Question

A solid cylinder of radius \(R\) rolls without slipping with a center of mass speed \[ v_0=\sqrt{\frac{gR}{3}} \] on a horizontal surface with a vertical edge, as shown in the figure. Here, \(g\) is the acceleration due to gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

• A. \(0\)
• B. \(\sqrt{\dfrac{5gR}{7}}\)
• C. \(\sqrt{\dfrac{gR}{15}}\)
• D. \(\sqrt{\dfrac{3gR}{7}}\)

Hint:

For rolling bodies about a pivot: \[ I_O=I_{CM}+MR^2 \] using parallel axis theorem.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When the cylinder hits the corner, it transitions from pure rolling to rotation about the corner. We use conservation of energy. Contact is lost when the normal force from the corner becomes zero.
Step 3: Detailed Explanation:
1) Initial and Final Energy:
Total energy on flat surface: \( E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{1}{2}mR^2)(v_0/R)^2 = \frac{3}{4}mv_0^2 \).
Let \( v \) be the speed at angle \( \theta \) from vertical.
\( E_f = \frac{3}{4}mv^2 + mgR(1 - \cos\theta) = E_i \)? No, it rotates about the corner, so \( E_f = \frac{1}{2}I_{corner}\omega^2 + \dots \).
Wait, the height of center of mass remains \( R \) relative to the corner at the start. As it rotates by \( \theta \), the height is \( R\cos\theta \).
\( \frac{3}{4}mv^2 = \frac{3}{4}mv_0^2 + mgR(1 - \cos\theta) \implies v^2 = v_0^2 + \frac{4}{3}gR(1 - \cos\theta) \).
2) Losing contact:
Equation: \( mg\cos\theta - N = mv^2/R \). Contact lost when \( N=0 \implies g\cos\theta = v^2/R \).
\( v^2 = v_0^2 + \frac{4}{3}gR - \frac{4}{3}v^2 \implies \frac{7}{3}v^2 = v_0^2 + \frac{4}{3}gR \).
Substitute \( v_0^2 = gR/3 \):
\( \frac{7}{3}v^2 = \frac{gR}{3} + \frac{4gR}{3} = \frac{5gR}{3} \implies v^2 = \frac{5gR}{7} \).
Step 4: Final Answer:
Matches (B).