Question

A solid sphere rolls without slipping on a horizontal plane. What is the ratio of translational kinetic energy to the rotational kinetic energy of the sphere?

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

Hint:

For rolling motion without slipping, the translational and rotational kinetic energies are related by the moment of inertia. For a solid sphere, the ratio of translational to rotational kinetic energy is \( \frac{3}{4} \).

Answer 1

The Correct Option is B

For a solid sphere rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energies. 1. The translational kinetic energy \( K_{\text{trans}} \) is given by: \[ K_{\text{trans}} = \frac{1}{2} m v^2 \] where \( m \) is the mass and \( v \) is the velocity of the center of mass. 2. The rotational kinetic energy \( K_{\text{rot}} \) is given by: \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia of the sphere and \( \omega \) is the angular velocity. For a solid sphere, the moment of inertia about the center is: \[ I = \frac{2}{5} m r^2 \] where \( r \) is the radius of the sphere. Since the sphere rolls without slipping, the relation between the linear velocity and angular velocity is: \[ v = r \omega \] Substituting \( \omega = \frac{v}{r} \) into the rotational kinetic energy formula: \[ K_{\text{rot}} = \frac{1}{2} \times \frac{2}{5} m r^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{5} m v^2 \] Now, the ratio of translational kinetic energy to rotational kinetic energy is: \[ \frac{K_{\text{trans}}}{K_{\text{rot}}} = \frac{\frac{1}{2} m v^2}{\frac{1}{5} m v^2} = \frac{5}{2} \times \frac{1}{2} = \frac{5}{4} \] Thus, the ratio of translational to rotational kinetic energy is \( \frac{3}{4} \).