Question

A solid sphere of mass \( m \) and radius \( r \) is allowed to roll without slipping from the highest point of an inclined plane of length \( L \) and makes an angle of \( 30^\circ \) with the horizontal. The speed of the particle at the bottom of the plane is \( v_1 \). If the angle of inclination is increased to \( 45^\circ \) while keeping \( L \) constant, the new speed of the sphere at the bottom of the plane is \( v_2 \). The ratio of \( v_1^2 : v_2^2 \) is:

• A. \( 1 : \sqrt{2} \)
• B. \( 1 : 3 \)
• C. \( 1 : 2 \)
• D. \( 1 : \sqrt{3} \)

Hint:

The speed of a rolling sphere is determined by the angle of inclination. The energy involved in pure rolling is partitioned between translational and rotational motion. The key idea is the relationship between the speed and the sine of the angle of inclination.

Answer 1

The Correct Option is A

Using the work-energy theorem (WET), we know that: \[ W_g = k_f - k_i \] where \( W_g \) is the work done by gravity, and \( k_f \) and \( k_i \) are the final and initial kinetic energies, respectively.
The gravitational potential energy is converted into kinetic energy. The kinetic energy in pure rolling is: \[ K.E. = \frac{1}{2} m v_{\text{cm}}^2 + \frac{1}{2} I_{\text{cm}} \omega^2 \] For a solid sphere, the moment of inertia about its center of mass is \( I_{\text{cm}} = \frac{2}{5} m r^2 \). Thus, the total kinetic energy becomes: \[ K.E. = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m r^2 \times \frac{v^2}{r^2} = \frac{7}{10} m v^2 \] From the conservation of mechanical energy: \[ m g L \sin \theta = \frac{7}{10} m v^2 \] Simplifying: \[ v^2 \propto \sin \theta \] Now, for two different angles \( \theta_1 = 30^\circ \) and \( \theta_2 = 45^\circ \), the ratio of the final speeds \( v_1^2 : v_2^2 \) becomes: \[ \frac{v_1^2}{v_2^2} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{1/2}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}} \] Thus, the correct answer is (1) \( 1 : \sqrt{2} \).