Rankers Physics
Topic: Rotational Motion
Subtopic: Rolling on Inclined Plane

A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
Solid cylinder
Solid sphere
Disc
Ring

Solution:

Given, rotational kinetic energy is 40% of total energy. so,

\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]

Solving ,

\[ I = \frac{2}{5}mR^{2} \]

Object is Solid Sphere.

 

2 responses to “”

  1. Solution is wrong but answer is correct

    note solution equates total energy of rolling with rotational kinetic energy, instead of kinetic energy of rotation and translation

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