Question

If the ring performs combined rotational and translational motion on a frictionless horizontal surface as shown in figure, find the magnitude of the velocity at a point $P$ on the ring.

• A. $2\sqrt{19}\text{m/s}$
• B. $\sqrt{38}\text{m/s}$
• C. $\sqrt{29}\text{m/s}$
• D. $\sqrt{19}\text{m/s}$

Answer 1

The correct option is D $\sqrt{19}\text{m/s}$

Radius, $R=1\text{m}$
Magnitude of velocity of $\text{COM}$ of the ring,
$|\overrightarrow{V_O}|=V_0=5\text{m/s}$
But, $\overrightarrow{V_P}=\overrightarrow{V_O}+\overrightarrow{V_{PO}}$
Magnitude of velocity of point $P$ w.r.t centre of ring $\left(O\right)$
$|\overrightarrow{V_{PO}}|=R\omega =1\times 2=2\text{m/s}$
${120}^{\circ }$ is the angle between $\overrightarrow{V_O}$ and $\overrightarrow{V_{PO}}$ as shown in figure:


Magnitude of velocity of point $P$ is,
$\left|\overrightarrow{V_P}\right|=\sqrt{|\overrightarrow{V_O}{|}^2+|\overrightarrow{V_{PO}}{|}^2+2\left(|\overrightarrow{V_O}|\right)\left(|\overrightarrow{V_{PO}}|\right)\times \cos {120}^{\circ }}$
where $\cos {120}^{\circ }=\frac{-1}{2}$
$\left|\overrightarrow{V_P}\right|=\sqrt{(5{)}^2+(2{)}^2+2\times 5\times 2\times \left(\frac{-1}{2}\right)}$

$\therefore \left|\overrightarrow{V_P}\right|=\sqrt{25+4-10}=\sqrt{19}\text{m/s}$