Question

As shown in figure, a wheel $(r=1\text{m})$ is performing combined translational and rotational motion with angular velocity $2\text{rad/s}$ and the velocity of its centre as $\overrightarrow{V_O}=4\hat{i}\text{m/s}$. Find the velocity of point $B$ in $\text{m/s}$ on the wheel.

• A. $\overrightarrow{V_B}=4\hat{i}+2\hat{j}$
• B. $\overrightarrow{V_B}=2\hat{i}-4\hat{j}$
• C. $\overrightarrow{V_B}=4\hat{i}-2\hat{j}$
• D. $\overrightarrow{V_B}=2\hat{i}+4\hat{j}$

Answer 1

The correct option is C $\overrightarrow{V_B}=4\hat{i}-2\hat{j}$

Angular velocity of ring $\overrightarrow{\omega }=-2\hat{k}\text{rad/s}$, from right hand thumb rule
$\Rightarrow$Velocity of $\text{COM}$ of the ring, $\overrightarrow{V_O}=4\hat{i}\text{m/s}$
For the velocity of point $B$ on ring,
$\overrightarrow{V_B}=\overrightarrow{V_O}+{\overrightarrow{V}}_{BO}...\left(i\right)$
Here,
${\overrightarrow{V}}_{BO}=\overrightarrow{\omega }\times {\overrightarrow{r}}_{BO}...\left(ii\right)$ $\because {\overrightarrow{r}}_{BO}=1\hat{i}$, it represents position vector of $B$ w.r.t centre $O$.
$\Rightarrow {\overrightarrow{V}}_{BO}=(-2\hat{k})\times \left(1\hat{i}\right)$
${\overrightarrow{V}}_{BO}=-2(\hat{k}\times \hat{i})=-2\hat{j}\text{m/s}$
Substituting in Eq $\left(i\right)$,
$\overrightarrow{V_B}=(4\hat{i}-2\hat{j})\text{m/s}$