Question

A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation motion, then the body is

• A. hollow cylinder
• B. ring
• C. solid disc
• D. solid sphere
• E. hollow sphere

Hint:

Always use \(\frac{K_{rot}}{K_{trans}} = \frac{I}{mR^2}\) for rolling bodies.

Answer 1

The Correct Option is D

Concept: For rolling motion: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] Also, \[ v = \omega R \Rightarrow \frac{K_{rot}}{K_{trans}} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2} = \frac{I}{mR^2} \]

Step 1: Given ratio. \[ \frac{K_{rot}}{K_{trans}} = 0.4 \] Using \( v = \omega R \): \[ \frac{K_{rot}}{K_{trans}} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2} = \frac{I\omega^2}{mv^2} = \frac{I}{mR^2} \]

Step 2: Compare with standard moments of inertia.
• Ring: \(I = mR^2 \Rightarrow 1\)
• Disc: \(I = \frac{1}{2}mR^2 = 0.5\)
• Sphere: \(I = \frac{2}{5}mR^2 = 0.4\)

Step 3: Match value. \[ \frac{2}{5} \Rightarrow \text{solid sphere} \]

Step 4: Conclusion. Correct answer should be solid sphere.