Question

A solid sphere and a hollow sphere of equal mass and radius are placed over a rough horizontal surface after rotating it about its mass centre with same angular velocity $ω_0$​. Once the pure rolling starts let $v_1$​ and $v_2$​ be the linear speeds of their centres of mass. Then
(A) ${v_1} = {v_2}$
(B) ${v_1} > {v_2}$
(C) ${v_1} < {v_2}$
(D) none of these

Answer 1

Hint
Frictional force derives the rotating object. When the velocity to the lowermost point such that it specifies a specific condition, its start pure rolling motion. The frictional force stays the same throughout the motion and this force is responsible for the velocity of the rotating object.
Acceleration due to force is given by, $F = Ma$
The relation between torque and angular acceleration is, $\tau = I\alpha$
Use the newton’s equation of motion i.e. $v = u + at$ and $\omega = {\omega _0} + \alpha t$

Complete step by step solution
Let’s have a look of the diagram and understand the diagram

Initially the object is rotating at the angular velocity $= ω_0$​
The frictional force $f$ starts acting in the forward direction. It does two things, it increases the linear velocity and decreases the angular velocity.
When the following condition is satisfied the lowermost point starts behaving like a fixed point with respect to the ground, then pure rolling motion starts.
$v = \omega t$
As we know that, $F = Ma$
$\Rightarrow a = \dfrac{F}{M}$
Where, $F$ is the frictional force
$M$ is the mass of the object
$a$ is the acceleration of the object
consider the object takes time $t$, to reach pure rolling motion.
Hence, the velocity of object will be
$v = 0 + \left( {\dfrac{F}{M}} \right)t$
$\Rightarrow t = \dfrac{{Mv}}{F}$
Now, frictional force also gives an opposite torque. Torque is given by
$\tau = Fr$
Hence, we can write it as
$I\alpha = Fr$
Where, $I$ is the moment of inertia of the object.
So, the angular acceleration will be,
$\alpha = \dfrac{{Fr}}{I}$
So, we can write
$\omega = {\omega _0} - \left( {\dfrac{{Fr}}{I}} \right)t \Rightarrow \dfrac{v}{r} = {\omega _0} - \left( {\dfrac{{Fr}}{I}} \right)\left( {\dfrac{{Mv}}{F}} \right) \Rightarrow v\left( {1 + \dfrac{{M{r^2}}}{I}} \right) = {\omega _0}r \Rightarrow v = \dfrac{{{\omega _0}r}}{{\left( {1 + \dfrac{{M{r^2}}}{I}} \right)}}$
Hence, the velocity of object in terms of the moment of inertia is given by
$v = \dfrac{{{\omega _0}}}{{\left( {1 + \dfrac{{M{r^2}}}{I}} \right)}}$ ………………… (1)
Moment of inertia of solid sphere is,
${I_{solid}} = \dfrac{2}{5}M{r^2}$
Moment of inertia of the hollow sphere is,
${I_{hollow}} = \dfrac{2}{3}M{r^2}$
Hence, ${I_{hollow}} > {I_{solid}}$
From equation (1), we get
${v_1} < {v_2}$
Hence, option (C) is correct

Note
We can compare the moment of inertia of the two objects without looking at the actual values. For example, if we have to compare the moment of inertia of solid sphere and hollow sphere, we can compare the distribution of the mass. If the mass is more at the periphery of the object then the moment of inertia will be more. So the moment of inertia of the hollow sphere is more than the solid sphere.