Question

A particle placed on top of a smooth sphere of radius $2\text{m}$ slides through a distance of $\frac{\pi }{3}\text{m}$ and falls off the sphere. What is the velocity of the particle when it is about to drop off the surface of the sphere?

• A. $0\text{m/s}$
• B. $5.16\text{m/s}$
• C. $4.16\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is C $4.16\text{m/s}$


Distance covered on the surface of the sphere $=s=\frac{\pi }{3}\text{m}$
$\therefore$ Angle subtended at the centre of the sphere $=\theta =\frac{s}{r}=\frac{\frac{\pi }{3}}{2}=\frac{\pi }{6}$
$\therefore$ The particle makes an angle of ${30}^{\circ }$ with the vertical as it falls off.
FBD of the particle as it falls off:


As the particle falls off, the normal force, $N=0$
Since $N=0$, from the free body diagram, we can see that,
$mg\cos \theta =\frac{m{v}^2}{r}\Rightarrow v=\sqrt{rg\cos \theta }$
$\Rightarrow v=\sqrt{2\times 10\times \cos {30}^{\circ }}=\sqrt{20\times \frac{\sqrt{3}}{2}}$
$\Rightarrow v=4.16\text{m/s}$