Question

As shown in figure, a ring is performing combined translational and rotational motion with angular velocity $2\text{rad/s}$ and the velocity of its $\text{COM}$ is $4\text{m/s}$. Find the magnitude of velocity of point $C$ on the ring.

• A. $2\text{m/s}$
• B. $4\text{m/s}$
• C. $4\sqrt{2}\text{m/s}$
• D. $2\sqrt{2}\text{m/s}$

Answer 1

The correct option is C $4\sqrt{2}\text{m/s}$


Angular velocity of ring $\overrightarrow{\omega }=-2\hat{k}\text{rad/s}$, from right hand thumb rule
Velocity of $\text{COM}$ of the ring, $\overrightarrow{V_O}=4\hat{i}\text{m/s}$
For the velocity of point $C$ on ring,
$\overrightarrow{V_C}=\overrightarrow{V_O}+{\overrightarrow{V}}_{CO}...\left(i\right)$
Here, ${\overrightarrow{V}}_{CO}=\overrightarrow{\omega }\times {\overrightarrow{r}}_{CO}...\left(ii\right)$
${\overrightarrow{r}}_{CO}=-2\hat{i}$ (it represents position vector of $C$ w.r.t centre $O$)
$\Rightarrow {\overrightarrow{V}}_{CO}=(-2\hat{k})\times (-2\hat{i})$
$=4(\hat{k}\times \hat{i})=4\hat{j}$
Substituting in Eq $\left(i\right)$,
$\overrightarrow{V_C}=4(\hat{i}+\hat{j})\text{m/s}$
$\therefore |\overrightarrow{V_C}|=\sqrt{4^2+4^2}=4\sqrt{2}\text{m/s}$