Rolling Motion

Hardphysics

A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down an incline. What is the ratio of its rotational kinetic energy to translational kinetic energy?

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2/5

2/7

1/2

1/3

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Subject: physics
Chapter: rotational motion
Topic: rolling motion
Difficulty: Hard
Year: 2025

Solution

Correct Answer:

2/5

Translational KE: $K_{t} = \frac{1}{2} M v^{2}$ .
Rotational KE: $K_{r} = \frac{1}{2} I ω^{2}$ .
Rolling Condition: For no slipping, $ω = v / R$ .
Sphere Inertia: $I = \frac{2}{5} M R^{2}$ .
Substitution:
- $K_{r} = \frac{1}{2} (\frac{2}{5} M R^{2}) (\frac{v}{R})^{2} = \frac{1}{5} M v^{2}$ .
Ratio: $\frac{K _{r}}{K _{t}} = \frac{( 1/5 ) M v ^{2}}{( 1/2 ) M v ^{2}} = \frac{2}{5}$ .

This hard difficulty physics question is from the chapter rotational motion, covering the topic of rolling motion. It appeared in the 2025 exam.

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Rolling Motion

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