Question

A solid sphere is set to rotate in the anticlockwise direction, on a rough horizontal surface with a linear speed $10\text{m/s}$ and an angular speed $5\text{rad/s}$ as shown in figure. Find the linear speed of the sphere when it stops rotating.

• A. $6\text{m/s}$
• B. $10\text{m/s}$
• C. $30\text{m/s}$
• D. Zero

Answer 1

The correct option is A $6\text{m/s}$

$V_0=10\text{m/s},\omega =5\text{rad/s},R=2\text{m}$
$\therefore V_0=R\omega ...\left(i\right)$
Since tangential velocity due to rotation is in the same direction as velocity of $\text{COM}$ at the point of contact $\left(P\right)$, kinetic friction$\left(f\right)$ will act at the point of contact in the opposite direction of velocity (leftward).

But ${\tau }_f=0$ about point of contact $\left(P\right)$, so there is no net torque due to any external forces about the point of contact. Hence, applying conservation of angular momentum about $\left(P\right)$
$L_i=L_f...\left(ii\right)$
Taking anticlockwise sense of rotation as $+ve$,
$L_i=+\left(I_{CM}\omega \right)-\left(m{V}_{0}R\right)...\left(iii\right)$
Here, $m{V}_{0}R$ represents moment of linear momentum of COM about $P$.
Substituting $\omega =\frac{V_0}{R}$ from Eq $\left(i\right)$,
$L_i=+\left(\frac{2}{5}m{R}^2\right)\frac{V_0}{R}-m{V}_{0}R$
$L_i=\frac{-3m{V}_{0}R}{5}$

When sphere stops rotating,


$L_f=I_{CM}-mvR=0-mvR$
$\Rightarrow L_f=-mvR$
Substituting in Eq $\left(ii\right)$,
$\frac{-3m{V}_{0}R}{5}=-mvR$
$\Rightarrow v=\frac{3V_0}{5}$
$\therefore v=\frac{3\times 10}{5}=6\text{m/s}$