Question

A uniform disc of radius $R=0.2\text{m}$ kept over a rough horizontal surface is given a translational velocity $V_0$ and angular velocity ${\omega }_0$. After some time, its kinetic energy become zero, then value of ${\omega }_0$ is:
Given: $V_0=10\text{m/s}$.

• A. $100\text{rad/s}$
• B. $50\text{rad/s}$
• C. $200\text{rad/s}$
• D. $5\text{rad/s}$

Answer 1

The correct option is A $100\text{rad/s}$

Initially, $V_0=10\text{m/s}$
$R=0.2\text{m}$
After some time, $KE=0$
$\Rightarrow K{E}_{Trans}+K{E}_{Rot}=0$
$\Rightarrow \frac{1}{2}m{V}^2+\frac{1}{2}I{\omega }^2=0$
$\Rightarrow \frac{1}{2}m{V}^2=0$ and $\frac{1}{2}I{\omega }^2=0$
[since sum of two $+ve$ quantities can be zero, only when each of them is zero]
$\therefore V=0\&\omega =0$


Kinetic friction $f_k$ will act as shown in figure to prevent slipping of contact point $P$.
$\Rightarrow$Torque of $f_k$ about point $P$ will be zero i.e ${\tau }_{f_k}=0$. Hence conserving angular momentum about point $P$,
$L_i=L_f$
$L_f=0(\because V=0,\omega =0)$
$\Rightarrow -m{V}_{0}R+(+I{\omega }_0)=0$
(Taking anticlockwise sense of rotation as $+ve$, as it gives $+vez-$axis by right hand thumb rule.)
$\Rightarrow m{V}_{0}R=\left(\frac{m{R}^2}{2}\right){\omega }_0$
$\Rightarrow {\omega }_0=\frac{2V_0}{R}=\frac{2\times 10}{0.2}$
$\therefore {\omega }_0=100\text{rad/s}$