Question

A smooth sphere of radius $R$ is set into translatory motion in a straight line with a constant acceleration $a=g{\text{m/s}}^2$. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. If the particle slides down, then the speed of the particle with respect to the sphere when $\theta ={45}^{\circ }$ is :(Here $g=10{\text{m/s}}^2$ is acceleration due to gravity and Radius, $R=0.8\text{m}$).

Answer 1

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
$\frac{1}{2}m{v}_r^2=ma\left(AB\right)+mg\left(AD\right)$

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

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

Given, $R=0.8\text{m}$ and $\theta ={45}^{\circ }$

$\Rightarrow v_r=\sqrt{2\times 10\times 0.8(1+\sin {45}^{\circ }-\cos {45}^{\circ })}$

$\therefore v_r=4\text{m/s}$

Accepted answer : $4,4.0,4.00$

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