Question

A vertical frictionless semicircular track of radius $1$ m is fixed on the edge of a movable trolley (figure). Initially the system is rest and a mass m is kept at the top of the track. The trolley starts moving to the right with a uniform horizontal acceleration $a=2g/9$. The mass slides down the track, eventually losing contact with it and dropping to the floor $1.3m$ below the trolley. This $1.3m$ is from the point where mass loses contact. $(g=10m/s^2)$ 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 $=$ $g\sin \theta +a\cos \theta$

Normal reaction will be zero at the contact loosing point, so

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

By work energy theorm

work done $W$ by all the forces on mass ($m$ ) will result in change of kinetic energy

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

due to perpendicular forces $\theta ={90}^{\circ }$

work done will be zero so total work done $W$ will be

$\Rightarrow W={\int }_0^{\theta }(mg\sin \theta +ma\cos \theta ).Rd\varphi$

$\Rightarrow W=mg-mg\cos \theta +ma\cos \theta =\frac{1}{2}m{v}^2$

putting $a=\frac{2g}{9}$ in eqn $\left(2\right)$

${\Rightarrow v}^2=2(g-g\cos \theta +\frac{2g}{9}\sin \theta )...\left(3\right)$

again putting the value of $v^2$ and $R=1m$ in eqn $\left(1\right)$

$\Rightarrow 3\cos \theta =\frac{2}{3}\sin \theta +2$

after solving this equation, $\theta =37°$