Question

A ring of radius $1\text{m}$ is performing combined translational and rotational motion on a frictionless horizontal surface with an angular velocity of $3\text{rad/s}$ as shown in figure. Find out velocity of the point $A$ in $\text{m/s}$, if the speed of the lowest point $V_P$ is $1\text{m/s}$.

• A. $4\hat{i}+3\hat{j}$
• B. $4\hat{i}-3\hat{j}$
• C. $-3\hat{i}+4\hat{j}$
• D. $-4\hat{i}-3\hat{j}$

Answer 1

The correct option is A $4\hat{i}+3\hat{j}$

From right hand thumb rule, $+vez-$axis comes out perpendicularly outward to $(x-y)$ plane.
$\Rightarrow \overrightarrow{\omega }=-3\hat{k}\text{rad/sec}$
and ${\overrightarrow{V}}_P=1\hat{i}\text{m/s}$
$R=1\text{m}\Rightarrow {\overrightarrow{R}}_{PO}=-1\hat{j}\text{m}$
For combined rotational and translational motion,
$\overrightarrow{V_P}=\overrightarrow{V_O}+{\overrightarrow{V}}_{PO}...\left(i\right)$


where ${\overrightarrow{V}}_{PO}$ is the velocity of point $P$ w.r.t centre of the ring $\left(O\right)$.
${\overrightarrow{V}}_{PO}=\overrightarrow{\omega }\times {\overrightarrow{R}}_{PO}$
Substituting in Eq $\left(i\right)$,
$\Rightarrow 1\hat{i}=\overrightarrow{V_O}+\left[\right(-3\hat{k})\times (-1\hat{j}\left)\right]$
$\Rightarrow 1\hat{i}=\overrightarrow{V_O}+3(\hat{k}\times \hat{j})=\overrightarrow{V_O}-3\hat{i}$
$\therefore \overrightarrow{V_O}=4\hat{i}$
Now,
$\overrightarrow{V_A}=\overrightarrow{V_O}+{\overrightarrow{V}}_{AO}$
$\overrightarrow{V_A}=\overrightarrow{V_O}+\overrightarrow{\omega }\times {\overrightarrow{R}}_{AO}$


$\because {\overrightarrow{R}}_{AO}=-1\hat{i}$
$\Rightarrow \overrightarrow{V_A}=4\hat{i}+\left[\right(-3\hat{k})\times (-1\hat{i}\left)\right]$
$\therefore \overrightarrow{V_A}=(4\hat{i}+3\hat{j})\text{m/s}$