important part of the areas of physics is mechanics, which forms a basis for other parts to be presented later. The quantities used here will reappear in many placer - there are many types of forces, but all follow the laws of Newton. Velocity is not only a quantity to be studied for its own sake, but for example the velocity of an electric charge affects how it reacts to magnetic fields. The mechanics studied in this topic is classical mechanics, developed mainly in the time period 1600-1800. A more precise modern theory of mechanics, involving Einstein's theory of relativity and quantum mechanics can be learned later. For most technical applications - including advanced technology like sending a spacecraft to the planet Mars - this classical mechanics is still sufficient.
2.2. Distance and displacement
We start the physics course with mechanics which deals with questions like where something is, how fast and in what direction it moves, how its motion changes, what causes it, and some consequences of the answers to these questions. All this fits the universal character of physics. An example of this is the quantity speed (described later): a car may drive at a speed, an animal may run or fly at a speed, blood can flow through your veins at a speed, a distant star or galaxy may move towards or away from us at some speed.
Before we get to the quantity speed, we need to describe something more fundamental: where something is. In physics, there are two ways to tell how far something has moved or how far from a certain point it is:
distance (etäisyys, avstånd) = how you went measured along the path you actually took. The tripmeter in a car measures distance. Since the road can be curved it is difficult to say what direction you took, and distance is then a scalar. Common symbol : s
displacement (siirtymä, lägesändring) = how far it is from where you started to where you stopped in a straight line. If you look at the map maybe you can find out that the town you drove to is 25 km to the northwest of where you started. This is a vector, which also often has the symbol s.
If only two directions are possible, it is convenient to used positive values for displacements in one direction and negative in the opposite.
Ex. The train moved 500 m forwards and then 200 m backwards.
* If we call the forwards direction positive, the total displacement is 500m + (-200m) = 500 m - 200 m = 300 m.
* If we call the backwards direction positive, we have - 500 m + 200 m = - 300 m.
Note:
· in both cases we could add the displacements, with their signs. The same formula could have been used: stotal = s1 + s2
· the answers are different although they represent the same motion in the real world. They must be interpreted using the chosen definition of which direction that is positive.
2.3. Speed and velocity
This can also be described either with or without a direction:
speed (vauhti, fart) = distance/time This is a scalar. Unit ms-1
velocity (nopeus, hastighet) = displacement/time This is a vector, same unit.
(note that 1 ms-1 = 3.6 kmh-1)
The same formula is used for both (v for velocity or speed, s for distance or displacement, t for time):
v = Ds/Dt [DB p. 4]
The symbol D stands for the change in something = the difference between what it is now and what it was before. In many situations it can be dropped - for example the time for something to happen is the difference between what the clock showed after it and when it started, but if we started a stopwatch from zero when the event started, then the reading on the stopwatch when the event is over equals the time it took. We then often use the formula in the form
v = s / t
If the velocity is constant (both magnitude and direction!), we have what is called uniform motion, UM.
Frames of reference and relative velocity
Example: A boat A moves with 5 ms-1 downstream, another boat B with 5 ms-1 upstream in a river flowing 2 ms-1 relative to the shore. Both move at 5 ms-1 in the "river's frame of reference", but their speeds in the "shore frame of reference" (or their speeds relative to the shore) are 3 ms-1 and 7 ms-1.
2.4. Acceleration (kiihtyvyys, acceleration)
If the velocity changes (magnitude and/or direction), we have an acceleration. We will first focus on cases where something moves along a straight line, but where the speed = the magnitude of the velocity changes. We use these symbols:
u = initial velocity
v = final velocity
t = time to change velocity from u to v
a = acceleration
Dv = v-u = change in velocity
The definition of acceleration is then
a = Dv/Dt [DB p. 4]
where we can write Dv/Dt = (v - u) / t (assuming that t = the time it took for the velocity to change from u to v). Acceleration is a vector and its unit is ms-2 (which means m/s2). The formula is often written in this form after solving for v:
· a = (v - u) / t multiply both sides with t so
· at = (v - u) = v - u move - u to the left side, letting it change sign
· at + u = v or as below:
v = u + at [DB p. 5]
If the acceleration a is constant, we have uniformly accelerated motion, UAM.
· Near earth, all things fall down with a gravity acceleration g = 9.81 ms-2 if we do not think of air resistance.
For UM we would have a = 0 and v = u + at would become v = u ; the velocity is constant
2.5. Graphs of UM and UAM
UM:
· The graph of velocity as a function of time (velocity on y-axis, time on x-axis) is a horizontal straight line (the velocity is constant). If an object has traveled for the time t with the velocity v, the displacement (how far it as moved) is given by v = s/t => s = vt. This is the of the area (rectangle) under the graph.
· The graph of displacement as a function of time is a straight line which is steeper the higher the velocity is. Compare this to the graphs of y = x, y = 2x, y = 3x etc where the graph y = kx is steeper the higher k is. Here we have s = vt with s instead of y, v instead of k and t instead of x. The velocity is now the gradient (slope) of the line. This means that you take any two points A and B on the curve and find how much higher B is than A, then divide it by how much further to the left B is than A.
m05a
UAM:
· The graph of v as a function of t is now a rising straight line starting from u (initial velocity) on the velocity axis. During the time t it reaches the level v (final velocity). The distance traveled is still the area under this graph - now a trapeze (like a triangle on top of a rectangle). This area can be found by adding the areas of the rectangle and the triangle or by finding the mean or average velocity vm which is (u+v)/2. Since s = vmt we then get:
s = [(u + v)/2]t [DB p. 5]
· The graph of displacement s as a function of time is now not a straight line but a curve bending upwards (getting steeper and steeper - the gradient or slope is still = the velocity, but since this increases all the time, we would need to draw a "help line" (called tangent) and find the gradient = slope of this by choosing two points on it)
m05b
Other types of motion (neither UM or UAM)
If the velocity is not constant (not UM) and the acceleration not constant (not UM) is still true that the travelled displacement is the area under the v-t curve (which may be found with geometry, numerical approximations on a computer, or other methods) and that the velocity at a certain time is the slope of the s-t curve.
m05c
Instantaneous and average values
If one quantity is the gradient (slope) of another (e.g. velocity from displacement or acceleration from velocity) we can graphically find either an average or an instantaneous value. The average value is the change in the vertical coordinate / the change in the horizontal coordinate. The instantaneous value is the "average" value for an infinitely small change in the horizontal coordinate.
m05d
2.6. The 4 equations of uniformly accelerated motion = UAM
We already have
v = u + at [DB p. 5]
and
s = vmt where vm = (u+v)/2 [DB p. 5]
We can now
· replace v in 2) by u +at and get vm = (u+u+at)/2 = (2u + at)/2
· simplify vm = (2u +at)/2 = 2u/2 + at/2 = u + ½at
· to get s = vmt multiply with t and have t(u + ½at) = ut + ½at2
so we have:
s = ut + ½at2 [DB p. 5]
Another possibility is to
· solve 1) for t which gives t = (v - u)/a
· replace t in 2) with this, so s =vmt = vm(v - u)/a
· use from 2) that vm = (u + v)/2 to get s = (u + v)(v - u)/2a
· let (u +v)(v - u) = (v + u)(v - u) = vv - vu + uv - uu = v2 -vu + vu - u2 = v2 - u2
which all gives us that
· s = (v2 - u2)/2a which we multiply with 2a to get v2 - u2 = 2as
· and finally v2 = u2 + 2as so :
v2 = u2 + 2as [DB p. 5]
Note that the equations are valid only for constant acceleration!
2.7. Force and mass
Force (voima, kraft) (a vector quantity)
is the cause of for example
· deformation (stretching, bending, compressing, other). It is measured with a forcemeter (dynamometer, newtonmeter) containing a spiral metal spring which is extended (stretched out) more the greater the force is. Unit : 1 newton = 1 N.
· acceleration = change in velocity per time.
Resultant force (resultant, total force, net force, sum of all forces)
Often several forces act on the same object. If you hold something in your hand, there is a force of gravity pulling it down and a force from your hand upwards which may balance out the downwards force so the resultant is zero. This can be handled by choosing one direction as positive (ex. up) and giving the forces signs accordingly.
Example:
Force of gravity = Fg = - 5.0 N Force of hand = Fh = 5.0 N
Resultant = Fg + Fh = - 5.0 N + 5.0 N = 0
Newton's 3 laws for forces :
Newton I
If the resultant force on an object is zero, its velocity will be constant.
This can mean either of two possibilities:
· the object is at rest and will remain so as long as the resultant is zero (like the object in your hand).
· the object has some velocity and will keep it (both direction and magnitude) as long as the resultant is zero. Example : A car comes to a curve where the road is extremelt slippery because of ice. The driver would like to either slow down or change direction, but because of the ice no force can be applied to it horizontally, so it continues out into the forest where forces from trees it collides with slows it down. (This was in the horizontal dimension - in the vertical dimension there is a force of gravity down which is balanced out by the force from the hard ice in the road keeping it from sinking into it).
A free-body diagram = sketch of an object showing the forces acting on it using arrows with a length proportional to the magnitude (if known). Forces (as other vectors) can using trigonometry be resolved into components in two dimensions perpendicular to each other, and the components added separately. The resultant force/ resultant/ total force/ net force can be found using Pythagoras.
Translational equilibrium = a situation where the net force in all dimensions is zero. Example: an object sliding down a slope at constant speed, when the component of the force of gravity down the slope and the force (ex. friction) balance out, and the same is true for the normal force (perpendicular to surface) and the component of the force of gravity perpendicular to the surface (draw diagram, choose labels, resolve into components!).
Newton II
If there is a resultant force F, then there will be a change in velocity = acceleration
which is greater the greater F is, but smaller the greater the mass of the object is.
a = F/m
A larger engine giving a larger net force will increase acceleration
A larger mass will decrease it.
F = ma [DB p. 5]
This means that the unit 1 N = 1 kgms-2 . Mass is a scalar, but acceleration is a vector, so the force is also a vector.
Newton III
If A acts on B with the force F then B acts back on A with - F
(-F is a force of the same magnitude but opposite direction to F). Examples:
· A rifle fires a bullet and acts with a force on it accelerating it forwards, but the bullet acts back on the rifle so it recoils
· A rocket engine in a space ship throws out gases acting with them, and then the gases act back on the rocket with a force forwards (note that the rocket does not "push against the air" to drive it forwards, it works out in empty space).
Mass and weight
· mass is a property of an object which it has whereever we take it - a 100 kg astronaut is a 100 kg astronaut here or on the moon
· weight is the force of gravity acting on something - on the moon where the force of gravity is weaker, the weight in newtons is lower.
The force of gravity is
Fg = mg
where g = the gravity constant or gravity acceleration = 9.81 ms-2 on earth, 1.6 ms-2 on the moon.
· Inertial mass = F/a (where F is resultant force, regardless of what kind of force this - force of gravity, force of hand, force of rocket engine, electrical forces or other).
· Gravitational mass = F/g (near earth) the property of an object which determines how large the force of gravity on the object is.
There is basically no "good" reason why the inertial and gravitational masses should be the same - why the quantity which says how much force of any kind is needed to accelerate an object should be the same as the one which says how strong one particular force (gravity) is. For other the three other fundamental forces (electromagnetic, strong and weak nuclear force) the strength of the force is determined by other quantities (ex. electric charge).
2.8. Work, energy, power
Work and energy
m09a
If the force or a component Fs of it is in the direction of its displacement, the work (a scalar) done is
W = (Fss =) Fs cosq [DB p. 5]
with the unit 1 joule = 1 J = 1 Nm.The amount of work done is the energy (same unit) converted from one form to another.
In a velocity-time diagram the displacement is the area under the graph since s =vt for UM, for other types of motion the area is not a rectangle but still equal to s. Similarly, in a graph of Fs as a function of s, the area under the graph - rectangle or other - is the work W.
Kinetic energy (liike-energia, rörelseenergi)
· if a car is accelerated from rest by the constant horizontal force F then the work done is W = Fs = mas; here q = 0
· from the equation for UAM v2 = u2 +2as we now get v2 = 2as and then a = v2/2s
· inserting this in W = mas gives W = ½mv2 which is "stored" in the moving car, so
Ek = ½mv2 [DB p. 5]
Gravitational potential energy (potentiaalinen energia, lägesenergi)
· if an object falls from the height h the force of gravity does a work W = Fs = mgs = mgh on it:
Ep = mgh [DB p. 5]
These sum of these is the total mechanical energy, which is constant (that is, conserved) unless energy is lost to do work against friction, air resistance or other.
Power (teho, effekt)
P (= E/t or W/t) = work/time = Fv [DB p. 5]
unit 1 watt = 1 W = 1 Js-1. Power is the amount of work done or energy transformed from one form to another per time; it can be called the rate of working. "The rate of X" means "how much X per time". Note that for an object moving at a constant speed v the power P = W/t = Fs/t = Fv where F is not the resultant force but the force keeping it in motion despite friction, air resistance etc. Note the older unit 1 horsepower = ca 735 W.
Efficiency (hyötysuhde, verkningsgrad)
e or h = Eout/Ein or Pout/Pin [not in DB but a similar definition is given in thermal physics, DB p.6]
where Ein is the work or energy supplied and Eout that which is converted to something "useful". What this is depends on the purpose of the device; for a light bulb where a certain amount of electric energy is supplied, the useful energy is that converted to light and the energy converted to heat wasted. For a bread toaster, it is the opposite. Power can be used instead of work or energy since the time t is canceled: Pout/Pin = (Eout/t)/(Ein/t) = Eout/Ein
2.9. Friction
Friction (kitka, friktion)
The force of friction is caused by interaction between atoms in the material of a surface and in an object in contact with it. For the force of friction we have
Ffr = mkN and Ffr < or = msN [DB p. 5]
m = positive friction coefficient, without unit, which can be
· kinetic (index k) or dynamic or sliding for moving object (force opposite to velocity)
· or static (index s) for object at rest (force opposite to net force trying to set it in motion). In this case the value is such that the force of friction balances any net force trying to set the object in motion until some maximum value, when the object "jumps" into motion and the force of friction then is kinetic (with a constant coefficient somewhat smaller than the maximum value of the static one)
N = normal force, the force with which the surface is pressing towards the object (on a horizontal surface N = -FG so it can be replaced by the force of gravity in a calculation where only magnitudes are involved.
Alternatively: We use different positive-negative directions in the horizontal and vertical dimensions. This means that N or FN (which is in the vertical dimension, balancing out the force of gravity G or FG) may be given a different sign when used to calculate the force of friction as the expression mN since m is always positive and the force of friction can be either positive or negative depending our choice of directions. The force of friction is, in principle, not affected by the area of the object which is in contact with the surface.
m08a
For an object on an incline (slope) it must be noted that the normal force is not the opposite of the force of gravity, but of the component of the force of gravity perpendicular to the slope.
m08b
For a moving object, Ffr is in the opposite direction to the velocity. For a static object, it is in the opposite direction to the resultant of all other forces acting on it.
2.10. Springs
Linear springs
If a spring is extended (pulled out) or compressed (pushed in) a displacement x it acts with a force according to
F = (-) ks [DB p. 5]
A force which follows this type of a formula is called a harmonic force.
m10a
where k = spring constant, unit Nm-1 (higher the stronger the spring is);
the minus sign shows that the force of the spring is in the opposite direction to the displacement s from the equilibrium position
Elastic potential energy
When a spring is extended or compressed, work is done on it which can be stored in it as an elastic potential energy. Since the force needed to overcome the force of the spring is not constant but increases linearly the work done = the area under the force graph = ½ * the base * the height = ½ * x * F = ½ * x * kx =
Eelas = ½kx2 [DB p. 5]
m10b
2.11.* Simple harmonic motion
Mass on spring
It can be shown that for a mass m oscillating on a spring with the spring constant k, the time period T for the oscillations follow the formula:
T = 2pÖ(k/m) [not in DB]
Simple pendulum
In a similar way it can be shown that for a mass m (sometimes called the pendulum "bob") swinging at the end of an assumedly massless pendulum of the length l has the time period
T = 2pÖ(l/g) [not in DB]
2.12. Momentum and impulse
(Linear) momentum (liikemäärä, rörelsemängd)
a vector quantity, unit 1 kgms-1 , is defined as:
p =mv [DB p. 5]
If we define momentum p = mv we can also write NII as F = Dp/t (meaning "net force is the rate of change in the momentum") since initial momentum = mu, final momentum = mv and change in momentum per time = (mv - mu)/t = m(v - u)/t = ma = F. Note that momentum = Fi. 'liikemäärä', Sw. 'rörelsemängd'. Fi. '(voiman) momentti' or 'vääntömomentti' and Sw. '(kraft)moment' or 'vridmoment' all = torque, a quantity to be presented later.
Note: here F is the resultant force
F = Dp/Dt [DB p. 5]
When two objects A and B collide or otherwise interact for the time t and no external force is acting (e.g. the force of friction can is neglected), the total moment is conserved (the same before and after the collision) since
· N III : A acts on B with F so B acts on A with - F
· no external forces, so these are the resultant forces on A and B
· N II for A: - F = maA = m(vA - uA)/t = (mvA - muA)/t = DpA / t
· N II for B: F = maB = m(vB - uB)/t = (mvB - muB)/t= DpB / t
· therefore DpA/t = - DpB/t and Dptotal = DpA + DpB = 0
· no change in total momentum means it is the same before and after
m11a
In calculations for problems with two objects colliding, the most useful form of this is
m1u1 + m2u2 = m1v1 + m2v2 [not in DB]
where the formula is adapted according to the situation, e.g. :
· if object 2 was at rest before the collision then u2 = 0 and the term m2u2 dropped
· if the objects stay together after the collision, then v1 = v2 = v and m1v1 + m2v2 = (m1 + m2)v
· one direction is chosen positive, and the velocities given positive or negative values accordingly. If a velocity is calculated, the sign shows its direction
Since momentum is a vector we can have collisions in two dimensions where the momentums and/or the velocities are split up into components in two perpenducular dimensions. These are then both conserved m1u1X + m2u2X = m1v1X + m2v2X and m1u1Y + m2u2Y = m1v1Y + m2v2Y). The components of the momentum are found using trigonometry like for velocities.
m11b
Another useful relation is the following: Since p = mv => p2 = m2v2 => p2/2m = ½mv2 so:
Ek = p2 / 2m [DB p. 5]
Impulse (impulssi, impuls)
I = FDt = Dp [DB p.5]
(unit 1 kgms-1 = 1 Ns) where F is the resultant force acting on an object, t the time during which the force acts (can be a very short time for a collision). If the force acting is not constant, the only way to find the impulse and with that the change in momentum is to find the area under the graph of F as a function t. If we find the impulse from the graph, then I = Dp = m(v-u).
m11c
Elastic collisions
In an elastic collision, e.g. two hard billiard balls colliding and bouncing apart, the total kinetic energy is also conserved.
Example: A billiard ball A with the mass m and velocity uA collides elastically with another identical ball B at rest. What will happen?
Conservation of momentum: muA + muB = mvA + mvB
=> muA = mvA + mvB
=> uA = vA + vB
Conservation of kinetic energy: ½muA2 + ½muB2 = ½mvA2 + ½mvB2
=> ½muA2 = ½mvA2 + ½mvB2
=> uA2 = vA2 + vB2
=> (vA + vB)2 = vA2 + vB2
=> vA2 + vB2 + 2vAvB = vA2 + vB2
=> 2vAvB = 0
which is possible only if vB or vA is = 0. The first would require that B is affected by a force without any change in velocity (impossible) so the latter is true.
Inelastic collision
If some kinetic energy is lost, only momentum is conserved (if no external forces act). We must assume that a collision is inelastic unless other information is given. In a completely inelastic collision, all kinetic energy is lost (like two identical cars colliding head with the same speed at forming a wreck at rest together. Since momentum is a vector, the total is conserved - it is zero both before and after!).
2.13. Projectile motion (heittoliike, kaströrelse)
Projectile motion = UM horizontally and UAM vertically at the same time
This can be a grenade shot from a cannon, a ball thrown or kicked. The horizontal and vertical motion can be separated - split the initial velocity vector u in such components uh and uv:
m12a
We get than uh = ucosq and uv = usinq. After that the UM horizontal part and the UAM vertical part are treated with the same equations as before:
Horizontally : uh = sh/t = constant = vh
Vertically : vv = uv + avt sv = ((uv + vv)/2)t sv = uvt + ½avt2 vv2 = uv2 + 2avsv
where the vertical acceleration av = g = 9.81 ms-2 downwards (given a positive or negative sign depending on whether you chose up or down as positive). The common variable is the time t which can be used to link results from the vertical and horizontal dimensions.
Example: A ball is kicked at the initial velocity u at an angle q on a horizontal field. What is its range?
· uv and uh are obtained as above
· the time is the same as the time would be for the ball to return to the ground if thrown vertically upwards with uv
· sv = uvt + ½avt2 = t(uv + ½avt)=with sv = 0 and uv and av having opposite signs gives t = 0 or (uv + ½avt) = 0 so t = - 2uv/av (positive)
· then the horizontal uh = sh/t gives sh = uht
At any time during the projectile motion the "final" velocity (the velocity after the object has travelled from the start to the point we are interested in) is found as v = (vh2 + vv2)½ and the angle q' to the horizon from vh/vv = tan q' giving q' = arctan(vh/vv).
m12b
The path followed by an object in projectile motion is part of an upside-down parabola (like the graph to y = -x2 ). The reason for this is that the sv as a function of time is second-degree equation
sv = uvt + ½avt2 or sv = ½avt2 + uvt (compare y = ax2 + bx)
and when changing the time values on the horizontal axis to displacement (positition) values with sh = uht for a constant uh the shape of the graph does not change (compare plotting y = -x2 with different scales on the horizontal axis).
2.14. Torque
Torque
t = Fr sin Q [DB p. 5]
The torque (turning moment, moment of a force) is F times the perpendicular distance r to the pivot (point around which we turn). If F is not at a 90 degree angle to r, we can either take the component of r which is (r sin Q) or the component of F which is (F sin Q). Both give the same formula. Torque is a vector quantity, the possible directions are clockwise and counter-clockwise. When Q = 90o we use the shorter formula
t = Fr
m13a
We now have two conditions for an object to be at rest:
· translational equilibrium : the resultant force acting on it is zero (in all dimensions, usually no more than two)
· rotational equilibrium : the resultant torque is zero (around all possible pivot points)
m13b
The center of gravity is a point where one can assume that the force of gravity is acting. For objects made of a homogenous material, it is the geometric center (ex. in the middle of a staff). The center of gravity may not be located in the object (e.g. for a ring it is in the center of the ring).
For problem solving:
· the forces must balance out in all directions (up/down, left/right, others)
· the torques must balance out around any pivot (choosing ones where the perpendicular distance for a force is zero makes a term disappear!)
Using these principles we try to form a number of equations which give us the values of all unknowns.
2.15. Circular motion
Angles in degrees and radians
One full turn (revolution) in a circle is 360o = 2p radians => 1 radian = (180/p)o. The time or period of a circular motion = T and the speed v = 2pr / T where r is the radius of the circle.
Centripetal and centrifugal force
"Uniform circular motion" = motion at a constant speed v in a circle It is not UM since the direction of the velocity is changing. To keep an object in circular motion we need
· a centripetal force directed in towards the center of the circle. (ex. whirling a ball in a string: you can not push with a string, only pull)
Because of Newton's III law we then also get
· a "centrifugal" force acting on what makes the object go in a circle, not on the object itself (ex. an outward force acts on the finger holding the string)
m14a
If we think of an equilateral triangle with two sides = r and between these a small angle, then third side is ≈ the distance (in a bent curve) traveled by a point on the circle. Since the vector v is always perpendicular to r, it turns the same angle as an imagined string with the length r would have. We note that the 'new' vector v is the first vector v plus the change in velocity Δv which only affects the direction of v, not its magnitude or length. Two equilateral triangles with the same angle between the equal sides are similar to each other in such a way that the ratio between corresponding sides is the same, so for example:
s/r = Δv/v but since v = s/t we get s = vt and then
vt/r = Δv/v so multiplying with v and dividing with t we get
v2/r = Δv/t = a = ac
which by inserting v = 2pr/T also gives
ac = (4p2r2/T2)/r = 4p2r/T2
a = v2/r = 4p2r/T2 [DB p. 5]
which together with F = ma gives the
centripetal force Fc = mv2/r
The centripetal force is not a new fundamental force (like gravity, electromagnetic force, nuclear forces) nor is it a particular force of any more specific type (friction, air resistance, tension in a string - all of which are consequences of mainly electromagnetic forces between atoms and molecules) but rather it is so that different forces (fundamental or their forms in specific cases) act as centripetal force in a certain sitation. Examples:
- for a planet around a sun or a moon or satellite around a planet : gravity
- for a car taking a curve: static fricition between wheels and ground
- for the ball whirled in a string : force of tension in the string
- later : electromagnetic force acting as centripetal force for particles in a magnetic field
2.16. Universal gravitation
Earlier we have always used FG = mg for the force of gravity. But if we go to another planet or moon g has a different value, and if we move far away from our own planet gravity also gets weaker. A more universal formula (valid everywhere in space) is that the force of gravity between two point masses (small masses) m1 and m2 at a distance r from each other is (Newton's law of universal gravitation)
F = Gm1m2/r2 [DB p. 5]
where G = the universal gravity constant = 6.67 x 10-11 Nm-2kg-2
and the minus sign means that gravity is always attractive
Strictly we should calculate the force of gravity between every possible pair of atom in two larger objects attracting each other, but it can be shown (using 3-dimensional integrals!) that
if the object is a homogenous sphere, for places outside the sphere we get the same result as if we assume that all the mass is in the center of the sphere.
Newton's III. law : if m1 attracts m2 with F then m2 attracts m1 with an equally big force in the opposite direction.
2.17. Gravitational field and potential
Gravitational field strength (= gravity acceleration)
The gravity acceleration or gravity "constant" g (9.81ms-2) is not always constant now, but can be calculated for a general case:
Let m1 be the mass of earth and m2 that of an object outside earth. FG = m2g and the law of universal gravitation give
m2g = (-)Gm1m2/r2 => (when m first stands for the mass of a small object and then for that of the planet or other large central body)
g = F/m = Gm/r2 [DB p. 4]
This can be called the gravitational field strength and is a vector, towards the center of the earth. Generally for any point in space, where more than one planet contributes,
g = Fresultant/m
Potential energy - the new way
Old way (still OK near earth or near planet with known g-values) : An object falls from rest from the height h2 down to h1. With what speed will it reach h1? The change in potential energy becomes kinetic energy, so
mgh2 - mgh1 = ½mv2 etc. But this was assuming a constant value for g, which is not correct if it falls from 5000 km to 3000 km above the surface of the earth.
Gravitational potential energy
It can be shown (with integrals) that the gravitational potential energy for an object m2 at a distance r from a point mass or from the center of a sphere (not the surface!) with the mass m1 is (same as force but r, not r2):
Ep = -Gm1m2/r [DB p. 5]
If we say that the zero level of the potential energy is infinitely far away and the minus sign shows that an object at this distance is bound to the planet m1 and if put there at rest soon will fall down to it, unless it has or gets energies of other kinds, e.g. kinetic or work done by a rocket engine.
Gravitational potential
To generally describe how much potential energy an object m2 would have if placed here we can give the potential energy per mass of the object, which is called gravitational potential. It has the unit Jkg-1 and is defined as V = Ep/m2 so (when m = mass of planet or large central body):
V = -Gm/r [DB p. 5]
This means that if we know the potential at some point, the Ep which is often useful in calculations is the potential times the mass of the object there. If air resistance can be skipped, it does not make any difference how we move between the levels h1 and h2 - the energy we get or which is required is the same (= the force of gravity is a conservative force; total mechanical energy is conserved regardless of how we move. The force of friction is non-conservative.)
Summary
We will now let the larger mass be called M and the smaller m:
Quantity
At planet surface
In general
Unit
force
F = mg
F = GMm/r2
N
field intensity
g = F/m
g = GM/r2
Nkg-1 (=ms-2)
potential energy
Ep = mgh
Ep = -GMm/r
J
potential
V = Ep/m = gh
V = -GM/r
Jkg-1
Where only one mass m is indicated in the "in space" versions, it indicates the mass of the planet or other massive central body. The quantity gravitational potential V = gh is rarely used in the "near earth" situation. It could be relevant if the same application could be used near the surfaces of two different planets. E.g. a pump which on earth can pump up water to a height h is more specifically able to move water through a certain gravitational potential difference which leads to different h-values depending on the g-value on the planet in question. This situation can be further complicated by differences in athmospheric pressure on the planets, if the pump mechanism depends on that.]
2.18. Orbital motion
For planets moving around a sun, moons or satellites around a planet, the force of gravity is acting as the centripetal force. We often start calculations by noting that for a satellite with mass m2 orbiting a planet with mass m1 at the distance r from the planets center, not its surface we have
Fc = FG => m2v2/r = Gm1m2/r2 => m2v2 = Gm1m2/r
For the satellite in a stable orbit we then have:
· the kinetic energy Ek = ½mv2 = Gm1m2/2r
· the potential energy Ep = -Gm1m2/r
· the total mechanical energy Ek + Ep = (Gm1m2 - 2Gm1m2)/2r = - Gm1m2/2r
m17a
Note: the so called free fall or "weightlessness" in a stable orbit does not mean that astronauts do not have any mass in space nor that the force of gravity has been shut off. The force of gravity has not even become very much weaker in an orbit near earth - e.g. 300 km above the planet surface the distance to the center has only increased from maybe 6370 km to 6670 km. The astronauts are "weightless" because the (slightly weaker) force of gravity is acting as a centripetal force, it is needed just to keep them circling the earth in this orbit instead of flying out in space in a straight line. There is no force left over to pull them towards the floor of the spacecraft as the force of gravity does when it stands on the ground.
Orbital speed
For an object m2 in a stable orbit around a planet m1 we have as above that
Fc = FG => m2v2/r = Gm1m2/r2 => m2v2 = Gm1m2/r => .... =>
vorbital = Ö(Gm1/r) [not in DB]
Escape speed
If a spacecraft is given a high enough speed from the surface of a planet, it may get a positive kinetic energy equal to or higher than the negative potential energy it has when it is "bound" to the planet. It could then move infinitely far away from the planet without ever being pulled back, unless it uses its engine to slow down. The minimum necessary speed for this (disregarding resistance in the planet's atmosphere) can be found using:
Ek + Ep = 0 (for the minimum escape speed, we just about reach infinity with the speed about 0 so the Ek = 0; the Ep is zero at infinity by definition).
so
Ek = - Ep => ½m2v2 = -(-Gm1m2/r) => v2 = 2Gm1/r => (when m1 = mass of planet or large central body, and m2 = the rocket)
vescape = Ö( 2Gm1/r) = vorbitalÖ2 [not in DB]
Note: Since the earth is rotating and spacecraft follows it, it has some kinetic energy before the start. To use this it is favourable to let the rocket start close to the equator (where the ground moves with a higher speed than close to the poles to make a revolution in 24 hours) and towards east, in the direction of rotation. One also wants to have some open sea under the first part of the trajectory (path) so that if the rocket explodes, the pieces do not fall on people. This has led to the choices of location (Florida for the USA, French Guayana for France).
2.19. Kepler's laws
Both Kepler I and II can mathematically be proven as necessary consequences of Newton's law of universal gravity, although this is very advanced.
Kepler I : the planets orbit the sun in ellipses with the sun in one focus
Kepler II : a line from the planet to the sun sweeps the same area in the same time
m18a
Consequence: it must move faster when it passes the focus where the sun is, and is near to it. For earth, this occurs when the axis of the earth is tilted so the southern hemisphere is towards the sun. For this reason the summer is slightly shorter down there and Antarctica colder than Greenland (there is a little more incoming sunlight when the sun is closer in the summer there, but this effect turns out to be less important than the length of the summer).
Kepler III : if a planet (now mass m2) orbits the sun (now mass m1) with the time period T at an average distances r to the (center of the) sun, we can by approximating the ellipses to circles get:
T2 is proportional to r3 <=> T2 µ r3 <=> T2 = a constant times r3 or r3 = another constant times T2
Why? The speed = distance/time = 2pr/T so Fc = FG gives as earlier p. 14
m2v2/r = Gm1m2/r2 => m1v2 = Gm1m2/r =>
m2(2pr/T)2 = Gm1m2/r => (4p2r2/T2) = Gm1/r =>
1/T2 = Gm1/4p2r3 => T2 = 4p2r3/Gm1 = kr3
which in the data booklet is described as:
T2 / R3 = constant [DB p. 5]
Note: Kepler III is valid only for objects rotating around the same central mass, e.g. different planets around a sun or different moons or satellites around the same planet. A convenient form of Kepler's III. law for two planets or other objects A and B for which it is valid is:
TA2 / rA3 = TB2 / rB3
using the more usual r instead of R for the distance between the centers of the bodies.
2.20* Rotational mechanics
For rotational motion a set of mechanics formulas similar to those for linear mechanics (objects moving in a straight line) can be developed. Instead of the distance or displacement s we can study the angle turned, or the angular displacement q. In radians we have by definition
q = s/r
where r = the radius of the circle and s = the distance covered along its circumference. In a similar way we can define an angular velocity w = q/t (the angle turned per time) and an angular acceleration a = w/t (the change in angular velocity per time).
The results on a rotational motion of a force depend on how far from the center of rotation it is applied, so force will be replaced by torque, t = Fr.
Without proof we will notice that mass also will be replaced by "moment of inertia", I or J where J = mr2 if all the mass is at the same distance from the center or axis of rotation. If not, then it can be shown that J follows certain formulas like J = 2/5*mr2 for a sphere, J = (1/3)ml2 for a bar of length l rotating around one end (like a baseball bat) or J = (1/12)ml2 for the bar rotating around its center (like a propeller). Time is the same for linear (translational) and rotational motion. Summary:
TRANSLATIONAL => ROTATIONAL
s => q = s/r
v => w = v/r
a => a = a/r
F => t = Fr
m => J = mr2
Using the "word list" above we can "translate" the known translational formulas into the corresponding rotational ones, for example:
v = u + at => wfinal = winitial + at
s = ut + ½at2 => q = winitialt + ½at2
F = ma => t = Ja
Ek = ½mv2 => Erotational = ½Jw2
p = mv => L = Jw
The rotational or angular momentum L will be relevant in Atomic physics later. We may notice that:
L = Jw = mr2(v/r) = mvr
for an electron in a circular orbit around the nucleus of an atom. ]
2.21* Fluid mechanics
Pressure, a scalar quantity, is defined as
p = F / A
where F = the force acting perpendicularly on a surface with the area A. Its unit is 1 pascal = 1 Pa = 1 Nm-2. Ordinary atmospheric pressure is ca 100 kPa.
The pressure at a depth h in a liquid = hydrostatic pressure p = rgh or
p = p0 + rgh
where g = 9.81 ms-2 and r ("rho") = the density = m/V in kgm-3 of the liquid. We may include the atmospheric pressure p0 acting on the surface of the liquid.
· Pascal's principle: pressure applied to a fluid is the same everywhere at the same depth in it, and "acts" in all directions (is a scalar quantity).
An application of this is the hydraulic lift (used in car brakes) where force is applied to a liquid (oil) on a large area and then spreads through the liquid where it is allowed to act on a much smaller area attached to the brake mechanism or other. Then with F = p/A => p = FA so
pin = pout gives FinAin = FoutAout and Fout = FinAin/Aout
which means that a small force in causes a larger force out (but to keep the volume of liquid constant, the Fin must move a piston or other a longer distance than Fout; therefore the work done is the same.
Archimede's law: the upwards force buoyancy (lyftkraft, nostovoima) on a submerged object is
F = rVg
= the force of gravity on the mass of the amount of water displaced by the object. r = density of the liquid the body is immersed in, V = its volume, g = 9.81 ms-2. The law can be proven by expressing the difference in pressure on the lower and upper side of an immersed body.
2.1. Mechanics - the foundation of physics
The first and most important part of the areas of physics is mechanics, which forms a basis for other parts to be presented later. The quantities used here will reappear in many placer - there are many types of forces, but all follow the laws of Newton. Velocity is not only a quantity to be studied for its own sake, but for example the velocity of an electric charge affects how it reacts to magnetic fields. The mechanics studied in this topic is classical mechanics, developed mainly in the time period 1600-1800. A more precise modern theory of mechanics, involving Einstein's theory of relativity and quantum mechanics can be learned later. For most technical applications - including advanced technology like sending a spacecraft to the planet Mars - this classical mechanics is still sufficient.
2.2. Distance and displacement
We start the physics course with mechanics which deals with questions like where something is, how fast and in what direction it moves, how its motion changes, what causes it, and some consequences of the answers to these questions. All this fits the universal character of physics. An example of this is the quantity speed (described later): a car may drive at a speed, an animal may run or fly at a speed, blood can flow through your veins at a speed, a distant star or galaxy may move towards or away from us at some speed.
Before we get to the quantity speed, we need to describe something more fundamental: where something is. In physics, there are two ways to tell how far something has moved or how far from a certain point it is:
distance (etäisyys, avstånd) = how you went measured along the path you actually took. The tripmeter in a car measures distance. Since the road can be curved it is difficult to say what direction you took, and distance is then a scalar. Common symbol : s
displacement (siirtymä, lägesändring) = how far it is from where you started to where you stopped in a straight line. If you look at the map maybe you can find out that the town you drove to is 25 km to the northwest of where you started. This is a vector, which also often has the symbol s.
If only two directions are possible, it is convenient to used positive values for displacements in one direction and negative in the opposite.
Ex. The train moved 500 m forwards and then 200 m backwards.
* If we call the forwards direction positive, the total displacement is 500m + (-200m) = 500 m - 200 m = 300 m.
* If we call the backwards direction positive, we have - 500 m + 200 m = - 300 m.
Note:
· in both cases we could add the displacements, with their signs. The same formula could have been used: stotal = s1 + s2
· the answers are different although they represent the same motion in the real world. They must be interpreted using the chosen definition of which direction that is positive.
2.3. Speed and velocity
This can also be described either with or without a direction:
speed (vauhti, fart) = distance/time This is a scalar. Unit ms-1
velocity (nopeus, hastighet) = displacement/time This is a vector, same unit.
(note that 1 ms-1 = 3.6 kmh-1)
The same formula is used for both (v for velocity or speed, s for distance or displacement, t for time):
v = Ds/Dt [DB p. 4]
The symbol D stands for the change in something = the difference between what it is now and what it was before. In many situations it can be dropped - for example the time for something to happen is the difference between what the clock showed after it and when it started, but if we started a stopwatch from zero when the event started, then the reading on the stopwatch when the event is over equals the time it took. We then often use the formula in the form
v = s / t
If the velocity is constant (both magnitude and direction!), we have what is called uniform motion, UM.
Frames of reference and relative velocity
Example: A boat A moves with 5 ms-1 downstream, another boat B with 5 ms-1 upstream in a river flowing 2 ms-1 relative to the shore. Both move at 5 ms-1 in the "river's frame of reference", but their speeds in the "shore frame of reference" (or their speeds relative to the shore) are 3 ms-1 and 7 ms-1.
2.4. Acceleration (kiihtyvyys, acceleration)
If the velocity changes (magnitude and/or direction), we have an acceleration. We will first focus on cases where something moves along a straight line, but where the speed = the magnitude of the velocity changes. We use these symbols:
u = initial velocity
v = final velocity
t = time to change velocity from u to v
a = acceleration
Dv = v-u = change in velocity
The definition of acceleration is then
a = Dv/Dt [DB p. 4]
where we can write Dv/Dt = (v - u) / t (assuming that t = the time it took for the velocity to change from u to v). Acceleration is a vector and its unit is ms-2 (which means m/s2). The formula is often written in this form after solving for v:
· a = (v - u) / t multiply both sides with t so
· at = (v - u) = v - u move - u to the left side, letting it change sign
· at + u = v or as below:
v = u + at [DB p. 5]
If the acceleration a is constant, we have uniformly accelerated motion, UAM.
· Near earth, all things fall down with a gravity acceleration g = 9.81 ms-2 if we do not think of air resistance.
For UM we would have a = 0 and v = u + at would become v = u ; the velocity is constant
2.5. Graphs of UM and UAM
UM:
· The graph of velocity as a function of time (velocity on y-axis, time on x-axis) is a horizontal straight line (the velocity is constant). If an object has traveled for the time t with the velocity v, the displacement (how far it as moved) is given by v = s/t => s = vt. This is the of the area (rectangle) under the graph.
· The graph of displacement as a function of time is a straight line which is steeper the higher the velocity is. Compare this to the graphs of y = x, y = 2x, y = 3x etc where the graph y = kx is steeper the higher k is. Here we have s = vt with s instead of y, v instead of k and t instead of x. The velocity is now the gradient (slope) of the line. This means that you take any two points A and B on the curve and find how much higher B is than A, then divide it by how much further to the left B is than A.
m05a
UAM:
· The graph of v as a function of t is now a rising straight line starting from u (initial velocity) on the velocity axis. During the time t it reaches the level v (final velocity). The distance traveled is still the area under this graph - now a trapeze (like a triangle on top of a rectangle). This area can be found by adding the areas of the rectangle and the triangle or by finding the mean or average velocity vm which is (u+v)/2. Since s = vmt we then get:
s = [(u + v)/2]t [DB p. 5]
· The graph of displacement s as a function of time is now not a straight line but a curve bending upwards (getting steeper and steeper - the gradient or slope is still = the velocity, but since this increases all the time, we would need to draw a "help line" (called tangent) and find the gradient = slope of this by choosing two points on it)
m05b
Other types of motion (neither UM or UAM)
If the velocity is not constant (not UM) and the acceleration not constant (not UM) is still true that the travelled displacement is the area under the v-t curve (which may be found with geometry, numerical approximations on a computer, or other methods) and that the velocity at a certain time is the slope of the s-t curve.
m05c
Instantaneous and average values
If one quantity is the gradient (slope) of another (e.g. velocity from displacement or acceleration from velocity) we can graphically find either an average or an instantaneous value. The average value is the change in the vertical coordinate / the change in the horizontal coordinate. The instantaneous value is the "average" value for an infinitely small change in the horizontal coordinate.
m05d
2.6. The 4 equations of uniformly accelerated motion = UAM
We already have
v = u + at [DB p. 5]
and
s = vmt where vm = (u+v)/2 [DB p. 5]
We can now
· replace v in 2) by u +at and get vm = (u+u+at)/2 = (2u + at)/2
· simplify vm = (2u +at)/2 = 2u/2 + at/2 = u + ½at
· to get s = vmt multiply with t and have t(u + ½at) = ut + ½at2
so we have:
s = ut + ½at2 [DB p. 5]
Another possibility is to
· solve 1) for t which gives t = (v - u)/a
· replace t in 2) with this, so s =vmt = vm(v - u)/a
· use from 2) that vm = (u + v)/2 to get s = (u + v)(v - u)/2a
· let (u +v)(v - u) = (v + u)(v - u) = vv - vu + uv - uu = v2 -vu + vu - u2 = v2 - u2
which all gives us that
· s = (v2 - u2)/2a which we multiply with 2a to get v2 - u2 = 2as
· and finally v2 = u2 + 2as so :
v2 = u2 + 2as [DB p. 5]
Note that the equations are valid only for constant acceleration!
2.7. Force and mass
Force (voima, kraft) (a vector quantity)
is the cause of for example
· deformation (stretching, bending, compressing, other). It is measured with a forcemeter (dynamometer, newtonmeter) containing a spiral metal spring which is extended (stretched out) more the greater the force is. Unit : 1 newton = 1 N.
· acceleration = change in velocity per time.
Resultant force (resultant, total force, net force, sum of all forces)
Often several forces act on the same object. If you hold something in your hand, there is a force of gravity pulling it down and a force from your hand upwards which may balance out the downwards force so the resultant is zero. This can be handled by choosing one direction as positive (ex. up) and giving the forces signs accordingly.
Example:
Force of gravity = Fg = - 5.0 N Force of hand = Fh = 5.0 N
Resultant = Fg + Fh = - 5.0 N + 5.0 N = 0
Newton's 3 laws for forces :
Newton I
If the resultant force on an object is zero, its velocity will be constant.
This can mean either of two possibilities:
· the object is at rest and will remain so as long as the resultant is zero (like the object in your hand).
· the object has some velocity and will keep it (both direction and magnitude) as long as the resultant is zero. Example : A car comes to a curve where the road is extremelt slippery because of ice. The driver would like to either slow down or change direction, but because of the ice no force can be applied to it horizontally, so it continues out into the forest where forces from trees it collides with slows it down. (This was in the horizontal dimension - in the vertical dimension there is a force of gravity down which is balanced out by the force from the hard ice in the road keeping it from sinking into it).
A free-body diagram = sketch of an object showing the forces acting on it using arrows with a length proportional to the magnitude (if known). Forces (as other vectors) can using trigonometry be resolved into components in two dimensions perpendicular to each other, and the components added separately. The resultant force/ resultant/ total force/ net force can be found using Pythagoras.
Translational equilibrium = a situation where the net force in all dimensions is zero. Example: an object sliding down a slope at constant speed, when the component of the force of gravity down the slope and the force (ex. friction) balance out, and the same is true for the normal force (perpendicular to surface) and the component of the force of gravity perpendicular to the surface (draw diagram, choose labels, resolve into components!).
Newton II
If there is a resultant force F, then there will be a change in velocity = acceleration
which is greater the greater F is, but smaller the greater the mass of the object is.
a = F/m
A larger engine giving a larger net force will increase acceleration
A larger mass will decrease it.
F = ma [DB p. 5]
This means that the unit 1 N = 1 kgms-2 . Mass is a scalar, but acceleration is a vector, so the force is also a vector.
Newton III
If A acts on B with the force F then B acts back on A with - F
(-F is a force of the same magnitude but opposite direction to F). Examples:
· A rifle fires a bullet and acts with a force on it accelerating it forwards, but the bullet acts back on the rifle so it recoils
· A rocket engine in a space ship throws out gases acting with them, and then the gases act back on the rocket with a force forwards (note that the rocket does not "push against the air" to drive it forwards, it works out in empty space).
Mass and weight
· mass is a property of an object which it has whereever we take it - a 100 kg astronaut is a 100 kg astronaut here or on the moon
· weight is the force of gravity acting on something - on the moon where the force of gravity is weaker, the weight in newtons is lower.
The force of gravity is
Fg = mg
where g = the gravity constant or gravity acceleration = 9.81 ms-2 on earth, 1.6 ms-2 on the moon.
· Inertial mass = F/a (where F is resultant force, regardless of what kind of force this - force of gravity, force of hand, force of rocket engine, electrical forces or other).
· Gravitational mass = F/g (near earth) the property of an object which determines how large the force of gravity on the object is.
There is basically no "good" reason why the inertial and gravitational masses should be the same - why the quantity which says how much force of any kind is needed to accelerate an object should be the same as the one which says how strong one particular force (gravity) is. For other the three other fundamental forces (electromagnetic, strong and weak nuclear force) the strength of the force is determined by other quantities (ex. electric charge).
2.8. Work, energy, power
Work and energy
m09a
If the force or a component Fs of it is in the direction of its displacement, the work (a scalar) done is
W = (Fss =) Fs cosq [DB p. 5]
with the unit 1 joule = 1 J = 1 Nm.The amount of work done is the energy (same unit) converted from one form to another.
In a velocity-time diagram the displacement is the area under the graph since s =vt for UM, for other types of motion the area is not a rectangle but still equal to s. Similarly, in a graph of Fs as a function of s, the area under the graph - rectangle or other - is the work W.
Kinetic energy (liike-energia, rörelseenergi)
· if a car is accelerated from rest by the constant horizontal force F then the work done is W = Fs = mas; here q = 0
· from the equation for UAM v2 = u2 +2as we now get v2 = 2as and then a = v2/2s
· inserting this in W = mas gives W = ½mv2 which is "stored" in the moving car, so
Ek = ½mv2 [DB p. 5]
Gravitational potential energy (potentiaalinen energia, lägesenergi)
· if an object falls from the height h the force of gravity does a work W = Fs = mgs = mgh on it:
Ep = mgh [DB p. 5]
These sum of these is the total mechanical energy, which is constant (that is, conserved) unless energy is lost to do work against friction, air resistance or other.
Power (teho, effekt)
P (= E/t or W/t) = work/time = Fv [DB p. 5]
unit 1 watt = 1 W = 1 Js-1. Power is the amount of work done or energy transformed from one form to another per time; it can be called the rate of working. "The rate of X" means "how much X per time". Note that for an object moving at a constant speed v the power P = W/t = Fs/t = Fv where F is not the resultant force but the force keeping it in motion despite friction, air resistance etc. Note the older unit 1 horsepower = ca 735 W.
Efficiency (hyötysuhde, verkningsgrad)
e or h = Eout/Ein or Pout/Pin [not in DB but a similar definition is given in thermal physics, DB p.6]
where Ein is the work or energy supplied and Eout that which is converted to something "useful". What this is depends on the purpose of the device; for a light bulb where a certain amount of electric energy is supplied, the useful energy is that converted to light and the energy converted to heat wasted. For a bread toaster, it is the opposite. Power can be used instead of work or energy since the time t is canceled: Pout/Pin = (Eout/t)/(Ein/t) = Eout/Ein
2.9. Friction
Friction (kitka, friktion)
The force of friction is caused by interaction between atoms in the material of a surface and in an object in contact with it. For the force of friction we have
Ffr = mkN and Ffr < or = msN [DB p. 5]
m = positive friction coefficient, without unit, which can be
· kinetic (index k) or dynamic or sliding for moving object (force opposite to velocity)
· or static (index s) for object at rest (force opposite to net force trying to set it in motion). In this case the value is such that the force of friction balances any net force trying to set the object in motion until some maximum value, when the object "jumps" into motion and the force of friction then is kinetic (with a constant coefficient somewhat smaller than the maximum value of the static one)
N = normal force, the force with which the surface is pressing towards the object (on a horizontal surface N = -FG so it can be replaced by the force of gravity in a calculation where only magnitudes are involved.
Alternatively: We use different positive-negative directions in the horizontal and vertical dimensions. This means that N or FN (which is in the vertical dimension, balancing out the force of gravity G or FG) may be given a different sign when used to calculate the force of friction as the expression mN since m is always positive and the force of friction can be either positive or negative depending our choice of directions. The force of friction is, in principle, not affected by the area of the object which is in contact with the surface.
m08a
For an object on an incline (slope) it must be noted that the normal force is not the opposite of the force of gravity, but of the component of the force of gravity perpendicular to the slope.
m08b
For a moving object, Ffr is in the opposite direction to the velocity. For a static object, it is in the opposite direction to the resultant of all other forces acting on it.
2.10. Springs
Linear springs
If a spring is extended (pulled out) or compressed (pushed in) a displacement x it acts with a force according to
F = (-) ks [DB p. 5]
A force which follows this type of a formula is called a harmonic force.
m10a
where k = spring constant, unit Nm-1 (higher the stronger the spring is);
the minus sign shows that the force of the spring is in the opposite direction to the displacement s from the equilibrium position
Elastic potential energy
When a spring is extended or compressed, work is done on it which can be stored in it as an elastic potential energy. Since the force needed to overcome the force of the spring is not constant but increases linearly the work done = the area under the force graph = ½ * the base * the height = ½ * x * F = ½ * x * kx =
Eelas = ½kx2 [DB p. 5]
m10b
2.11.* Simple harmonic motion
Mass on spring
It can be shown that for a mass m oscillating on a spring with the spring constant k, the time period T for the oscillations follow the formula:
T = 2pÖ(k/m) [not in DB]
Simple pendulum
In a similar way it can be shown that for a mass m (sometimes called the pendulum "bob") swinging at the end of an assumedly massless pendulum of the length l has the time period
T = 2pÖ(l/g) [not in DB]
2.12. Momentum and impulse
(Linear) momentum (liikemäärä, rörelsemängd)
a vector quantity, unit 1 kgms-1 , is defined as:
p =mv [DB p. 5]
If we define momentum p = mv we can also write NII as F = Dp/t (meaning "net force is the rate of change in the momentum") since initial momentum = mu, final momentum = mv and change in momentum per time = (mv - mu)/t = m(v - u)/t = ma = F. Note that momentum = Fi. 'liikemäärä', Sw. 'rörelsemängd'. Fi. '(voiman) momentti' or 'vääntömomentti' and Sw. '(kraft)moment' or 'vridmoment' all = torque, a quantity to be presented later.
Note: here F is the resultant force
F = Dp/Dt [DB p. 5]
When two objects A and B collide or otherwise interact for the time t and no external force is acting (e.g. the force of friction can is neglected), the total moment is conserved (the same before and after the collision) since
· N III : A acts on B with F so B acts on A with - F
· no external forces, so these are the resultant forces on A and B
· N II for A: - F = maA = m(vA - uA)/t = (mvA - muA)/t = DpA / t
· N II for B: F = maB = m(vB - uB)/t = (mvB - muB)/t= DpB / t
· therefore DpA/t = - DpB/t and Dptotal = DpA + DpB = 0
· no change in total momentum means it is the same before and after
m11a
In calculations for problems with two objects colliding, the most useful form of this is
m1u1 + m2u2 = m1v1 + m2v2 [not in DB]
where the formula is adapted according to the situation, e.g. :
· if object 2 was at rest before the collision then u2 = 0 and the term m2u2 dropped
· if the objects stay together after the collision, then v1 = v2 = v and m1v1 + m2v2 = (m1 + m2)v
· one direction is chosen positive, and the velocities given positive or negative values accordingly. If a velocity is calculated, the sign shows its direction
Since momentum is a vector we can have collisions in two dimensions where the momentums and/or the velocities are split up into components in two perpenducular dimensions. These are then both conserved m1u1X + m2u2X = m1v1X + m2v2X and m1u1Y + m2u2Y = m1v1Y + m2v2Y). The components of the momentum are found using trigonometry like for velocities.
m11b
Another useful relation is the following: Since p = mv => p2 = m2v2 => p2/2m = ½mv2 so:
Ek = p2 / 2m [DB p. 5]
Impulse (impulssi, impuls)
I = FDt = Dp [DB p.5]
(unit 1 kgms-1 = 1 Ns) where F is the resultant force acting on an object, t the time during which the force acts (can be a very short time for a collision). If the force acting is not constant, the only way to find the impulse and with that the change in momentum is to find the area under the graph of F as a function t. If we find the impulse from the graph, then I = Dp = m(v-u).
m11c
Elastic collisions
In an elastic collision, e.g. two hard billiard balls colliding and bouncing apart, the total kinetic energy is also conserved.
Example: A billiard ball A with the mass m and velocity uA collides elastically with another identical ball B at rest. What will happen?
Conservation of momentum: muA + muB = mvA + mvB
=> muA = mvA + mvB
=> uA = vA + vB
Conservation of kinetic energy: ½muA2 + ½muB2 = ½mvA2 + ½mvB2
=> ½muA2 = ½mvA2 + ½mvB2
=> uA2 = vA2 + vB2
=> (vA + vB)2 = vA2 + vB2
=> vA2 + vB2 + 2vAvB = vA2 + vB2
=> 2vAvB = 0
which is possible only if vB or vA is = 0. The first would require that B is affected by a force without any change in velocity (impossible) so the latter is true.
Inelastic collision
If some kinetic energy is lost, only momentum is conserved (if no external forces act). We must assume that a collision is inelastic unless other information is given. In a completely inelastic collision, all kinetic energy is lost (like two identical cars colliding head with the same speed at forming a wreck at rest together. Since momentum is a vector, the total is conserved - it is zero both before and after!).
2.13. Projectile motion (heittoliike, kaströrelse)
Projectile motion = UM horizontally and UAM vertically at the same time
This can be a grenade shot from a cannon, a ball thrown or kicked. The horizontal and vertical motion can be separated - split the initial velocity vector u in such components uh and uv:
m12a
We get than uh = ucosq and uv = usinq. After that the UM horizontal part and the UAM vertical part are treated with the same equations as before:
Horizontally : uh = sh/t = constant = vh
Vertically : vv = uv + avt sv = ((uv + vv)/2)t sv = uvt + ½avt2 vv2 = uv2 + 2avsv
where the vertical acceleration av = g = 9.81 ms-2 downwards (given a positive or negative sign depending on whether you chose up or down as positive). The common variable is the time t which can be used to link results from the vertical and horizontal dimensions.
Example: A ball is kicked at the initial velocity u at an angle q on a horizontal field. What is its range?
· uv and uh are obtained as above
· the time is the same as the time would be for the ball to return to the ground if thrown vertically upwards with uv
· sv = uvt + ½avt2 = t(uv + ½avt)=with sv = 0 and uv and av having opposite signs gives t = 0 or (uv + ½avt) = 0 so t = - 2uv/av (positive)
· then the horizontal uh = sh/t gives sh = uht
At any time during the projectile motion the "final" velocity (the velocity after the object has travelled from the start to the point we are interested in) is found as v = (vh2 + vv2)½ and the angle q' to the horizon from vh/vv = tan q' giving q' = arctan(vh/vv).
m12b
The path followed by an object in projectile motion is part of an upside-down parabola (like the graph to y = -x2 ). The reason for this is that the sv as a function of time is second-degree equation
sv = uvt + ½avt2 or sv = ½avt2 + uvt (compare y = ax2 + bx)
and when changing the time values on the horizontal axis to displacement (positition) values with sh = uht for a constant uh the shape of the graph does not change (compare plotting y = -x2 with different scales on the horizontal axis).
2.14. Torque
Torque
t = Fr sin Q [DB p. 5]
The torque (turning moment, moment of a force) is F times the perpendicular distance r to the pivot (point around which we turn). If F is not at a 90 degree angle to r, we can either take the component of r which is (r sin Q) or the component of F which is (F sin Q). Both give the same formula. Torque is a vector quantity, the possible directions are clockwise and counter-clockwise. When Q = 90o we use the shorter formula
t = Fr
m13a
We now have two conditions for an object to be at rest:
· translational equilibrium : the resultant force acting on it is zero (in all dimensions, usually no more than two)
· rotational equilibrium : the resultant torque is zero (around all possible pivot points)
m13b
The center of gravity is a point where one can assume that the force of gravity is acting. For objects made of a homogenous material, it is the geometric center (ex. in the middle of a staff). The center of gravity may not be located in the object (e.g. for a ring it is in the center of the ring).
For problem solving:
· the forces must balance out in all directions (up/down, left/right, others)
· the torques must balance out around any pivot (choosing ones where the perpendicular distance for a force is zero makes a term disappear!)
Using these principles we try to form a number of equations which give us the values of all unknowns.
2.15. Circular motion
Angles in degrees and radians
One full turn (revolution) in a circle is 360o = 2p radians => 1 radian = (180/p)o. The time or period of a circular motion = T and the speed v = 2pr / T where r is the radius of the circle.
Centripetal and centrifugal force
"Uniform circular motion" = motion at a constant speed v in a circle It is not UM since the direction of the velocity is changing. To keep an object in circular motion we need
· a centripetal force directed in towards the center of the circle. (ex. whirling a ball in a string: you can not push with a string, only pull)
Because of Newton's III law we then also get
· a "centrifugal" force acting on what makes the object go in a circle, not on the object itself (ex. an outward force acts on the finger holding the string)
m14a
If we think of an equilateral triangle with two sides = r and between these a small angle, then third side is ≈ the distance (in a bent curve) traveled by a point on the circle. Since the vector v is always perpendicular to r, it turns the same angle as an imagined string with the length r would have. We note that the 'new' vector v is the first vector v plus the change in velocity Δv which only affects the direction of v, not its magnitude or length. Two equilateral triangles with the same angle between the equal sides are similar to each other in such a way that the ratio between corresponding sides is the same, so for example:
s/r = Δv/v but since v = s/t we get s = vt and then
vt/r = Δv/v so multiplying with v and dividing with t we get
v2/r = Δv/t = a = ac
which by inserting v = 2pr/T also gives
ac = (4p2r2/T2)/r = 4p2r/T2
a = v2/r = 4p2r/T2 [DB p. 5]
which together with F = ma gives the
centripetal force Fc = mv2/r
The centripetal force is not a new fundamental force (like gravity, electromagnetic force, nuclear forces) nor is it a particular force of any more specific type (friction, air resistance, tension in a string - all of which are consequences of mainly electromagnetic forces between atoms and molecules) but rather it is so that different forces (fundamental or their forms in specific cases) act as centripetal force in a certain sitation. Examples:
- for a planet around a sun or a moon or satellite around a planet : gravity
- for a car taking a curve: static fricition between wheels and ground
- for the ball whirled in a string : force of tension in the string
- later : electromagnetic force acting as centripetal force for particles in a magnetic field
2.16. Universal gravitation
Earlier we have always used FG = mg for the force of gravity. But if we go to another planet or moon g has a different value, and if we move far away from our own planet gravity also gets weaker. A more universal formula (valid everywhere in space) is that the force of gravity between two point masses (small masses) m1 and m2 at a distance r from each other is (Newton's law of universal gravitation)
F = Gm1m2/r2 [DB p. 5]
where G = the universal gravity constant = 6.67 x 10-11 Nm-2kg-2
and the minus sign means that gravity is always attractive
Strictly we should calculate the force of gravity between every possible pair of atom in two larger objects attracting each other, but it can be shown (using 3-dimensional integrals!) that
if the object is a homogenous sphere, for places outside the sphere we get the same result as if we assume that all the mass is in the center of the sphere.
Newton's III. law : if m1 attracts m2 with F then m2 attracts m1 with an equally big force in the opposite direction.
2.17. Gravitational field and potential
Gravitational field strength (= gravity acceleration)
The gravity acceleration or gravity "constant" g (9.81ms-2) is not always constant now, but can be calculated for a general case:
Let m1 be the mass of earth and m2 that of an object outside earth. FG = m2g and the law of universal gravitation give
m2g = (-)Gm1m2/r2 => (when m first stands for the mass of a small object and then for that of the planet or other large central body)
g = F/m = Gm/r2 [DB p. 4]
This can be called the gravitational field strength and is a vector, towards the center of the earth. Generally for any point in space, where more than one planet contributes,
g = Fresultant/m
Potential energy - the new way
Old way (still OK near earth or near planet with known g-values) : An object falls from rest from the height h2 down to h1. With what speed will it reach h1? The change in potential energy becomes kinetic energy, so
mgh2 - mgh1 = ½mv2 etc. But this was assuming a constant value for g, which is not correct if it falls from 5000 km to 3000 km above the surface of the earth.
Gravitational potential energy
It can be shown (with integrals) that the gravitational potential energy for an object m2 at a distance r from a point mass or from the center of a sphere (not the surface!) with the mass m1 is (same as force but r, not r2):
Ep = -Gm1m2/r [DB p. 5]
If we say that the zero level of the potential energy is infinitely far away and the minus sign shows that an object at this distance is bound to the planet m1 and if put there at rest soon will fall down to it, unless it has or gets energies of other kinds, e.g. kinetic or work done by a rocket engine.
Gravitational potential
To generally describe how much potential energy an object m2 would have if placed here we can give the potential energy per mass of the object, which is called gravitational potential. It has the unit Jkg-1 and is defined as V = Ep/m2 so (when m = mass of planet or large central body):
V = -Gm/r [DB p. 5]
This means that if we know the potential at some point, the Ep which is often useful in calculations is the potential times the mass of the object there. If air resistance can be skipped, it does not make any difference how we move between the levels h1 and h2 - the energy we get or which is required is the same (= the force of gravity is a conservative force; total mechanical energy is conserved regardless of how we move. The force of friction is non-conservative.)
Summary
We will now let the larger mass be called M and the smaller m:
Quantity
At planet surface
In general
Unit
force
F = mg
F = GMm/r2
N
field intensity
g = F/m
g = GM/r2
Nkg-1 (=ms-2)
potential energy
Ep = mgh
Ep = -GMm/r
J
potential
V = Ep/m = gh
V = -GM/r
Jkg-1
Where only one mass m is indicated in the "in space" versions, it indicates the mass of the planet or other massive central body. The quantity gravitational potential V = gh is rarely used in the "near earth" situation. It could be relevant if the same application could be used near the surfaces of two different planets. E.g. a pump which on earth can pump up water to a height h is more specifically able to move water through a certain gravitational potential difference which leads to different h-values depending on the g-value on the planet in question. This situation can be further complicated by differences in athmospheric pressure on the planets, if the pump mechanism depends on that.]
2.18. Orbital motion
For planets moving around a sun, moons or satellites around a planet, the force of gravity is acting as the centripetal force. We often start calculations by noting that for a satellite with mass m2 orbiting a planet with mass m1 at the distance r from the planets center, not its surface we have
Fc = FG => m2v2/r = Gm1m2/r2 => m2v2 = Gm1m2/r
For the satellite in a stable orbit we then have:
· the kinetic energy Ek = ½mv2 = Gm1m2/2r
· the potential energy Ep = -Gm1m2/r
· the total mechanical energy Ek + Ep = (Gm1m2 - 2Gm1m2)/2r = - Gm1m2/2r
m17a
Note: the so called free fall or "weightlessness" in a stable orbit does not mean that astronauts do not have any mass in space nor that the force of gravity has been shut off. The force of gravity has not even become very much weaker in an orbit near earth - e.g. 300 km above the planet surface the distance to the center has only increased from maybe 6370 km to 6670 km. The astronauts are "weightless" because the (slightly weaker) force of gravity is acting as a centripetal force, it is needed just to keep them circling the earth in this orbit instead of flying out in space in a straight line. There is no force left over to pull them towards the floor of the spacecraft as the force of gravity does when it stands on the ground.
Orbital speed
For an object m2 in a stable orbit around a planet m1 we have as above that
Fc = FG => m2v2/r = Gm1m2/r2 => m2v2 = Gm1m2/r => .... =>
vorbital = Ö(Gm1/r) [not in DB]
Escape speed
If a spacecraft is given a high enough speed from the surface of a planet, it may get a positive kinetic energy equal to or higher than the negative potential energy it has when it is "bound" to the planet. It could then move infinitely far away from the planet without ever being pulled back, unless it uses its engine to slow down. The minimum necessary speed for this (disregarding resistance in the planet's atmosphere) can be found using:
Ek + Ep = 0 (for the minimum escape speed, we just about reach infinity with the speed about 0 so the Ek = 0; the Ep is zero at infinity by definition).
so
Ek = - Ep => ½m2v2 = -(-Gm1m2/r) => v2 = 2Gm1/r => (when m1 = mass of planet or large central body, and m2 = the rocket)
vescape = Ö( 2Gm1/r) = vorbitalÖ2 [not in DB]
Note: Since the earth is rotating and spacecraft follows it, it has some kinetic energy before the start. To use this it is favourable to let the rocket start close to the equator (where the ground moves with a higher speed than close to the poles to make a revolution in 24 hours) and towards east, in the direction of rotation. One also wants to have some open sea under the first part of the trajectory (path) so that if the rocket explodes, the pieces do not fall on people. This has led to the choices of location (Florida for the USA, French Guayana for France).
2.19. Kepler's laws
Both Kepler I and II can mathematically be proven as necessary consequences of Newton's law of universal gravity, although this is very advanced.
Kepler I : the planets orbit the sun in ellipses with the sun in one focus
Kepler II : a line from the planet to the sun sweeps the same area in the same time
m18a
Consequence: it must move faster when it passes the focus where the sun is, and is near to it. For earth, this occurs when the axis of the earth is tilted so the southern hemisphere is towards the sun. For this reason the summer is slightly shorter down there and Antarctica colder than Greenland (there is a little more incoming sunlight when the sun is closer in the summer there, but this effect turns out to be less important than the length of the summer).
Kepler III : if a planet (now mass m2) orbits the sun (now mass m1) with the time period T at an average distances r to the (center of the) sun, we can by approximating the ellipses to circles get:
T2 is proportional to r3 <=> T2 µ r3 <=> T2 = a constant times r3 or r3 = another constant times T2
Why? The speed = distance/time = 2pr/T so Fc = FG gives as earlier p. 14
m2v2/r = Gm1m2/r2 => m1v2 = Gm1m2/r =>
m2(2pr/T)2 = Gm1m2/r => (4p2r2/T2) = Gm1/r =>
1/T2 = Gm1/4p2r3 => T2 = 4p2r3/Gm1 = kr3
which in the data booklet is described as:
T2 / R3 = constant [DB p. 5]
Note: Kepler III is valid only for objects rotating around the same central mass, e.g. different planets around a sun or different moons or satellites around the same planet. A convenient form of Kepler's III. law for two planets or other objects A and B for which it is valid is:
TA2 / rA3 = TB2 / rB3
using the more usual r instead of R for the distance between the centers of the bodies.
2.20* Rotational mechanics
For rotational motion a set of mechanics formulas similar to those for linear mechanics (objects moving in a straight line) can be developed. Instead of the distance or displacement s we can study the angle turned, or the angular displacement q. In radians we have by definition
q = s/r
where r = the radius of the circle and s = the distance covered along its circumference. In a similar way we can define an angular velocity w = q/t (the angle turned per time) and an angular acceleration a = w/t (the change in angular velocity per time).
The results on a rotational motion of a force depend on how far from the center of rotation it is applied, so force will be replaced by torque, t = Fr.
Without proof we will notice that mass also will be replaced by "moment of inertia", I or J where J = mr2 if all the mass is at the same distance from the center or axis of rotation. If not, then it can be shown that J follows certain formulas like J = 2/5*mr2 for a sphere, J = (1/3)ml2 for a bar of length l rotating around one end (like a baseball bat) or J = (1/12)ml2 for the bar rotating around its center (like a propeller). Time is the same for linear (translational) and rotational motion. Summary:
TRANSLATIONAL => ROTATIONAL
s => q = s/r
v => w = v/r
a => a = a/r
F => t = Fr
m => J = mr2
Using the "word list" above we can "translate" the known translational formulas into the corresponding rotational ones, for example:
v = u + at => wfinal = winitial + at
s = ut + ½at2 => q = winitialt + ½at2
F = ma => t = Ja
Ek = ½mv2 => Erotational = ½Jw2
p = mv => L = Jw
The rotational or angular momentum L will be relevant in Atomic physics later. We may notice that:
L = Jw = mr2(v/r) = mvr
for an electron in a circular orbit around the nucleus of an atom. ]
2.21* Fluid mechanics
Pressure, a scalar quantity, is defined as
p = F / A
where F = the force acting perpendicularly on a surface with the area A. Its unit is 1 pascal = 1 Pa = 1 Nm-2. Ordinary atmospheric pressure is ca 100 kPa.
The pressure at a depth h in a liquid = hydrostatic pressure p = rgh or
p = p0 + rgh
where g = 9.81 ms-2 and r ("rho") = the density = m/V in kgm-3 of the liquid. We may include the atmospheric pressure p0 acting on the surface of the liquid.
· Pascal's principle: pressure applied to a fluid is the same everywhere at the same depth in it, and "acts" in all directions (is a scalar quantity).
An application of this is the hydraulic lift (used in car brakes) where force is applied to a liquid (oil) on a large area and then spreads through the liquid where it is allowed to act on a much smaller area attached to the brake mechanism or other. Then with F = p/A => p = FA so
pin = pout gives FinAin = FoutAout and Fout = FinAin/Aout
which means that a small force in causes a larger force out (but to keep the volume of liquid constant, the Fin must move a piston or other a longer distance than Fout; therefore the work done is the same.
Archimede's law: the upwards force buoyancy (lyftkraft, nostovoima) on a submerged object is
F = rVg
= the force of gravity on the mass of the amount of water displaced by the object. r = density of the liquid the body is immersed in, V = its volume, g = 9.81 ms-2. The law can be proven by expressing the difference in pressure on the lower and upper side of an immersed body.
Deadline
This is to remind all the DP1 students the Lab report are expected on the 9th of October. We shall clear the circular motion section on the same day and in the second section i will share with you all the syllabus as well as the options.
The other tuesdays will be lab work. Please note down the date, title and hours you have spent for each lab work you have done.
Mr. G
The other tuesdays will be lab work. Please note down the date, title and hours you have spent for each lab work you have done.
Mr. G
Subscribe to:
Comments (Atom)