Question

A spherical ball of mass $m$ begins to slide down a fixed smooth sphere from the top with negligible initial velocity. What is its tangential acceleration when it breaks off the sphere?

• A. $\frac{2g}{3}$
• B. $\frac{g}{3}$
• C. $\frac{\sqrt{5}}{3}g$
• D. Ball never leaves contact

Answer 1

The correct option is C $\frac{\sqrt{5}}{3}g$

Let radius of smooth sphere be $R$.


If a spherical ball starts from rest from top,
applying conservation of energy for the spherical ball,
Loss in potential energy = gain in KE
i.e $mgR(1-\cos \theta )=\frac{1}{2}m{v}^2$ ... (i)

Since it breakes off at this point, so normal force will be zero,
Therefore, $mg\cos \theta =\frac{m{v}^2}{R}+N$
$\Rightarrow \frac{m{v}^2}{R}=mg\cos \theta$ ... (ii)
On equating (i) and (ii) $mg\cos \theta =2mg(1-\cos \theta )$
$\Rightarrow \cos \theta =\frac{2}{3}\&\sin \theta =\sqrt{1-{\cos }^2\theta }=\frac{\sqrt{5}}{3}$
From diagram, tangential acceleration, $a_t=\frac{mg\sin \theta }{m}$
$\Rightarrow a_t=g\sin \theta =g\times \frac{\sqrt{5}}{3}=\frac{\sqrt{5}}{3}g$