Question

A small object of uniform density rolls up a curved surface with an initial velocity . It reaches up to a maximum height of $\frac{3v^2}{4g}$with respect to the initial position. The object is

• A. Ring
• B. Solid sphere
• C. Hollow sphere
• D. Disc

Answer 1

The Correct Option is D

To solve this problem, we need to understand the motion of objects with different shapes and how their moments of inertia affect their rolling motion.

The given problem involves a small object rolling up a curved surface with an initial velocity \( v \) and reaching a maximum height of \(\frac{3v^2}{4g}\) with respect to its initial position. We need to identify what shape the object could be to reach this height.

Let's consider the energy conservation principle for a rolling object. The total mechanical energy (kinetic energy + potential energy) remains constant if there's no slipping or external force. The object initially has kinetic energy due to both translational and rotational motion and zero potential energy:

Initial total energy, \( E_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)

At the maximum height, all kinetic energy is converted into potential energy:

Potential Energy, \( E_f = mgh \)

Equating initial and final energy gives us: \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh .

For rolling without slipping, the relation between linear velocity and angular velocity is \( v = r\omega \), where \( r \) is the radius. Also, the height reached is given as \( h = \frac{3v^2}{4g} \).

Let's calculate the potential energy at the maximum height: mgh = mg \left(\frac{3v^2}{4g}\right) = \frac{3mv^2}{4}.

Substitute this into the energy conservation equation: \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3mv^2}{4}.

Rearrange the equation to solve for the moment of inertia \( I \): \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v^2}{r^2}\right) = \frac{3mv^2}{4} .

Substitute \( \omega = \frac{v}{r} \) into the equation: \frac{1}{2}I\frac{v^2}{r^2} = \frac{3mv^2}{4} - \frac{1}{2}mv^2 \Rightarrow \frac{1}{2}I\frac{v^2}{r^2} = \frac{mv^2}{4} .

Solve for \( I \): I = \frac{mr^2}{2} .

Taking into account standard moments of inertia:

• Ring: \( I = mr^2 \)
• Solid sphere: \( I = \frac{2}{5}mr^2 \)
• Hollow sphere: \( I = \frac{2}{3}mr^2 \)
• Disc: \( I = \frac{1}{2}mr^2 \)

The calculated \( I \) matches the moment of inertia of a Disc. Therefore, the correct answer is a Disc.