Question

A solid sphere is rolling on a horizontal surface without slipping. The ratio of the translational and rotational kinetic energies of the sphere is

• A. \( 3:2 \)
• B. \( 7:2 \)
• C. \( 5:2 \)
• D. \( 7:5 \)

Hint:

For rolling motion, remember the relation between translational and rotational kinetic energy: the rotational part is often a fraction of the translational kinetic energy depending on the object's moment of inertia.

Answer 1

The Correct Option is C

For a solid sphere rolling without slipping, the total kinetic energy is the sum of translational kinetic energy and rotational kinetic energy. The total kinetic energy \( K \) is given by: \[ K = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \] where \( m \) is the mass, \( v \) is the velocity, \( I \) is the moment of inertia of the sphere, and \( \omega \) is the angular velocity.
For a solid sphere, the moment of inertia \( I = \frac{2}{5} m r^2 \), and the relationship between linear and angular velocity for rolling without slipping is \( v = r \omega \).
Thus, the rotational kinetic energy is \( \frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2 \).
The total kinetic energy is: \[ K = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2 \] The translational kinetic energy is \( \frac{1}{2} m v^2 \). Therefore, the ratio of translational to rotational kinetic energy is: \[ \frac{\frac{1}{2} m v^2}{\frac{1}{5} m v^2} = \frac{5}{2} \] Thus, the ratio of translational to rotational kinetic energy is \( 5:2 \).