AP Physics 1- 6.1 Rotational Kinetic Energy - Exam Style questions - FRQs- New Syllabus
Rotational Kinetic Energy AP Physics 1 FRQ
Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
Question
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 \( I=\dfrac{1}{2}MR^2 \). The pulley can rotate about its center with negligible friction. The string does not slip on the pulley as the block falls.
Scenarios \( 1 \) and \( 2 \) show two different pulleys. In Scenario \( 1 \), the pulley is the same solid disk referenced in part \( \mathrm{(a)} \). In Scenario \( 2 \), the pulley is a hoop that has the same mass \( M \) and radius \( R \) as the disk. Each pulley has a lightweight string wrapped around it several turns and is mounted on a horizontal axle, as shown. Each pulley is free to rotate about its center with negligible friction.
Most-appropriate topic codes (AP Physics 1):
• Topic \( 6.1 \) — Rotational Kinetic Energy (Part \( \mathrm{(b)} \))
• Topic \( 6.3 \) — Angular Momentum and Angular Impulse (Part \( \mathrm{(b)} \))
• Topic \( 5.4 \) — Torque and the Second Condition of Equilibrium (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \))
▶️ Answer/Explanation
(a)
Use Newton’s second law for rotation:
\( \tau = I\alpha \)
The torque on the disk is produced by the string tension:
\( \tau = F_T R \)
For the solid disk,
\( I=\dfrac{1}{2}MR^2 \)
Substitute into the rotational equation:
\( F_T R = \left(\dfrac{1}{2}MR^2\right)\alpha_D \)
Solve for \( \alpha_D \):
\( \alpha_D = \dfrac{F_T R}{\frac{1}{2}MR^2} \)
\( \boxed{\alpha_D = \dfrac{2F_T}{MR}} \)
The result makes sense dimensionally: \( \dfrac{F}{MR} \) has units of \( \mathrm{rad/s^2} \).
(b)
The torque on each pulley is the same because the same force \( F_A \) is applied at the same radius \( R \). Therefore each pulley experiences the same torque, \( \tau = F_A R \). Since the force is applied for the same time interval \( \Delta t \), the angular impulse is the same for both pulleys:
\( \Delta L = \tau \Delta t \)
so the change in angular momentum is the same for the disk and the hoop.
However, the rotational inertias are different. The solid disk has \( I_{\text{disk}}=\dfrac{1}{2}MR^2 \), while the hoop has \( I_{\text{hoop}}=MR^2 \). Thus the hoop has the larger rotational inertia.
For a given torque, \( \alpha = \dfrac{\tau}{I} \), so the disk has the larger angular acceleration. Starting from rest and acted on for the same time interval, the disk reaches a greater angular speed than the hoop.
Rotational kinetic energy is
\( K_{\text{rot}} = \dfrac{1}{2}I\omega^2 \)
Even though the hoop has larger \( I \), the disk’s angular speed is large enough that its rotational kinetic energy increase is greater. Another equivalent way to say this is that, for the same change in angular momentum, energy depends on how that angular momentum is distributed relative to the rotational inertia. The smaller rotational inertia of the disk allows a larger \( \omega \), which gives a greater increase in \( K_{\text{rot}} \).