Question

A smooth sphere of radius $R$ is made to translate in a straight 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. The speed of particle with respect to the sphere as a function of angle $\theta$ as it slides down is :
Here $g$ is acceleration due to gravity.

• A. $\frac{\sqrt{Rg(\sin \theta +\cos \theta )}}{2}$
• B. $\sqrt{Rg(1+\cos \theta -\sin \theta )}$
• C. $\sqrt{4Rg\sin \theta }$
• D. $\sqrt{\left(2Rg\right(1+\sin \theta -\cos \theta )}$

Answer 1

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

Three forces are acting on the particle with respect to the sphere,
$\left(i\right)$ pseudo force $=ma=mg$ (as $a=g$)
$\left(ii\right)$ weight $=mg$
$\left(iii\right)$ normal reaction $=N$
Of these, first two forces are constant and will do work. The third force is not constant and it does not perform any work.


From work-energy theorem $($in non-inertial frame$)$

$\frac{1}{2}m{v}_r^2=$ Work done by pseudo force $+$ Work done by force of gravity

Here,
$v_r=$ speed of particle relative to sphere
$\frac{1}{2}m{v}_r^2=ma\left(AB\right)+mg\left(AD\right)$

$\Rightarrow \frac{1}{2}{v}_r^2=g[R\sin \theta +R(1-\cos \theta \left)\right]\because [a=g]$

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

Hence, option (d) is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{This question is very useful in understanding the application of work-energy theorem in non-inertial frame}\end{array}$