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AP Physics 1 Unit 6: Energy and Momentum of Rotating Systems

Connect rotational kinetic energy, angular momentum, angular impulse, conservation laws, and rolling motion in rotating systems.

Syllabus
Effective Fall 2025
Course
AP Physics 1: Algebra-Based

6 Energy and Momentum of Rotating Systems question 1

[Maximum number: 1]

(7 points, suggested time 13 minutes)

A rod with a sphere attached to the end is connected to a horizontal mounted axle and carefully balanced so that it rests in a position vertically upward from the axle. The center of mass of the rod-sphere system is indicated with a ⊗, as shown in Figure 1. The sphere is lightly tapped, and the rod-sphere system rotates clockwise with negligible friction about the axle due to the gravitational force.

A student takes a video of the rod rotating from the vertically upward position to the vertically downward position. Figure 2 shows five frames (still shots) that the student selected from the video.
Note: these frames are not equally spaced apart in time.

Figure 2

Figure 2

Use the frames of the video shown in Figure 2 to answer the following questions.

In which frame is the rotational kinetic energy of the rod-sphere system the greatest? Briefly justify your answer.

Figure 3

Figure 3

6 Energy and Momentum of Rotating Systems question 2

[Maximum number: 1]

34.

Figure for Question 6 Energy and Momentum of Rotating Systems question 2 — AP Physics 1: Algebra-Based

One end of a string is attached to the ceiling and the other end is attached to a small sphere traveling in a circular path, as shown above, with a constant speed of 1 m/s. The string makes an angle of 30° to the vertical. If the tension in the string is 6 N and the circumference of the circle is 1 m, how much work is done on the sphere by the string as the sphere travels through one revolution?

International Exam 2021 MCQ

A

0 J

B

3 J

C

6 J

D

12 J

6 Energy and Momentum of Rotating Systems question 3

[Maximum number: 12]

(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 mdisk m_{\text {disk }} toward the rod with velocity v0v_{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.

Question (a)

(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 ]

Question (b)

(b)

On the Internet, a student finds the following equation for the postcollision angular speed ω\omega of the rod in this situation: ω=mdisk xv0I\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 ω\omega have the expected dependence as reasoned in part (a) ? Yes No
Briefly explain your reasoning without deriving an equation for ω\omega.

[ 2 ]

Question (c)

(c)

Another student deriving an equation for the postcollision angular speed ω\omega of the rod makes a mistake and comes up with ω=Ixv0mdisk d4\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 ]

Question (d)

(d)

Immediately before colliding with the rod, the disk's rotational inertia about the pivot is mdisk x2m_{\text {disk }} x^{2} and its angular momentum with respect to the pivot is mdisk v0xm_{\text {disk }} v_{0} x. Derive an equation for the postcollision angular speed ω\omega of the rod. Express your answer in terms of d,mdisk ,I,x,v0d, m_{\text {disk }}, I, x, v_{0}, and physical constants, as appropriate.

[ 4 ]

Question (e)

(e)

Consider the collision for which your equation in part (d) was derived, except now suppose the disk bounces backward off the rod instead of sticking to the rod. Is the postcollision angular speed of the rod when the disk bounces off it greater than, less than, or equal to the postcollision angular speed of the rod when the disk sticks to it? Greater than Less than Equal to
Briefly explain your reasoning.

Figure for Question (e) — AP Physics 1: Algebra-Based
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