Question

A disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points equidistant from $C$ as shown in figure. Let $v_P,v_Q$ and $v_C$ be the magnitude of velocities of points $P,Q$ and $C$ respectively, then

• A. $v_Q>v_C>v_P$
  
• 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 A $v_Q>v_C>v_P$


Since $P$ and $Q$ are equidistant from the centre $C$, so it's $\perp$ distance from the axis of rotation is also same i.e $r$


Considering rightwards as $+vex-$ direction, the velocity of point $P$ and $Q$ on the disc can be written as:
${\overrightarrow{v}}_P=(v-r\omega )\hat{i}$
${\overrightarrow{v}}_Q=(v+r\omega )\hat{i}$
${\overrightarrow{v}}_C=v\hat{i}$
Hence magnitude of the velocity is in order:
$\Rightarrow v_P<v_C<v_Q$