Question

A vertical frictionless semicircular track of radius $1\text{m}$ is fixed on the edge of a movable trolley (figure). Initially the system is at rest and a mass $m$ is kept at the the top of the track. The trolley starts moving to the right with a uniform horizontal acceleration $a=\frac{2g}{9}$. The mass slides down the track, eventually losing contact with it and dropping on to the floor. Calculate the angle $\theta$ at which it loses contact with the trolley and the track.

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

Answer 1

The correct option is B ${37}^{\circ }$


From the diagram,
Tangential acceleration of mass $m$,

$a_t=g\sin \theta +a\cos \theta$

Normal reaction will be zero at the point where contact is lost, so

$\frac{m{v}^2}{R}+ma\sin \theta =mg\cos \theta ...\left(1\right)$

As, $R=1\text{m}$ and $a=\frac{2g}{9}$ so, from $\left(1\right)$

$v^2=g\cos \theta -\frac{2g}{9}\sin \theta ...\left(2\right)$

By work energy theorem, work done $W$ by all the forces on mass $\left(m\right)$ will result in change of kinetic energy ,

$\Rightarrow W=F.dS\cos \theta =\frac{1}{2}m{v}^2...\left(3\right)$

Due to perpendicular forces $\theta ={90}^{\circ }$,
work done will be zero so total work done $W$ will be due to tangential forces
$\Rightarrow W=\underset{0^{\circ }}{\overset{\theta }{\int }}(mg\sin \theta +ma\cos \theta ).Rd\theta$

$\Rightarrow W=mgR{[-\cos \theta ]}_{0^{\circ }}^{\theta }+maR{\left[\sin \theta \right]}_{0^{\circ }}^{\theta }$

$\Rightarrow W=mgR[-\cos \theta -(-\cos 0^{\circ }\left)\right]+maR[\sin \theta -\sin 0^{\circ }]$

$\Rightarrow W=-mgR\cos \theta +mgR+maR\sin \theta$

$\Rightarrow W=R(mg-mg\cos \theta +ma\sin \theta )...\left(5\right)$

From $\left(3\right)$ and $\left(4\right)$, we have
$R(mg-mg\cos \theta +ma\sin \theta )=\frac{1}{2}m{v}^2$

putting $a=\frac{2g}{9}$ and $R=1\text{m}$ in the above equation,
$\Rightarrow v^2=2(g-g\cos \theta +\frac{2g}{9}\sin \theta )...\left(4\right)$

From equation $\left(2\right)$ and $\left(5\right)$, we have

$2(g-g\cos \theta +\frac{2g}{9}\sin \theta )=g\cos \theta -\frac{2g}{9}\sin \theta$

$\Rightarrow 3\cos \theta =\frac{2}{3}\sin \theta +2$
$\Rightarrow 9{\cos }^2\theta =\frac{4}{9}{\sin }^2\theta +4+\frac{8}{3}\sin \theta$
$\Rightarrow 9(1-{\sin }^2\theta )=\frac{4}{9}{\sin }^2\theta +4+\frac{8}{3}\sin \theta$
$\Rightarrow \frac{85}{9}{\sin }^2\theta -5+\frac{8}{3}\sin \theta =0$
$\Rightarrow 85{\sin }^2\theta -45+24\sin \theta =0$
$\Rightarrow \sin \theta =\frac{-24\pm \sqrt{{24}^2+4\left(45\right)\left(85\right)}}{170}=\frac{3}{5}$

after solving this equation, $\theta ={37}^{\circ }$

Hence, option (b) is correct answer.