Question:
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :
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For rolling, \(K_{rot}/K_{trans} = I/(mR^2)\); set it equal to 1/2.
Updated On: Nov 4, 2026
- Solid sphere
- Solid cylinder
- Hollow cylinder
- Ring
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The Correct Option is B
Solution and Explanation
For rolling without slipping, \(v = \omega R\).
\(K_{rot} = \dfrac12 I\omega^2 = \dfrac12 I\dfrac{v^2}{R^2}\), \(K_{trans} = \dfrac12 m v^2\).
\(\dfrac{K_{rot}}{K_{trans}} = \dfrac{I}{mR^2} = \dfrac{50}{100} = \dfrac12\).
So \(I = \dfrac12 mR^2\), which is the moment of inertia of a solid cylinder (sphere: \(\frac25 mR^2\); hollow cylinder and ring: \(mR^2\)).
Answer: (2) Solid cylinder
\(K_{rot} = \dfrac12 I\omega^2 = \dfrac12 I\dfrac{v^2}{R^2}\), \(K_{trans} = \dfrac12 m v^2\).
\(\dfrac{K_{rot}}{K_{trans}} = \dfrac{I}{mR^2} = \dfrac{50}{100} = \dfrac12\).
So \(I = \dfrac12 mR^2\), which is the moment of inertia of a solid cylinder (sphere: \(\frac25 mR^2\); hollow cylinder and ring: \(mR^2\)).
Answer: (2) Solid cylinder
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