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

We need to identify the object undergoing pure rolling motion, given the relationship between its total energy ($ E $) and the square of the speed of its center of mass ($ V_{cm}^2 $), represented by a graph. The mass of the object is 1 kg.

1. Understanding Total Kinetic Energy Equations:

Total Kinetic Energy (E) in pure rolling = Translational Kinetic Energy + Rotational Kinetic Energy: $$ E = \frac{1}{2}mV_{cm}^2 + \frac{1}{2}I\omega^2 $$
For pure rolling, $ V_{cm} = R\omega $, where $ R $ is the radius, and $ \omega = \frac{V_{cm}}{R} $. Substituting $ \omega $ into the equation for $ E $: $$ E = \frac{1}{2}mV_{cm}^2 + \frac{1}{2}I\left(\frac{V_{cm}}{R}\right)^2 $$
Simplify further: $$ E = \frac{1}{2}mV_{cm}^2 + \frac{1}{2}\left(\frac{I}{R^2}\right)V_{cm}^2 $$
Combine terms: $$ E = \frac{1}{2}V_{cm}^2 \left(m + \frac{I}{R^2}\right) $$
Using $ I = kmR^2 $, where $ k $ is a constant depending on the geometry of the object: $$ E = \frac{1}{2}V_{cm}^2 \left(m + \frac{kmR^2}{R^2}\right) $$
Simplify: $$ E = \frac{1}{2}V_{cm}^2 \left(m + km\right) $$
Factor out $ m $: $$ E = \frac{1}{2}m(1 + k)V_{cm}^2 $$

2. Analyzing the Graph:

The graph shows a linear relationship between $ E $ and $ V_{cm}^2 $. The equation of the line is of the form: $$ E = \text{slope} \cdot V_{cm}^2 $$

From the graph, when $ V_{cm}^2 = 4 $, $ E = 3 $. Therefore, the slope is: $$ \text{slope} = \frac{3}{4} $$

Thus, the equation becomes: $$ E = \frac{3}{4}V_{cm}^2 $$

3. Comparing the Equations:

From the derived equation: $$ E = \frac{1}{2}m(1 + k)V_{cm}^2 $$ and the graph equation: $$ E = \frac{3}{4}V_{cm}^2, $$ equate the coefficients: $$ \frac{1}{2}m(1 + k) = \frac{3}{4}. $$

Given $ m = 1 \, \text{kg} $: $$ \frac{1}{2}(1 + k) = \frac{3}{4}. $$

Solve for $ k $: $$ 1 + k = \frac{3}{2}, $$ $$ k = \frac{3}{2} - 1 = \frac{1}{2}. $$

4. Finding the Object:

The object must have $ k = \frac{1}{2} $, or $ I = \frac{1}{2}MR^2 $. Looking at the moments of inertia:

• Solid Sphere: $ I = \frac{2}{5}MR^2 $,
• Ring: $ I = MR^2 $,
• Disc: $ I = \frac{1}{2}MR^2 $,
• Hollow Cylinder: $ I = MR^2 $.

The only object with the required $ k $ value is the Disc.

Final Answer: The correct answer is (C) disc.

Answer 2

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.