Question

A particle is projected along the inner surface of a smooth vertical circle of radius $R$, its velocity at the lowest point being $\frac{\sqrt{95Rg}}{5}$. It will leave the circle at an angular distance of______from the vertical line joining highest and lowest points.

• A. ${37}^{\circ }$
• B. ${53}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${53}^{\circ }$

For Force balance
$mgcos\theta +N=\frac{m{v}^2}{R}$
Assuming body leaves the surface at $B$ $(\because N=0)$
$mgcos\theta =\frac{m{v}^2}{R}$
$mgRcos\theta =m{v}^2$

By conservation of mechanical energy [between point $A$ and $B$]
$\frac{1}{2}m{u}^2=mgh+\frac{1}{2}m{v}^2$
$\frac{1}{2}m{u}^2=mgR(1+cos\theta )+\frac{1}{2}m{v}^2$

Substituting values of $u$ and $m{v}^2$, we get

$\left(\frac{1}{2}\right)m{\left(\frac{1}{5}\sqrt{95Rg}\right)}^2=mgR(1+cos\theta )+\frac{1}{2}mgRcos\theta$

$\Rightarrow \frac{95}{25}=2+2cos\theta +cos\theta \Rightarrow 3cos\theta =\frac{45}{25}\Rightarrow cos\theta =\frac{15}{25}=\frac{3}{5}\Rightarrow \theta ={53}^o$