Question

A ball of mass $m$ slides down a fixed sphere. Find the velocity of the ball when it makes an angle $\theta$ with the vertical as shown in the figure, if it is released from rest from the topmost position. The sphere has radius $R$ and ball has radius $r$. Assume the sphere to be smooth.

• A. $\sqrt{2gR(1-\cos \theta )}$
• B. $\sqrt{2gr(1-\cos \theta )}$
• C. $\sqrt{2(R+r)g(1-\cos \theta )}$
• D. $\sqrt{2g(R+r)}$

Answer 1

The correct option is C $\sqrt{2(R+r)g(1-\cos \theta )}$

Reduction in height of centre of mass of the ball is
$h=(R+r)-(R+r)\cos \theta =(R+r)(1-\cos \theta )$



Since velocity is tangential, hence $v\perp$ Normal reaction ($N$) which means $W_N=0$.
Also, gravity is a conservative force.

Applying mechanical energy conservation from initial position to final position:
$\therefore \text{Loss in PE}=\text{Gain in KE}$
$\Rightarrow mgh=\frac{1}{2}m{v}^2$
$\Rightarrow mg(R+r)(1-\cos \theta )=\frac{1}{2}m{v}^2$
$\therefore v=\sqrt{2(R+r)g(1-\cos \theta )}$