Question

In figure E and V_cm represent the total energy and speed of centre of mass of an object of mass 1kg in pure rolling. The object is :

• A. sphere
• B. ring
• C. disc
• D. Hollow Cylinder

Answer 1

The Correct Option is C

The total energy (E) of an object rolling without slipping is given by:

$E = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$

where:

• $m$ is the mass,
• $v_{cm}$ is the velocity of the center of mass,
• $I$ is the moment of inertia,
• $\omega$ is the angular velocity.

For pure rolling, $v_{cm} = R\omega$, where $R$ is the radius. Thus, $\omega = \frac{v_{cm}}{R}$. Substituting this into the energy equation:

$E = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\left(\frac{v_{cm}}{R}\right)^2$

$E = \frac{1}{2}mv_{cm}^2\left(1 + \frac{I}{mR^2}\right)$

From the graph, we see that when $v_{cm}^2 = 4$, $E = 3$. Since the mass is 1 kg:

$3 = \frac{1}{2}(1)(4)\left(1 + \frac{I}{mR^2}\right)$

$3 = 2\left(1 + \frac{I}{mR^2}\right)$

$1 + \frac{I}{mR^2} = \frac{3}{2}$

$\frac{I}{mR^2} = \frac{1}{2}$

This means $I = \frac{1}{2}mR^2$, which is the moment of inertia of a disc.

The correct answer is (C) disc.