5 Steps to a 5: AP Physics C (2016)
Review the Knowledge You Need to Score High
IN THIS CHAPTER
Summary: The mechanics of rotating objects make up a significant chunk of the AP Physics C exam. But, rotational motion has many direct analogs from linear motion. If you understand the mechanics unit from your first-year physics course, rotational motion can be learned by analogy.
Rotational kinematics uses essentially the same equations as kinematics, but distances, speeds, and accelerations are rotational quantities expressed with radians instead of with meters.
The rotational inertia defines an object”s resistance to rotation. It is the rotational analog of mass.
An object can possess rotational kinetic energy in addition to translational kinetic energy.
Angular momentum is conserved anytime no external torque acts on a system of objects. This includes planets in orbit.
Rotational kinematics equations.
First, for constant angular acceleration:
And, in all cases:
Conversion between linear and rotational variables:
Newton”s second law for rotation:
Rotational kinetic energy:
Okay, you now have thoroughly studied how an object moves; that is, if the object can be treated as a point particle. But what happens when an object is spinning? How does that affect motion? Now is the time to learn.
You”ll find clear similarities between the material in this chapter and what you already know. In fact, you”ll see that every concept, every variable, every equation for rotational motion has an analog in our study of translational motion. The best way to understand rotational motion is simply to apply the concepts of linear motion to the case of a spinning object.
For an object moving in a straight line, we defined five variables.
Now consider a fixed object spinning, like a compact disc. The relevant variables become the following:
These variables are related via the following three equations. Obviously, these equations differ from the “star equations” used for kinematics … but they”re nonetheless very similar:
So try this example:
A bicycle has wheels with radius 50 cm. The bike starts from rest, and the wheels speed up uniformly to 200 revolutions per minute in 10 seconds. How far does the bike go?
In any linear kinematics problem the units should be in meters and seconds; in rotational kinematics, the units MUST be in RADIANS and seconds. So convert revolutions per minute to radians per second. To do so, recall that there are 2π radians in every revolution:
200 rev/min × 2π rad/rev × 1 min/60 s = 21 rad/s.
Now identify variables in a chart:
We want to solve for Δθ because if we know the angle through which the wheel turns, we can figure out how far the edge of the wheel has gone. We know we can solve for Δθ , because we have three of the five variables. Plug and chug into the rotational kinematics equations:
What does this answer mean? Well, if there are 2π (that is, 6.2) radians in one revolution, then 105 radians is about 17 revolutions through which the wheel has turned.
Now, because the wheel has a radius of 0.50 m, the wheel”s circumference is 2πr = 3.1 m; every revolution of the wheel moves the bike 3.1 meters forward. And the wheel made 17 revolutions, giving a total distance of about 53 meters.
Is this reasonable? Sure—the biker traveled across about half a football field in 10 seconds.
There are a few other equations you should know. If you want to figure out the linear position, speed, or acceleration of a spot on a spinning object, or an object that”s rolling without slipping, use these three equations:
where r represents the distance from the spot you”re interested in to the center of the object.
So in the case of the bike wheel above, the top speed of the bike was v = (0.5 m) (21 rad/s) = 11 m/s, or about 24 miles per hour—reasonable for an average biker. Note: To use these equations, angular variable units must involve radians, not degrees or revolutions!!!
The rotational kinematics equations, just like the linear kinematics equations, are only valid when acceleration is constant. If acceleration is changing, then the same calculus equations that were used for linear kinematics apply here:
Newton”s second law states that F net = ma ; this tells us that the greater the mass of an object, the harder it is to accelerate. This tendency for massive objects to resist changes in their velocity is referred to as inertia.
Well, spinning objects also resist changes in their angular velocity. But that resistance, that rotational inertia, depends less on the mass of an object than on how that mass is distributed. For example, a baseball player often warms up by placing a weight on the outer end of the bat—this makes the bat more difficult to swing. But he does not place the weight on the bat handle, because extra weight in the handle hardly affects the swing at all.
The rotational inertia, I , is the rotational equivalent of mass. It tells how difficult it is for an object to speed up or slow its rotation. For a single particle of mass m a distance r from the axis of rotation, the rotational inertia is
To find the rotational inertia of several masses—for example, two weights connected by a thin, light rod—just add the I due to each mass.
For a complicated, continuous body, like a sphere or a disk, I can be calculated through integration:
Exam tip from an AP Physics veteran:
On the AP exam, you will only very occasionally have to use calculus to derive a rotational inertia. Usually you will either be able to sum the I due to several masses, or you will be given I for the object in question.
—Joe, college physics student and Physics C alumnus
Newton”s Second Law for Rotation
For linear motion, Newton says F net = ma ; for rotational motion, the analog to Newton”s second law is
where t net is the net torque on an object. Perhaps the most common application of this equation involves pulleys with mass.
A 2.0-kg block on a smooth table is connected to a hanging 3.0-kg block with a light string. This string passes over a pulley of mass 0.50 kg, as shown in the diagram below. Determine the acceleration of the masses. (The rotational inertia for a solid disc of mass m and radius r is ½mr 2 .)
We learned how to approach this type of problem in Chapter 12 —draw a free-body diagram for each block, and use F net = ma . So we start that way.
The twist in this problem is the massive pulley, which causes two changes in the problem-solving approach:
- We have to draw a free-body diagram for the pulley as well as the blocks. Even though it doesn”t move, it still requires torque to accelerate its spinning speed.
- We oversimplified things in Chapter 12 when we said, “One rope = one tension.” The Physics C corollary to this rule says, “… unless the rope is interrupted by a mass.” Because the pulley is massive, the tension in the rope may be different on each side of the pulley.
The Physics C corollary means that the free-body diagrams indicate T 1 and T 2 , which must be treated as different variables. The free-body diagram for the pulley includes only these two tensions:
Now, we write Newton”s second law for each block:
For the pulley, because it is rotating , we write Newton”s second law for rotation. The torque provided by each rope is equal to the tension in the rope times the distance to the center of rotation; that is, the radius of the pulley. (We”re not given this radius, so we”ll just call it R for now and hope for the best.)
The acceleration of each block must be the same because they”re connected by a rope; the linear acceleration of a point on the edge of the pulley must also be the same as that of the blocks. So, in the pulley equation, replace a by a /R. Check it out, all the R terms cancel! Thank goodness, too, because the radius of the pulley wasn”t even given in the problem.
The pulley equation, thus, simplifies to
Now we”re left with an algebra problem: three equations and three variables (T 1 , T 2 , and a ). Solve using addition or substitution. Try adding the first two equations together—this gives a T 1 − T 2 term that meshes nicely with the third equation.
The acceleration turns out to be 5.6 m/s2 . If you do the problem neglecting the mass of the pulley (try it!) you get 5.9 m/s2 . This makes sense—the more massive the pulley, the harder it is for the system of masses to speed up.
Rotational Kinetic Energy
The pulley in the last example problem had kinetic energy—it was moving, after all—but it didn”t have linear kinetic energy, because the velocity of its center of mass was zero. When an object is rotating, its rotational kinetic energy is found by the following equation:
Notice that this equation is very similar to the equation for linear kinetic energy. But, because we”re dealing with rotation here, we use rotational inertia in place of mass and angular velocity in place of linear velocity.
If an object is moving linearly at the same time that it”s rotating, its total kinetic energy equals the sum of the linear KE and the rotational KE.
Let”s put this equation into practice. Try this example problem.
A ball of mass m sits on an inclined plane, with its center of mass at a height h above the ground. It is released from rest and allowed to roll without slipping down the plane. What is its velocity when it reaches the ground? I ball = ( )mr 2 .
This is a situation you”ve seen before, except there”s a twist: this time, when the object moves down the inclined plane, it gains both linear and rotational kinetic energy. However, it”s still just a conservation of energy problem at heart. Initially, the ball just has gravitational potential energy, and when it reaches the ground, it has both linear kinetic and rotational kinetic energy.
A bit of algebra, and we find that
If the ball in this problem hadn”t rolled down the plane—if it had just slid—its final velocity would have been . (Don”t believe us? Try the calculation yourself for practice!) So it makes sense that the final velocity of the ball when it does roll down the plane is less than ; only a fraction of the initial potential energy is converted to linear kinetic energy.
Angular Momentum and Its Conservation
It probably won”t surprise you by this point that momentum, too, has a rotational form. It”s called angular momentum (abbreviated, oddly, as L ), and it is found by this formula:
This formula makes intuitive sense. If you think of angular momentum as, roughly, the amount of effort it would take to make a rotating object stop spinning, then it should seem logical that an object with a large rotational inertia or with a high angular velocity (or both) would be pretty tough to bring to rest.
For a point particle, this formula can be rewritten as
where v is linear velocity, and r is either (1) the radius of rotation, if the particle is moving in a circle, or (2) distance of closest approach if the particle is moving in a straight line. (See Figure 16.1 .)
Figure 16.1 Angular momentum.
Wait a minute! How can an object moving in a straight line have angular momentum?!? Well, for the purposes of the AP exam, it suffices just to know that if a particle moves in a straight line, then relative to some point P not on that line, the particle has an angular momentum. But if you want a slightly more satisfying—if less precise—explanation, consider this image. You”re standing outside in an open field, and an airplane passes high overhead. You first see it come over the horizon far in front of you, then it flies toward you until it”s directly over where you”re standing, and then it keeps flying until it finally disappears beneath the opposite horizon. Did the plane fly in an arc over your head or in a straight path? It would be hard for you to tell, right? In other words, when a particle moves in a straight line, an observer who”s not on that line would think that the particle sort of looked like it were traveling in a circle.
As with linear momentum, angular momentum is conserved in a closed system; that is, when no external torques act on the objects in the system. Among the most famous examples of conservation of angular momentum is a satellite”s orbit around a planet. As shown in Figure 16.2 , a satellite will travel in an elliptical orbit around a planet. This means that the satellite is closer to the planet at some times than at others.
Figure 16.2 Elliptical orbit.
Obviously, at point A , the satellite is farther from the center of rotation than at point B . Conservation of angular momentum tells us that, correspondingly, the angular speed at point A must be less than at point B . 1
The other really famous example of conservation of angular momentum involves a spinning figure skater. When a skater spinning with his or her arms outstretched suddenly brings the arms in close to the body, the speed of rotation dramatically increases. Rotational inertia decreased, so angular speed increased.
You can demonstrate this phenomenon yourself! Sit in a desk chair that spins, and with your legs outstretched, push off forcefully and start spinning. Then tuck in your feet. Dizzying, isn”t it?
1 . All of the objects mentioned in the choices below have the same total mass and length. Which has the greatest rotational inertia about its midpoint?
(A) a very light rod with heavy balls attached at either end
(B) a uniform rod
(C) a nonuniform rod, with the linear density increasing from one end to the other
(D) a nonuniform rod, with the linear density increasing from the middle to the ends
(E) a very light rod with heavy balls attached near the midpoint
2 . A pool ball is struck at one end of the table; it moves at constant speed to the far end of the table. A camera is mounted at the side pocket at the table”s midpoint, as shown. From the camera”s point of view, the pool ball travels from right to left. At which point in its motion does the ball have the greatest angular momentum about the camera”s position?
(A) when the ball was first struck
(B) at the midpoint of the table
(C) the angular momentum is the same throughout the motion
(D) at the far end of the table
(E) one-quarter of the way across the table, and then again three-quarters of the way across the table
3 . A ladder of length L leans against a wall at an angle of θ from the horizontal, as shown above. The normal force FN applied from the ground on the ladder applies what torque about the ladder”s center of mass?
(A) FN ·(L /2)
(B) FN ·L cos θ
(C) FN ·L sin θ
(D) FN ·(L /2) cos θ
(E) FN ·(L /2) sin θ
4 . The front wheel on an ancient bicycle has a radius of 0.5 m. It moves with angular velocity given by the function ω (t ) = 2 + 4t 2 , where t is in seconds. About how far does the bicycle move between t = 2 and t = 3 seconds?
(A) 36 m
(B) 27 m
(C) 21 m
(D) 14 m
(E) 7 m
5 . A stick of mass M and length L is pivoted at one end. A small mass m << M is attached to the right-hand end of the stick. The stick is held horizontally and released from rest.
(a) Given that the rotational inertia of a uniform rod pivoted around one end is (1/3)ML 2 , determine the rotational inertia of the described contraption.
(b) Calculate the angular velocity of the contraption when it reaches a vertical position.
(c) Calculate the linear velocity of the small mass m when it is at its lowest position.
(d) The figure below represents the stick at its lowest position. On this figure, draw a vector to represent the net force on the rod”s center of mass at this point. Justify your answer.
Solutions to Practice Problems
1 . A— The farther the mass from the midpoint, the larger its contribution to the rotational inertia. In choice A the mass is as far as possible from the midpoint; because all items have the same mass, A must have the largest I .
2 . C— The angular momentum of a point particle moving in a straight line about a position near the particle is mvr , where r is the distance of closest approach. This value is constant as long as the particle keeps going in a straight line, which our pool ball does.
3 . D— Torque is force times the distance to the fulcrum. The force in question is FN , and acts straight up at the base of the ladder. The distance used is the distance perpendicular to the normal force; this must be (L /2) cos θ , as shown below:
4 . D— The angular position function is given by the integral of the angular velocity function with respect to time. The limits on the integral are 2 and 3 seconds:
this evaluates to approximately 27 radians. Using x = rθ , the distance traveled is closest to 14 m.
5 . (a) The rotational inertia of the entire contraption is the sum of the moments of inertia for each part. I for the rod is given; I for a point mass a distance L from the pivot is mL 2 . So, I total = (1/3)ML 2 + m L2 . Be sure to differentiate between M and m .
(b) Rotational kinematics won”t work here because angular acceleration isn”t constant. We must use energy.
U1 + E1 = U2 + K2
Define U = 0 at the bottom of the contraption when it hangs vertically. Then, U 2 is only caused by the rod”s mass, which is concentrated L /2 above the zero point, so U 2 = MgL /2. U 1 is due to all of the mass, concentrated Labove the zero point: U 1 = (M + m )gL . K1 = 0, and K2 is unknown.
(M + m )gL + 0 = MgL /2 + ½ Iω )2.
Plug in I from part (a ) and solve for ω to get
(c) Just use v = rω . Here r = L because the center of rotation is L away from the mass.
(d) At this position, the mass is instantaneously in uniform circular motion. So, acceleration (and therefore net force) must be centripetal. Net force is straight up, toward the center of rotation.
- Rotational kinematics is very similar to linear kinematics. But instead of linear velocity, you work with angular velocity (in radians/s); instead of linear acceleration, you work with angular acceleration (in radians/s2); and instead of linear displacement, you work with angular displacement (in radians).
- When doing rotation problems, work in radians, not degrees.
- Rotational inertia is the rotational equivalent of mass—it”s a measure of how difficult it is to start or stop an object spinning.
- The rotational equivalent of Newton”s second law says that the NET torque on an object equals that object”s rotational inertia multiplied by its angular acceleration.
- To solve problems involving a massive pulley, make sure you draw a free-body diagram of the pulley. Also, when a rope passes over a massive pulley, the tension in the rope on one side of the pulley won”t be the same as the tension on the other side.
- The total kinetic energy of a rolling object is equal to the sum of its linear kinetic energy and its rotational kinetic energy.
- Angular momentum in a closed system is conserved. An object doesn”t necessarily need to travel in a circle to have angular momentum.
1 Note the consistency with Kepler”s law of equal areas in equal times, as discussed in Chapter 15 .