5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

STEP 4

Review the Knowledge You Need to Score High

CHAPTER 14

Rotation

IN THIS CHAPTER

Summary : When an object rotates, the rotation obeys rules similar to those of a moving object. Most equations and concepts have analogues for rotation. Whereas mass describes an object’s resistance to a change in speed, rotational inertia describes an object’s resistance to a change in rotational speed. Rotational inertia depends on mass as well as on the distribution of that mass.

Definitions

   Centripetal acceleration is the name given to an object’s acceleration toward the center of a circle. “Centripetal” simply means “toward the center.”

   Torque occurs when a force applied to an object could cause the object to rotate.

   The lever arm for a force is the closest distance from the fulcrum, pivot, or axis of rotation to the line on which that force acts.

   Everything covered in the previous review chapters is sufficient to describe “translational” motion. When an object rotates around a central point, or when an object is itself rotating as it moves, then we need some additional concepts. Just know that each of these rotational quantities is not truly new. Each rotational quantity should be treated exactly the same way as its translational analogue: for example, if you know how to deal with linear momentum, then angular momentum applies the same ideas to rotating objects.

The ball rolling down the ramp in the preceding figure has translational kinetic energy because it is moving. It also has rotational kinetic energy because it is spinning. The ball is losing gravitational energy because it is changing its vertical height; that loss of potential energy is converted into a gain in kinetic energy because no nonconservative forces act on the ball.

Circular Motion

Example 1: A car of mass 1,000 kg travels at constant speed around a flat curve that has a radius of curvature of 100 m. The car is going as fast as it can go without skidding.

Does this car have an acceleration? Why yes, it does, even though it moves at constant speed. Its acceleration along its direction of motion is zero, because the car isn’t speeding up or slowing down. However, its direction of motion is always changing; acceleration is technically the change in an object’s vector velocity each second, and a change in the direction of motion is a change in velocity.

FACT: When an object moves in a circle, it has an acceleration directed toward the center of the circle. The amount of that acceleration is

Strategy: When an object is moving in a circle, often a standard Newton’s second law approach is correct. Draw a free-body diagram, and then write F net = ma in each direction.

Draw a free-body diagram of this car as it moves. It’s most useful to view this car from behind; let’s say the car is turning to the right, so that the center of the circular motion is in the place indicated.

The forces are as follows:

F N is the normal force of the road on the car.

Weight is the force of the Earth on the car.

F f is the frictional force of the road on the car.

Wait, why is friction acting to the right? Shouldn’t friction act opposite the direction of motion?

Ah. There certainly could be friction or air drag acting backwards, opposite the direction of motion; but with the car moving at constant speed, that would have to be canceled by a forward engine force. 1 Since we know the car moves at constant speed, that’s probably not particularly relevant to the problem.

When a car goes around a flat curve, some sort of force must act toward the center of the circle—otherwise, the centripetal acceleration couldn’t exist. How do we know it’s friction in this case? Imagine the car were moving forward on a slick, flat sheet of ice. The car couldn’t go around a curve at all, then; turning the wheels would do nothing. On an asphalt road, it’s the static frictional force of the asphalt on the tires that pushes the car toward the center of the circle.

No matter what kind of question you’re asked about this situation, the next step is to use Newton’s second law in both the vertical and horizontal directions. Vertically, the car’s acceleration is zero; the car isn’t burrowing into the road or lifting off the road. Horizontally, we don’t have a numerical value for the acceleration, but we know its equation: 

The first equation tells us that the normal force on the car is equal to the car’s weight of 10,000 N. In order to calculate the friction force on the car, we’d need to know one of two pieces of additional information. On one hand, if we know the car’s speed, we can use the second equation to calculate F f . On the other hand, since we know the normal force, if we knew the coefficient of static friction between the car’s tires and the road, we could calculate the friction force using F f = μF n .

Strategy: You cannot allow yourself to become angry or frustrated when you don’t have enough information to complete a calculation. Sometimes, an AP Physics 1 problem will be deliberately concocted to ask, “What additional information would you need to solve this problem?” (Remember, actual calculation on the exam will be rare.) Or, often, some seemingly necessary information will be omitted, because it will turn out that the omitted information is irrelevant.

For example, an excellent problem using this situation might not give the mass of the car but instead give the car’s speed and ask what minimum coefficient of friction would be necessary for the car to round the curve. Pretend, say, that the car’s speed is 20 m/s.

I can’t do that. I need the mass to calculate the force of friction.

Well, true, if you needed a value for the force of friction, but that’s not the question. We want the coefficient of friction.

Yeah, I know. The equation is F f = μF n . I need the mass to calculate the friction force, and since the normal force is equal to the weight, I need the mass to calculate that, too. This isn’t possible.

When you’re stuck with a calculation that you think needs a value that wasn’t given, try just making up that value. It’s likely that the unknown value will cancel out. Or, if you want to be more elegant, assign a variable to the unknown value.

Let the mass be m . The equations we wrote from the free-body diagram show that the friction force is  . The normal force is the weight of the car, or the car’s mass times the gravitational field g of 10 N/kg. Now use the equation for friction force:

Look at that: solve for μ , and the masses cancel. You can plug in the 20-m/s speed and the 100-m radius to get μ = 0.4. A car of any mass can go around this curve, as long as the coefficient of friction is at least 0.4. 2

Torque

FACT: The torque τ provided by a force is given by the equation

I write the symbol “⊥” by the d to emphasize that the distance we want is the perpendicular distance from the line of the force to the fulcrum. 3 Usually that’s an easy distance to visualize.

What if a force isn’t acting perpendicular to an extended object, like the force F on the pivoted bar that follows?

Example 2:

The easiest way to find the torque applied by this force is to break the force F into vertical and horizontal components.

The vertical component of F applies a torque of (F sin θ )x . The horizontal component of F does not apply any torque, because it could not cause the bar to rotate. The total torque provided by the force F is just (F sin θ )x .

Lever Arm

This distance d  is sometimes referred to as the “lever arm” for a force. By definition, the lever arm for a force is the closest distance from the fulcrum to the line on which that force acts.

An alternate method of determining the torque applied by the force in Example 2 would be to find the lever arm instead of breaking F into components. Extend the line of the force in the diagram—now it’s easy to label the lever arm as the closest distance from the pivot to the line of the force. By trigonometry, you can figure out that the lever arm distance is equal to x sin θ . No matter how you look at it, then, the torque provided by F is still (F sin θ )x .

Calculations with Torque

You may be asked to calculate a force or a torque when an extended object experiences multiple forces, but generally only when that object is in equilibrium—that is, when up forces equal down forces, left forces equal right forces, and counterclockwise torques equal clockwise torques.

Example 3: Bob is standing on a bridge. The bridge itself weighs 10,000 N. The span between pillars A and B is 80 m. Bob, whose mass is 100 kg, stands 20 m from the center of the bridge as shown.

Generally, a problem with a bridge, plank, or some sort of extended object will ask you to describe or solve for the forces supporting the bridge. The approach is to make a list of torques acting in each direction, clockwise and counterclockwise, and then set the counterclockwise torques equal to the clockwise torques.

Aarrgh. Where’s the fulcrum? This bridge isn’t rotating anywhere!

Exactly. Since the bridge is not actually rotating, you can choose anywhere you like as the fulcrum. It’s easiest in this case, to choose one of the supports as the fulcrum, because then that support provides zero torque, and the lever arm for that force would be zero.

Let’s choose support A as the fulcrum. What torques do we see?

The force of Support B (I’ll call it F B ) provides a torque equal to F B (80 m), because support B is 80 m from support A. This torque is counterclockwise , because pushing up on the bridge pivoted at A would rotate the bridge this way  .

The weight of Bob provides a clockwise torque of (1,000 N)(20 m) = 20,000 m · N. (We don’t use 100 kg, because that’s a mass, not a force; the force acting on the bridge is due to Bob’s weight.)

Exam Tip from an AP Physics Veteran

In a torque problem with a heavy extended object, just pretend that the object’s weight is all hanging at the object’s center of mass.

The 10,000-N weight of the bridge itself provides a torque. Pretend that all 10,000 N act at the center of the bridge, 40 m away from each support. A weight pulling down at the bridge’s center would tend to rotate the bridge clockwise. The torque we want here is (10,000 N)(40 m) = 400,000 m · N, clockwise.

Now, set counterclockwise torques equal to clockwise torques:

F B (80 m) = 20,000 m · N + 400,000 m · N

Solve for F B to get 5,300 N. This is reasonable because pillar B is supporting less than half of the 11,000-N weight of the bridge and Bob. Because Bob is closer to pillar A , and otherwise the bridge is symmetric, A should bear the majority of the weight—and it does.

Rotational Kinematics

An object’s “rotational speed” says how fast the object rotates—that is, how many degrees or radians it rotates through per second. Rotational speed is generally given by the lowercase Greek variable omega, ω . An object’s “rotational acceleration” α describes how much the rotational speed changes in one second. The variable θ represents the total angle through which an object rotates in some time period.

Just like position x , speed v , and acceleration a are related through the kinematics formulas given in Chapter 10 , rotational angle, speed, and acceleration are related by the same formulas:

It is highly unlikely that you’ll be asked to actually make much of a calculation with these equations. Rather, you might be asked to rank rotating objects by their angular speed or acceleration; or you might be asked, “Is it possible to calculate….” The general approach to a calculation should be identical to that for nonrotational kinematics: Make a chart with the five variables in it. If you can identify a value for three of the five variables, the problem is solvable.

Rotational Inertia

Just like “inertia” refers to an object’s ability to resist changes in its motion, “rotational inertia” 4 refers to an object’s ability to resist changes in its rotational motion. Two things affect an object’s ability to resist rotational motion changes: the object’s mass and how far away that mass is from the center of rotation.

There are three ways to figure out something’s rotational inertia:

  1. For a single “point” particle that is moving in a circle around an axis, its rotational inertia is given by the equation Imr 2 . Here, m is the mass of the particle, and r is the radius of the circle.
  2. For an object with some kind of structure, like a spinning ball, a disk, or a rod, a formula for its rotational inertia will generally be given if you need it.
  3. For a system consisting of several objects, you can add together the rotational inertia of each of the objects to find the total rotational inertia of a system.

Example 4: Three meter-long, uniform 200-g bars each have a small 200-g mass attached to them in the positions shown in the diagram. A person grips the bars in the locations shown and attempts to rotate the bars in the directions shown.

It’s possible that you might be asked to calculate the rotational inertia of one of these gripped rods. If so, in the problem statement you’d be given the formula to calculate the rotational inertia of a rod:  when pivoted in the center, and  when pivoted at the end. 5 You know that the rotational inertia of the small mass is mr 2 , where r is the distance from the mass to the grip. In each situation, add the rotational inertia of the rod to the rotational inertia of the small mass.

For Grip A, the rod’s rotational inertia is  . The small mass contributes nothing to the rotational inertia, because it is not rotating—the r term in I = mr 2 is zero. Thus, the total rotational inertia is 0.067 kg · m2 .

For Grip B, the rod is pivoted in the center, so its rotational inertia is  . The small mass isn’t rotating. So the total rotational inertia is just 0.017 kg · m2 .

Finally, for Grip C, the rod’s rotational inertia is  . The small mass’s rotational inertia is (0.2 kg)(1 m)2 = 0.20 kg · m2 . Thus, the total rotational inertia is the sum of both contributions, 0.27 kg · m2 .

Calculations aren’t usually the point, though. This situation is just begging to become a ranking task: without any specific values for the masses or lengths of the items, rank the grips by their rotational inertia. As long as we know that the small mass is equal to the mass of the rod, and that the rods are equal in length, then the ranking can be done. You can see that the small mass contributes nothing to the rotational inertia in A and B without calculation.

You can see that Grip A provides a greater rotational inertia than Grip B, even without knowing the formulas  and  . Reason from the properties of an object that contribute to its rotational inertia: mass, and how far away that mass is from the axis of rotation. The rods have the same mass in A and B. But Rod A has much more mass that is far away from the grip; Rod B has more mass closer to the grip. Therefore, Rod B will be easier to rotate, and Rod A will have more rotational inertia.

Then, of course, Grip C combines the “worst” of both worlds: just the rod by itself provides the same rotational inertia as in Grip A, but the mass is also contributing to the rotational inertia. The final ranking would be I C > I A > I B . 6

Newton’s Second Law for Rotation

Just as linear acceleration is caused by a net force, angular acceleration is caused by a net torque:

Only the net torque can cause an angular acceleration. If more than one force is applying a torque, then use the sum 7 of the torques to find the angular acceleration.

Example 5: A turntable of known mass and radius is attached to a motor that provides a known torque. Using the torque of the motor and the rotational inertia of the turntable in Newton’s second law for rotation, then, using rotational kinematics, a student predicts that it should take 5.0 s for the turntable to speed up from rest to its maximum rotational speed. When the student measures the necessary time, though, he discovers that it takes 6.8 s to reach maximum rotational speed.

First, you should be able to describe how to perform such a measurement in your laboratory. There’s a bazillion ways of doing so: the idea is to make many measurements of rotational speed until that rotational speed doesn’t change. Rotational speed could be measured with a video camera and a protractor, by running the video frame-by-frame to see how many degrees the turntable advances per frame. Or tape a tiny piece of paper to the edge of the turntable, and have that paper trigger a few photogates; you can measure the angle between the photogates, and the photogates will tell you how much time it took for the turntable to traverse that angle.

This problem is setting up for you to figure out why the prediction didn’t match the measurement. 8 The most obvious issue is that the torque provided by the motor might not be the net torque on the turntable. Friction in the bearings of the turntable could easily provide a torque in the opposite direction to that provided by the motor. Thus, the real value for net torque will be lower than the value the student used. And, by τnet  , the real angular acceleration will be lower; finally, by ω f ω 0 + αt , a smaller angular acceleration to get to the same final speed means that the time t will be longer than predicted.

Angular Momentum

The rotational analogue of the impulse-momentum theorem involves torque and angular momentum rather than force and linear momentum:

Here τ is the net torque acting on an object, and Δt is the time during which that torque acts. The change in the object’s angular momentum is ΔL .

An object’s angular momentum can be calculated using three methods:

  1. For a single “point” particle that is moving in a circle around an axis, its angular momentum is given by Lmvr . Here, r represents the radius of that circle.
  2. For a single “point” particle that is moving in a straight line,9 its angular momentum is also given by L = mvr ; but in this case r represents the “distance of closest approach” from the line of the particle’s motion to the position about which angular momentum is calculated, as shown below.
  3. For an extended object with known rotational inertia I, angular momentum is given by L =  .

For most problems, you can reason by analogy to the linear impulse-momentum theorem. Just as a force applied for some time will change an object’s linear momentum, a torque applied for some time will change an object’s angular momentum. Just as the area under a force versus time graph gives the change in an object’s linear momentum, the area under a torque versus time graph gives the change in an object’s angular momentum.

Conservation of Angular Momentum

FACT: In any system in which the only torques acting are between objects in that system, angular momentum is conserved. This effectively means that angular momentum is conserved in all collisions, but also in numerous other situations.

Example 6: A uniform rod is at rest on a frictionless table. A ball of putty, whose mass is half that of the rod, is moving to the left, as shown. The ball of putty collides with and sticks to the rod.

Exam Tip from an AP Physics Veteran

Usually, the fulcrum or axis of rotation is obvious. But when an object is not forced to pivot at some specific position, if it rotates it will most likely rotate about its center of mass.

Start with what quantities are conserved for the putty-rod system. It’s a collision, in which the only forces involved are the force of the putty-on-rod and rod-on-putty. Therefore, linear momentum is conserved. Similarly, the putty applies a torque to the rod because it pushes on the rod at a position away from its center of mass. But the only torques involved are provided by objects in the system, so angular momentum is also conserved.

This collision cannot be elastic because the putty sticks to the rod; so kinetic energy was not conserved. Mechanical energy was also not conserved because the kinetic energy of the putty was not stored as potential energy in a spring or gravitational field. Of course, the sum of all forms of energy was conserved, because whatever kinetic energy was lost by the putty-rod system was converted to microscopic internal energy, and thus the temperature of the putty-rod will increase.

Where’s the center of mass? Using the equation from Chapter 12 , call x = 0 the top of the rod. Pretend the rod is 1-m long and 1 kg in mass. Then the putty is 0.5 kg in mass. So m putty (0) + m rod (0.5 m) = m total (x cm ). 10Plugging in the masses, you get x = 0.33 m.

But the problem emphatically did not say that the rod was 1-m long.

Right. Whatever the rod’s length, its center of mass is one-third of the way down the rod.

Exam Tip from an AP Physics Veteran

When you’re asked about the center of mass speed, you can ignore all angular stuff. In that case, just treat the collision as if these were carts colliding.

Again, you can make up values to find the speed of the center of mass after collision. If the putty’s initial speed were 1 m/s, then the total momentum before collision is 0.5 N · s. By conservation of momentum, that’s also the total momentum after collision, but the mass of the combined objects after collision is 1.5 kg. The speed of the center of mass would be 0.33 m/s. But since the initial speed wasn’t given, you can only definitively say that the speed of the center of mass after collision will be one-third of the speed of the putty before collision.

Conservation of Angular Momentum Without Collisions

FACT: Angular momentum is conserved any time an object, or system of objects, experiences no net torque.

Example 7: A person stands on a frictionless turntable. She and the turntable are spinning at one revolution every two seconds.

Sure, the person can wiggle and exert a torque on the turntable. But this torque is internal to the person-turntable system. The person thus cannot change the angular momentum of the person-turntable system.

This doesn’t mean that she can’t change her angular speed. What if she throws her arms way out away from her body? 11 Her rotational inertia would change, because she’d have the same total mass but more of that mass would be far away from the center of rotation. Her angular momentum can’t change. By L =  , to keep a constant L with a bigger I , the angular speed ω must decrease. This is the physical basis for how figure skaters can control their spinning.

Example 8: A planet orbits a sun in an elliptical orbit.

You won’t have to deal with elliptical orbits in the sense of making calculations, or using Kepler’s Laws and Physics C-style calculations. But you can understand that angular momentum of a planet orbiting a sun must be conserved. Since the force of the sun on a planet is always on a line toward the sun itself, this force cannot provide any torque—there’s no lever arm. With no torque exerted on a planet, that planet cannot change its angular momentum. Treat the planet as a point particle; its angular momentum is L = mvr , where r is the distance from the sun. Whenever r is big, v must be small to keep L constant. The farther away the planet is from the sun, the slower it moves.

Rotational Kinetic Energy

FACT: When an object is rotating, its rotational kinetic energy is

In Example 6, the putty-rod system has both rotational and linear kinetic energy after the collision. The total kinetic energy is then the sum of ½mv 2 , where v is the speed of the center of mass, plus the rotational kinetic energy.

When an object rotates, the work-energy theorem still applies. The kinetic energy terms each include the addition of rotational and linear kinetic energy. In Example 6, work is done by a nonconservative force (the force of the rod on the putty). To find out how much work was done, set W NC = ΔKE + ΔPE. This situation doesn’t involve a spring or a changing vertical height, so ΔPE = 0. The kinetic energy was originally just ½mv 2 for the putty; after the collision, the kinetic energy is as discussed in the previous paragraph, the speed of the center of mass plus the rotational kinetic energy.

 Practice Problems

1 .    A uniform wooden block has a mass m . On it is resting half of an identical block, as shown above. The blocks are supported by two table legs, as shown.

(a)   Which table leg, if either, should provide a larger force on the bottom block? Answer with specific reference to the torque equation.

(b)   In terms of given variables and fundamental constants, what is the force of the right-hand table leg on the bottom mass?

2 .    A small ball of mass m moving on a frictionless horizontal surface is attached to a rubber band whose other end is fixed at point P . The ball moves along the dotted line in the preceding figure, stretching the rubber band. When it passes Point A , its velocity is v directed as shown.

(a)   Is the angular momentum of the ball about Point P conserved between positions A and B?

(b)   Is the linear momentum of the ball conserved between positions A and B?

(c)   Describe a system in this problem for which mechanical energy is conserved as the ball moves from A to B.

(d)   Explain why the net force on the ball at Point B is not  .

3 .    A smooth, solid ball is released from rest from the top of an incline, whose surface is very rough. The ball rolls down the incline without slipping.

(a)   Describe in words the energy conversion for the ball from its release until it reaches the bottom of the incline.

(b)   Is the mechanical energy of the ball-Earth system conserved during its roll?

(c)   This ball is replaced by a new ball, whose surface and mass are identical to the first ball, but which is predominantly hollow inside. Describe any differences in its roll down the incline without slipping, with explicit reference to forms of energy.

 Solutions to Practice Problems

1 .    (a)   Call the force of the left support F L , and the force of the right support F R . Consider the middle of the bottom block as the fulcrum. Then one clockwise torque acts: F L ·(L/ 2). Two counterclockwise torques act, though: FR ·(L/ 2) and (1/2)m ·(1/4)L . The point is that if you have to add something to the torque provided by the right support to get the torque provided by the left support, the left support thus provides more torque. Because the supports are the same distance from the center, the left support provides more force, too.

(b)   You certainly could use the reasoning in Part (a) with the fulcrum in the center, along with the total support force equaling 1.5Mg (vertical equilibrium of forces). However, it’s much easier mathematically to just call the left end of the rod the fulcrum. Then the counterclockwise torque is F R · L . The clockwise torque is (1/2)mg · (L /4) + mg · (L/ 2). Set these equal and play with the fractions to get  .

2 .    (a)   Angular momentum is conserved when no torques external to the system act. Here the system is just the ball. The only force acting on the ball is the rubber band, which is attached to Point P . The force applied by the rubber band can’t have any lever arm with respect to P and thus provides no torque about point P , so the ball’s angular momentum about point P can’t change. Angular momentum is conserved.

(b)   Linear momentum is conserved when no forces external to the system act. Here the system is just the ball. The rubber band is external to the system and applies a force; therefore, linear momentum is not conserved.

(c)   Mechanical energy is conserved when no force external to the system does work. The rubber band does work on the ball, because it applies a force and stretches in a direction parallel to the force it produces. Consider the rubber band part of the system. The post at Point P still applies a force to the ball–rubber band system, but since the post doesn’t move, that force does no work on the ball–rubber band system. Any kinetic energy lost by the ball will be stored as elastic energy in the rubber band. The mechanical energy of the ball–rubber band system is conserved.

(d)   The general form of this equation is fine—the ball’s path at Point P is, at least in the neighborhood of P , approximately circular. The ball experiences a centripetal acceleration at Point B , and centripetal acceleration is v 2 /r. The problem is that if the r term is 1.0 m, then the v term must represent the speed at Point B . With angular momentum conserved, the total of mvr must always be the same. The ball’s mass doesn’t change. The distance rfrom Point P gets bigger from A to B , so the speed must get smaller. The equation given uses the given variable v which represents the speed of the ball at Point A , not the speed at B , and so is invalid.

3 .    (a)   Gravitational energy at the top (because the ball is some vertical height above its lowest position) is converted to both rotational and translational kinetic energy at the bottom—rotational because the ball will be spinning, and translational because the ball’s center of mass will move down the incline.

(b)   Mechanical energy is conserved when no non-conservative forces act. Here the Earth’s gravitational field can give the ball kinetic energy, but since the Earth is part of the system and since the gravitational force is conservative, that still allows for conservation of mechanical energy. Friction is a nonconservative force, but here friction does no work.

(c)   The hollow ball of the same mass will have greater rotational inertia, because the mass is concentrated farther from the center of rotation. The ball’s gravitational energy before the rolling begins is the same as the previous scenario, because the height of the incline is the same. The total kinetic energy at the bottom will not change; the question is how much of that kinetic energy will be rotational, and how much will be translational.

Rotational KE is ½ 2 ; the angular speed ω depends on the translational speed v . (The faster the ball is moving, the more it’s rotating, too.) Therefore, rotational kinetic energy depends on v 2 . Translational kinetic energy also depends on v 2 in the formula ½mv 2 . The hollow ball has bigger I . The speed v must be lower for the hollow ball so that ½ 2 + ½mv 2 adds to the same value for both balls.

 Rapid Review

  • When an object moves in a circle, it has an acceleration directed toward the center of the circle. The amount of that acceleration is  .
  • The torque τprovided by a force is given by the equation τ = Fd  .
  • In any system in which the only torques acting are between objects in that system, angular momentum is conserved. This effectively means that angular momentum is conserved in allcollisions, but also in numerous other situations.
  • Angular momentum is conserved any time an object, or system of objects, experiences no net torque.
  • When an object is rotating, its rotational kinetic energy is ½2 .

1 Technically, this would be a force of static friction between the tires and the road, but that’s a different day’s lesson.

2 That makes perfect sense—ever see those yellow signs warning of the appropriate speed for going around a curve? They just say, “Curve: 40 mph.” They certainly don’t say something silly like “Curve: go 10 mph for every 500 kg in your vehicle.”

3 The “fulcrum” is the point about which an object rotates, or could rotate.

4 The AP Physics 1, Algebra-Based Exam will always use the term “rotational inertia,” usually represented by the variable I . Many textbooks and teachers will use the older term “moment of inertia” to refer to the same quantity. Don’t be confused.

5 Here, L represents the length of the bar.

6 Okay, you do need to get comfortable with this sort of verbal explanation of concepts that refers to equations and facts but doesn’t make direct calculation. If you are confused on this sort of problem, it is okay to make up values for whatever you need, and calculate. I don’t at all recommend memorizing all the different formulas for rotational inertia of a rod, sphere, hoop, disk, etc. But if you happen to remember them, it’s fine to use them.

7 Or, if the torques are acting in opposite directions, use the difference.

8 Please don’t automatically say “human error.” There’s no such thing as “human error,” and using that phrase is basically an automatic wrong on the AP Exam.

9 Angular momentum must always be defined with respect to some central axis of rotation. For most rotating objects, that axis is obvious. For a particle moving in a straight line, you have to say what position you’re calculating angular momentum for, but the particle can still have angular momentum.

10 It’s (0.5 m) because the center of mass of the rod by itself must be halfway down the rod.

11 AP reader Matt Sckalor quite reasonably asks, “After she throws the first arm, what part of her body does she use to throw the other arm?” Perhaps I should say she “extends” her arms.