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 solve the problem of finding the ratio of the squares of the speeds \(v_1^2\) and \(v_2^2\), we need to analyze the dynamics of a rolling sphere on an inclined plane.

The speed at the bottom of the inclined plane results from the conversion of potential energy to kinetic energy. For a solid sphere rolling without slipping on an incline:

\(v = \sqrt{\frac{10gh}{7}}\) 

where \(g\) is the acceleration due to gravity, and \(h\) is the height of the incline.

The height \(h\) of the incline can be calculated using the length \(L\) and the angle of inclination \(\theta\):

\(h = L \sin \theta\)

So, the velocity at the bottom can be expressed as:

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

Let's consider the two scenarios:

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

Now, to find the ratio \(v_1^2 : v_2^2\), calculate the squares of both expressions:

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

Therefore, the ratio is:

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

This matches the given correct answer: \(1 : \sqrt{2}\). Thus, the correct option is \(1 : \sqrt{2}\).

Answer 2

To find the ratio of v₁² : v₂², we start by using the principles of energy conservation. When the sphere rolls down without slipping, both translational and rotational kinetic energies are involved.

Step 1: Calculate the speed when inclined at 30°.
The potential energy at the top is converted into translational and rotational kinetic energy at the bottom.
\( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
For a solid sphere, \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \).
Substitute these values:
\( mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \frac{v^2}{r^2} \)
\( mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \)
\( mgh = \frac{7}{10}mv^2 \)
\( v^2 = \frac{10}{7}gh \)
For a plane inclined at 30°: \( h = L \sin(30^\circ) = \frac{L}{2} \)
\( v_1^2 = \frac{10}{7}g \cdot \frac{L}{2} = \frac{5}{7}gL \)

Step 2: Calculate the speed when inclined at 45°.
For an incline of 45°: \( h = L \sin(45^\circ) = \frac{\sqrt{2}}{2}L \)
\( v_2^2 = \frac{10}{7}g \cdot \frac{\sqrt{2}}{2}L = \frac{5\sqrt{2}}{7}gL \)

Step 3: Find the ratio of v₁² : v₂².
\( \frac{v_1^2}{v_2^2} = \frac{\frac{5}{7}gL}{\frac{5\sqrt{2}}{7}gL} = \frac{1}{\sqrt{2}} \)
Hence, the ratio is \( 1: \sqrt{2} \).