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:
Show 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.
When a solid sphere undergoes rolling without slipping, its total kinetic energy \( K_{\text{total}} \) is the sum of its linear kinetic energy and rotational kinetic energy.
- The linear kinetic energy of the center of mass is \( K_{\text{linear}} = \frac{1}{2} m v^2 \), where \( m \) is the sphere's mass and \( v \) is the linear velocity of the center of mass.
- The rotational kinetic energy is \( K_{\text{rotational}} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, \( I = \frac{2}{5} m r^2 \), with \( r \) being the sphere's radius.
Given the rolling without slipping condition, \( v = r \omega \). Substituting this into the rotational kinetic energy equation yields \( K_{\text{rotational}} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2 \).
The ratio of linear to rotational kinetic energy is:
\[\text{Ratio} = \frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{\frac{1}{2} m v^2}{\frac{1}{5} m v^2} = \frac{5}{2}.\]
Final Answer: The ratio is \( \frac{5}{2} \).
