Question

A particle is projected along the inner surface of a smooth vertical circle of radius $R$, its velocity at the lowest point being $\frac{1}{5}\sqrt{95Rg}$. It will leave the circle at angle $\theta$ with the verticle. The angle $\theta$ is

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

Answer 1

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

Let $\text{B}$ be the point where the particle leaves the circle and its velocity at point $\text{B}$ be $v$

Using conservation of mechanical energy between points $\text{A}$ and $\text{B}$
$\frac{1}{2}m{u}^2=mgR(1+cos\theta )+\frac{1}{2}m{v}^2$

Using second law at point $\text{B}$,

$mgcos\theta =m\frac{v^2}{R}(\because N=0)$

$\Rightarrow v^2=Rgcos\theta$

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

$\Rightarrow \frac{95}{25}=2+2cos\theta +cos\theta$

$\Rightarrow 3cos\theta =\frac{45}{25}$

$\Rightarrow cos\theta =\frac{15}{25}=\frac{3}{5}$

or, $\theta ={53}^{\circ }$

Hence, option $\left(b\right)$ is the correct answer.