Question

A smooth sphere of radius $R$ is made to translate line with a constant acceleration $a=g$. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of angle $\theta$ as it slides down.

• A. $\sqrt{Rg(1+\sin \theta -\cos \theta )}$
• B. $\sqrt{2Rg(2+\sin \theta -\cos \theta )}$
• C. $\sqrt{2Rg(1+\sin \theta -\cos \theta )}$
• D. $\sqrt{2Rg(1-\sin \theta -\cos \theta )}$

Answer 1

The correct option is C $\sqrt{2Rg(1+\sin \theta -\cos \theta )}$

A force $f$ acts on the sphere towards left, so the particle experience pseudo force $f=ma$ towards right.

Initial kinetic energy of the particle is zero.

Let the speed of the particle at point C be $v$.

From work-energy theorem, $W_g+W_f=\Delta K.E=K.E_f$

where $W_g$ is work done by gravity and $W_f$ is work done by pseudo force.

From figure, we get $AB=R-R\cos \theta =R(1-\cos \theta )$

Also $AC=R\sin \theta$

$\therefore$ $mg\left(AB\right)+f\left(AC\right)=\frac{1}{2}m{v}^2$

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

Or $mgR(1+\sin \theta -\cos \theta )=\frac{1}{2}m{v}^2$ $(\because a=g)$

$⟹$ $v=\sqrt{2gR(1+\sin \theta -\cos \theta )}$