Question

A disc is rolling without slipping on a horizontal surface. $P$ and $Q$ are two points equidistant from $C$. Then, identify the correct statement regarding the velocity of the points:

• 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. $v_Q=v_P<v_C$

Answer 1

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

Let the linear velocity of C.O.M be $v$ and angular velocity of the disc about an axis passing through C.O.M be $\omega$.


In pure rolling, velocity of any point is the resultant of translational velocity and the tangential velocity due to rotation about C.O.M,
$\overrightarrow{v}={\overrightarrow{v}}_{c.o.m}+{\overrightarrow{\omega }}_{c.o.m}\times \overrightarrow{r}$
i.e $\overrightarrow{v}={\overrightarrow{v}}_1+\overrightarrow{v_2}...\left(1\right)$
$[{\overrightarrow{v}}_{c.o.m}=\overrightarrow{v_1}\text{and}{\overrightarrow{\omega }}_{c.o.m}\times \overrightarrow{r}=\overrightarrow{v_2}]$

where, $v_1\&v_2$ are representing the translational velocity and tangential velocity( due to rotation) for points.
$\Rightarrow$As the points $P$ and $Q$ are equidistant from centre.
$(v_2=\omega r)$ for both points $P$ and $Q$

For point $Q$, ${\theta }_Q<{90}^{\circ }$
$\therefore \cos \theta \text{is}+ve...\left(2\right)$
For point $P$, $Q_P>{90}^{\circ }$
$\therefore \text{cos}\theta \text{is}-ve...\left(3\right)$
For point $C$, tangential velocity due to rotation is zero about $C$.
$\Rightarrow v_2=0,v_1=v$
Hence, net velocity of point $C$,
$v_C=v_1=v$

From Eq.$\left(1\right)$ for vectorial addition, we can see that:
$\left(a\right)$. The component of tangential velocity $\left(\omega r\right)$ will be opposite to translational velocity $\left(v\right)$ in case of point $P$.
and
$\left(b\right)$. The component of tangential velocity$\left(\omega r\right)$ will be in the direction of translational velocity $\left(v\right)$ in case of point $Q$.

Hence the resultant velocity will be in the order,
$v_Q>v_C>v_P$