Question

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:

• A. \( \frac{4}{3} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{5}{2} \)

Hint:

For rolling motion, the total kinetic energy is the sum of the translational and rotational kinetic energies. Use the relation between the linear velocity and angular velocity to derive expressions for both energies.

Answer 1

The Correct Option is C

When a solid sphere is rolling without slipping, the total kinetic energy \( K_{\text{total}} \) is the sum of the linear kinetic energy and rotational kinetic energy. - The linear kinetic energy of the centre of mass is given by: \[ K_{\text{linear}} = \frac{1}{2} m v^2, \] where \( m \) is the mass of the sphere and \( v \) is the linear velocity of the centre of mass. - The rotational kinetic energy is given by: \[ K_{\text{rotational}} = \frac{1}{2} I \omega^2, \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, the moment of inertia about the centre of mass is: \[ I = \frac{2}{5} m r^2, \] where \( r \) is the radius of the sphere. Since the sphere is rolling without slipping, the relation between the linear velocity and angular velocity is \( v = r \omega \). Therefore, the rotational kinetic energy becomes: \[ K_{\text{rotational}} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2. \] Now, we find the ratio of the linear kinetic energy to the rotational kinetic energy: \[ \text{Ratio} = \frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{\frac{1}{2} m v^2}{\frac{1}{5} m v^2} = \frac{5}{2}. \] Final Answer: The ratio is \( \frac{5}{2} \).