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

To determine the ratio of the squares of the speeds \(v_1^2\) and \(v_2^2\), we examine the dynamics of a solid sphere rolling without slipping down an inclined plane. The speed at the base is derived from the conservation of energy, specifically the conversion of potential energy to kinetic energy. The formula for the speed at the bottom is:

\(v = \sqrt{\frac{10gh}{7}}\)

Here, \(g\) represents the acceleration due to gravity, and \(h\) is the vertical height of the incline. The height \(h\) can be expressed in terms of the incline's length \(L\) and angle \(\theta\) as:

\(h = L \sin \theta\)

Substituting this into the velocity equation yields:

\(v = \sqrt{\frac{10gL \sin \theta}{7}}\)

Consider two distinct scenarios:

• For an angle of \(30^\circ\), the speed is \(v_1\): \(v_1 = \sqrt{\frac{10gL \sin 30^\circ}{7}} = \sqrt{\frac{10gL \times \frac{1}{2}}{7}} = \sqrt{\frac{5gL}{7}}\)
• For an angle of \(45^\circ\), the speed is \(v_2\): \(v_2 = \sqrt{\frac{10gL \sin 45^\circ}{7}} = \sqrt{\frac{10gL \times \frac{\sqrt{2}}{2}}{7}} = \sqrt{\frac{5\sqrt{2}gL}{7}}\)

To find the ratio \(v_1^2 : v_2^2\), we square these velocities:

• \(v_1^2 = \frac{5gL}{7}\)
• \(v_2^2 = \frac{5\sqrt{2}gL}{7}\)

The resulting ratio is:

\(v_1^2 : v_2^2 = \frac{5gL/7}{5\sqrt{2}gL/7} = 1 : \sqrt{2}\)

This result, \(1 : \sqrt{2}\), aligns with the provided correct answer.