Question

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy , $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is

• A. 2 : 5
• B. 7 : 10
• C. 10 : 7
• D. 5 : 7

Answer 1

The Correct Option is D

To find the ratio $K_t : (K_t + K_r)$ for a solid sphere in rolling motion, we need to consider both the translational and rotational kinetic energy components of the sphere.

1. K_t, or translational kinetic energy, is given by: K_t = \frac{1}{2} m v^2 where m is the mass of the sphere and v is the linear velocity of its center of mass.
2. The rotational kinetic energy, K_r, for a solid sphere is: K_r = \frac{1}{2} I \omega^2 where I = \frac{2}{5} m r^2 is the moment of inertia for a solid sphere and \omega is the angular velocity.
3. In rolling motion without slipping, the relation between linear velocity v and angular velocity \omega is: \omega = \frac{v}{r}
4. Substituting \omega = \frac{v}{r} into the equation for K_r, we get: K_r = \frac{1}{2} \cdot \frac{2}{5} m r^2 \left( \frac{v}{r} \right)^2 = \frac{1}{5} m v^2
5. The total kinetic energy is the sum of translational and rotational kinetic energy: K_t + K_r = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2
6. Now, the ratio K_t : (K_t + K_r) is: \frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7}

Therefore, the ratio of translational kinetic energy to the total kinetic energy for the sphere is 5:7.