Question

A disc of radius $R$ is rolling (without slipping) on a frictionless surface as shown in figure. $C$ is it's centre and $Q$ and $P$ are two points equidistant from $C$. Let $v_P,v_Q$ and $v_C$ be the magnitudes of velocities of points $P$, $Q$ and $C$ respectively, then

• A. $v_Q>v_P>v_C$
  
• B. $v_Q<v_C<v_P$
  
• C. $v_Q=v_P,v_C=\frac{1}{2}{v}_P$
  
• D. $v_Q>v_C>v_P$
  

Answer 1

The correct option is D $v_Q>v_C>v_P$


Points $P$ and $Q$ are equidistant from $C$, then
$CQ=CP=r$
Let $\omega$ be the angular speed of disc.
For pure rolling, $v=R\omega$,
where $v$ is the translational velocity of COM of the disc.
$\Rightarrow$Net velocity at any point on the disc will be due to vector sum of translational velocity and tangential velocity due to rotational motion.


$\Rightarrow {\overrightarrow{v}}_Q$ will be greater in magnitude compared to ${\overrightarrow{v}}_P$, since taking resultant of velocities, $\theta <{90}^{\circ },\cos \theta \text{is}+ve$ hence component vectors will add up together at point $Q$
$\Rightarrow$ At point $P$, angle between velocity vectors is $\theta >{90}^{\circ },\cos \theta \text{is}-ve$ so component vectors will get substracted at point $P$ to give resultant ${\overrightarrow{v}}_P$.
and at point $C$, $|{\overrightarrow{v}}_C|=v$.

Order of magnitude of velocity,
$\therefore v_Q>v_C>v_P$