AP Physics C Mechanics - 6.1 Rotational Kinetic Energy- Exam Style questions- FRQs

Rotational Kinetic Energy AP  Physics C Mechanics FRQ

Unit 6: Energy and Momentum of Rotating Systems

Weightage : 10-15%

AP Physics C Mechanics Exam Style Questions – All Topics

Question

A system consists of a small sphere of mass \(m\) and radius \(R\) at rest on a horizontal surface and a uniform rod of mass \(M=2m\) and length \(\ell\) attached at one end to a pivot with negligible friction, where \(R\ll \ell\). There is negligible friction between the surface and the sphere to the right of Point A and nonnegligible friction to the left of Point A. The rod is held horizontally as shown in Figure 1, then is released from rest.
The total rotational inertia of the rod about the pivot is \(\frac{1}{3}M\ell^2\) and the rotational inertia of the sphere about its center is \(\frac{2}{5}mR^2\).
After the rod is released, the rod swings down and strikes the sphere head-on. As a result of this collision, the rod is stopped, and the ball initially slides without rotating to the left across the horizontal surface.
(a) Derive an expression for the angular speed of the rod just before striking the sphere in terms of the length \(\ell\) and physical constants as appropriate.
(b) Derive an expression for the linear speed \(v_0\) of the sphere immediately after colliding with the rod in terms of the length \(\ell\) and physical constants as appropriate.
After sliding a short distance, at time \(t=0\) the sphere encounters a region of the horizontal surface with a coefficient of kinetic friction \(\mu\), beginning at Point A as indicated in Figure 1. The sphere begins rotating while sliding and eventually begins rolling without sliding at Point B, also as indicated.
(c) In the following diagram, which represents the sphere while the sphere is traveling between Points A and B, draw and label the forces (not components) that act on the sphere. Each force must be represented by a distinct arrow starting on, and pointing away from, the point of application on the sphere.
(d) Derive an expression for each of the following as the sphere is rotating and sliding between points A and B in terms of \(v_0\), \(\mu\), \(R\), \(t\), and physical constants as appropriate.
i. The linear velocity \(v\) of the center of mass of the sphere as a function of time \(t\)
ii. The angular velocity \(\omega\) of the sphere as a function of time \(t\)
(e)
i. Derive an expression for the time it takes the sphere to travel from Point A to Point B in terms of \(v_0\), \(\mu\), and physical constants as appropriate.
ii. Derive an expression for the linear velocity of the sphere upon reaching Point B in terms of \(v_0\).

Most-appropriate topic codes (AP Physics C: Mechanics):

• Topic \(6.1\) — Rotational Kinetic Energy (Part \( \mathrm{a} \))
• Topic \(6.4\) — Conservation of Angular Momentum (Part \( \mathrm{b} \))
• Topic \(2.2\) — Forces and Free-Body Diagrams (Part \( \mathrm{c} \))
• Topic \(5.6\) — Newton’s Second Law in Rotational Form (Part \( \mathrm{d} \))
• Topic \(6.5\) — Rolling (Part \( \mathrm{e} \))
▶️ Answer/Explanation

We solve this sequence by first applying conservation of mechanical energy as the rod swings downward, converting potential energy into rotational kinetic energy. During the perfectly inelastic collision with the sphere, the net external torque is zero, so the system’s angular momentum is conserved. Once the sphere begins sliding across the rough surface, kinetic friction acts against its motion, resulting in a constant linear deceleration. Simultaneously, this frictional force produces a torque that creates an angular acceleration, causing the sphere’s rotation to increase. These dual kinematic changes persist until the sphere’s linear and rotational velocities perfectly synchronize into rolling without slipping.

(a)
\(\Delta U + \Delta K = 0\)
\((0 – Mgh_{CM}) + \left(\frac{1}{2}I\omega_f^2 – 0\right) = 0\)
Substitute \(M = 2m\), \(h_{CM} = \frac{\ell}{2}\), and \(I = \frac{1}{3}(2m)\ell^2\):
\((2m)g\left(\frac{\ell}{2}\right) = \frac{1}{2}\left(\frac{2}{3}m\ell^2\right)\omega_f^2\)
\(mg\ell = \frac{1}{3}m\ell^2 \omega_f^2\)
\(\boxed{\omega_f = \sqrt{\frac{3g}{\ell}}}\)

(b)
\(L_i = L_f\)
\(I_{rod}\omega_f = m v_0 \ell\)
\(\left(\frac{2}{3}m\ell^2\right)\sqrt{\frac{3g}{\ell}} = m v_0 \ell\)
\(v_0 = \frac{2}{3}\ell\sqrt{\frac{3g}{\ell}}\)
\(\boxed{v_0 = \sqrt{\frac{4}{3}g\ell}}\)

(c)


• The normal force \(F_N\) points directly upward from the bottom of the sphere.
• The force due to gravity \(F_g\) points directly downward from the center of mass.
• The frictional force \(F_f\) points horizontally to the right from the bottom of the sphere.

(d)(i)
\(\Sigma F = ma\)
\(-\mu m g = m a\)
\(a = -\mu g\)
\(v = v_0 + at\)
\(\boxed{v = v_0 – \mu g t}\)

(d)(ii)
\(\tau_{net} = I\alpha\)
\(F_f R = \frac{2}{5}mR^2 \alpha\)
\(\mu m g R = \frac{2}{5}mR^2 \alpha\)
\(\alpha = \frac{5\mu g}{2R}\)
\(\omega = \omega_0 + \alpha t\)
\(\boxed{\omega = \frac{5\mu g}{2R} t}\)

(e)(i)
Rolling without sliding occurs when \(v = R\omega\).
\(v_0 – \mu g t = R\left(\frac{5\mu g}{2R}t\right)\)
\(v_0 – \mu g t = \frac{5}{2}\mu g t\)
\(v_0 = \frac{7}{2}\mu g t\)
\(\boxed{t = \frac{2v_0}{7\mu g}}\)

(e)(ii)
Substitute \(t\) into the linear velocity equation:
\(v = v_0 – \mu g\left(\frac{2v_0}{7\mu g}\right)\)
\(v = v_0 – \frac{2}{7}v_0\)
\(\boxed{v = \frac{5}{7}v_0}\)

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