AP Physics 1: Algebra-Based 6.3 Angular Momentum and Angular Impulse Questions

Describe angular momentum about a chosen axis and connect angular impulse to changes in angular momentum during rotational interactions.

Exam points

calculate and compare angular momentum using rotational inertia, angular speed and axis choice
analyse impact geometry and test proposed postcollision relations for physical plausibility
use torque–time products to compare angular impulses and calculate angular-speed changes
interpret angular-momentum time graphs through net-torque slopes and interval changes
calculate collision impulse or stopping torque from a system's angular-momentum change

Covered syllabus points

6.3.A—Describe the angular momentum of an object or rigid system 6.3.B—Describe the angular impulse delivered to an object or rigid system by a torque 6.3.C—Relate an object’s or rigid system’s change in angular momentum to the angular impulse given to that object or…

[Maximum number: 6]

(12 points, suggested time 25 minutes)
The left end of a rod of length d and rotational inertia I is attached to a frictionless horizontal surface by a frictionless pivot, as shown above. Point C marks the center (midpoint) of the rod. The rod is initially motionless but is free to rotate around the pivot. A student will slide a disk of mass $m_{\text {disk }}$ toward the rod with velocity $v_{0}$ perpendicular to the rod, and the disk will stick to the rod a distance x from the pivot. The student wants the rod

disk system to end up with as much angular speed as possible.

(a)

Suppose the rod is much more massive than the disk. To give the rod as much angular speed as possible, should the student make the disk hit the rod to the left of point C, at point C, or to the right of point C ? To the left of C At C To the right of C
Briefly explain your reasoning without manipulating equations.

[ 1 ]

(b)

On the Internet, a student finds the following equation for the postcollision angular speed $\omega$ of the rod in this situation: $\omega=\frac{m_{\text {disk }} x v_{0}}{I}$ . Regardless of whether this equation for angular speed is correct, does it agree with your qualitative reasoning in part (a) ? In other words, does this equation for $\omega$ have the expected dependence as reasoned in part (a) ? Yes No
Briefly explain your reasoning without deriving an equation for $\omega$ .

[ 2 ]

2 points
Correct answer: "Yes"
If "No" is selected, the explanation may still earn full credit if an incorrect selection was made in part (a).
For a selection consistent with the selection from part (a)
For indicating that the equation shows that $\omega$ increases with increasing x
Example: According to the equation, $\omega$ increases with x;
a bigger x produces a bigger angular speed. This agrees with my reasoning from part (a), where I said a bigger x creates a bigger angular speed after the collision.

(c)

Another student deriving an equation for the postcollision angular speed $\omega$ of the rod makes a mistake and comes up with $\omega=\frac{I x v_{0}}{m_{\text {disk }} d^{4}}$ . Without deriving the correct equation, how can you tell that this equation is not plausible-in other words, that it does not make physical sense? Briefly explain your reasoning.

For parts (d) and (e), do NOT assume that the rod is much more massive than the disk.

[ 3 ]

3 points
For focusing on functional dependence (instead of, for example, considering units/dimensions)
For addressing $m_{\text {disk }}, I$ , or both
For correctly concluding that the equation is wrong because of the dependence on $m_{\text {disk }}, I$ , or both
Example: If $m_{\text {disk }}$ is large, then more angular momentum will be transferred during the collision. But the equation shows the angular speed decreasing with increasing $m_{\text {disk }}$ , because it is in the denominator.

Distribution

[Maximum number: 1]

Figure for Question 2 — AP Physics 1: Algebra-Based

A string of negligible mass is wrapped around the rim of a solid disk that can rotate without friction about its center, as shown above. The string is pulled down with a force of constant magnitude F for an interval of time $\Delta t$ . If the disk starts from rest, which combination of F and $\Delta t$ will result in the greatest angular speed of the disk?

Figure for Question 2 — AP Physics 1: Algebra-Based — Option A

Figure for Question 2 — AP Physics 1: Algebra-Based — Option B

Figure for Question 2 — AP Physics 1: Algebra-Based — Option C

Figure for Question 2 — AP Physics 1: Algebra-Based — Option D

[Maximum number: 6]

(7 points, suggested time 13 minutes)

A block of unknown mass is attached to a long, lightweight string that is wrapped several turns around a pulley mounted on a horizontal axis through its center, as shown. The pulley is a uniform solid disk of mass M and radius R. The rotational inertia of the pulley is described by the equation $I=\frac{1}{2} M R^{2}$ . The pulley can rotate about its center with negligible friction. The string does not slip on the pulley as the block falls.

When the block is released from rest and as the block travels toward the ground, the magnitude of the tension exerted on the block by the string is $F_{\mathrm{T}}$ .

Consider scenarios 1 and 2 at the end of time interval $\Delta t$ . In a clear, coherent paragraph-length response that may also contain equations and drawings, explain why the change in angular momentum of both pulleys is the same but the change in rotational kinetic energy is greater for the disk.

Figure 1

For indicating that the torque, $\tau$ , is the same for both pulleys

1 point

For indicating that the impulse, $\tau \Delta t$ , (or change in momentum $\Delta L$ ) is the same for both
pulleys because $\tau$ and $\Delta t$ are the same

1 point

For indicating that the rotational inertia, I, of the disk and hoop are different

1 point

For providing reasoning that because the rotational inertia, I, are different for the disk and
hoop, the kinematic quantities ( $\Delta \theta, \omega, \alpha$ ) are also different for the disk and hoop

1 point

For one of the following:
- Indicating that because $\Delta \theta$ is greater for the disk the work done on the disk is
greater

1 point

For a logical, relevant, and internally consistent argument that follows the guidelines
described in the published requirements for the paragraph-length response

1 point

Example Response
The rotational inertia, I, of the hoop is larger than the rotational inertia of the disk because the hoop's mass is all on the outside instead of distributed throughout like the disk. Equal forces are applied to both pulleys at the same distance, which means that the torques exerted on the pulleys will also be equal. Since the same torque is applied to both pulleys for the same time period, the change in angular momentum will be the same for the disk and hoop. The magnitude of the angular velocity for the hoop will be smaller than that of the disk since angular velocity is inversely proportional to the rotational inertia $\left(\omega=\frac{L}{I}\right)$ . Since kinetic energy is proportional to rotational inertia and the square of angular velocity $\left(K_{R}=\frac{1}{2} I \omega^{2}\right)$ , the difference in angular velocity more greatly affects the rotational kinetic energy. That means the disk will have a greater rotational kinetic energy than the hoop.

Total for part (b) for question 47 points

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question 3