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.
For a solid sphere undergoing pure rolling motion, its total kinetic energy \( K_{\text{total}} \) is the sum of its translational kinetic energy and rotational kinetic energy.
- The translational kinetic energy of the center of mass is given by:
\[
K_{\text{linear}} = \frac{1}{2} m v^2,
\]
where \( m \) denotes the mass of the sphere and \( v \) represents the linear velocity of the center of mass.
- The rotational kinetic energy is expressed as:
\[
K_{\text{rotational}} = \frac{1}{2} I \omega^2,
\]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, the moment of inertia about its center of mass is:
\[
I = \frac{2}{5} m r^2,
\]
where \( r \) signifies the radius of the sphere.
Given that the sphere is rolling without slipping, the relationship between linear and angular velocity is \( v = r \omega \). Consequently, the rotational kinetic energy can be rewritten as:
\[
K_{\text{rotational}} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2.
\]
The ratio of the translational kinetic energy to the rotational kinetic energy is calculated as follows:
\[
\text{Ratio} = \frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{\frac{1}{2} m v^2}{\frac{1}{5} m v^2} = \frac{5}{2}.
\]
Final Answer: The ratio is \( \frac{5}{2} \).
