Question

A disc rotating about its axis with angular speed ${\omega }_0$ is placed smoothly (without any translational velocity) on a perfectly frictionless table. Select the correct statement(s):

• A. Speed of point $A$ is $v_A={\omega }_{0}R$
• B. Speed of point $B$ is zero.
• C. Speed of point $C$ is $v_C=\frac{{\omega }_{0}R}{2}$
• D. Speed of point $C$ is $v_C={\omega }_{0}R$

Answer 1

Answer: (A), (C)

The correct options are
A Speed of point $A$ is $v_A={\omega }_{0}R$
C Speed of point $C$ is $v_C=\frac{{\omega }_{0}R}{2}$

Given,
Angular velocity of the disc $={\omega }_0(-\hat{k})$
Linear velocity $=0$
Disc is placed on a frictionless table.
Hence, ${\tau }_{com}=0$,
i.e $L=I{\omega }_0=$ constant
$\Rightarrow {\omega }_0$=constant

As we know, velocity of any point on the disc
$v_P=v_{com}+v_{P,com}$
Here, $v_{com}=0$
${\overrightarrow{v}}_{p,com}$ = velocity of point w.r.t. COM
$={\overrightarrow{\omega }}_0\times \overrightarrow{R}$
$|{\overrightarrow{\omega }}_0\times \overrightarrow{R}|={\omega }_{0}R\sin \theta$

Here, direction of ${\omega }_0$ is into the plane and $R$ is on the plane. Thus, angle between ${\overrightarrow{\omega }}_0$ and $\overrightarrow{R}$ is ${90}^{\circ }$

For point $A:{\nu }_A={\omega }_{0}R\sin {90}^{\circ }={\omega }_{0}R$
For point $B:{\nu }_B={\omega }_{0}R$
For point $C:{\nu }_C={\omega }_0\left(\frac{R}{2}\right)=\frac{{\omega }_{0}R}{2}$
Only option A and C are correct.