Question

A stone tied to an inextensible string of length $1\text{m}$ is released from horizontal position. The stone is free to revolve in a vertical plane around the other end of the string. Find the angular speed of the stone when the string makes an angle $\theta ={30}^{\circ }$ with horizontal. Take $g=10{\text{m/s}}^2$.

• A. $\omega =4\text{rad/ s}$
• B. $\omega =\sqrt{15}\text{rad/ s}$
• C. $\omega =\sqrt{10}\text{rad/ s}$
• D. $\omega =2.5\text{rad/ s}$

Answer 1

The correct option is C $\omega =\sqrt{10}\text{rad/ s}$


On applying energy conservation between intial and final position i.e $A$ and $B$, taking datum for $PE=0$ at point $A$:
$P{E}_A+K{E}_A=P{E}_B+K{E}_B$
$\Rightarrow 0+\frac{1}{2}m{u}^2=-mgh+\frac{1}{2}m{v}^2$
$\Rightarrow v^2=u^2+2gh$
where $h=l\sin \theta ,u=0$
$\Rightarrow v^2=0+2g\left(l\sin \theta \right)$
or $v=\sqrt{2\times 10\times 1\times \sin {30}^{\circ }}=\sqrt{10}\text{m/s}$

Now angular speed of body at $B$:
$\omega =\frac{v}{r}$
$\because$ radius of circular path about $\text{O}$ is $r=l=1\text{m}$
$\omega =\frac{\sqrt{10}}{1}=\sqrt{10}\text{rad/s}$