Question

A particle of mass $m$ is kept on a fixed, smooth sphere of radius $R$ at a position, where the radius through the particle makes an angle of ${30}^{\circ }$ with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release? (b) Find the distance traveled by the particle before it leaves contact with the sphere.

Answer 1

(a)Just after the release, the particle will experience two forces: Gravitational force ($mg$) and normal reaction from the surface of the fixed, smooth sphere ($N$).

As per the FBD, we can clearly see that upon resolving along the line of action of $N$, we get:

$N-mg\cos {30}^{\circ }=0$

$\Rightarrow N=mg\cos {30}^{\circ }=\frac{\sqrt{3}}{2}mg\to \text{(A)}$

(B)Now, assuming that the particle leaves contact with the sphere, at a position whose angle with the vertical is $\theta$. In this situation, the normal reaction due to the sphere ($N_{\theta }$) becomes zero, as this is the instant the particle leaves the sphere. Let's assume the velocity of the particle here is $v$.

$\Rightarrow N_{\theta }=0$

The resultant of $N_{\theta }$ and the radial component of $mg$ (i.e. $mg\cos \theta$) provide the necessary centripetal force for the particle to continue on a circular path right before it leaves contact with the sphere.

$\Rightarrow m\frac{v^2}{R}=mg\cos \theta -N_{\theta }$

$\Rightarrow v^2=gR\cos \theta$

Applying work-energy theorem on the system:

$W_{mg}=K{E}_f-K{E}_i$

$\Rightarrow mgh=\frac{1}{2}m{v}^2-0$

$\Rightarrow mgR(\cos {30}^{\circ }-\cos \theta )=\frac{1}{2}m{v}^2$

Replacing $v^2=gR\cos \theta$:

$\Rightarrow gR(\frac{\sqrt{3}}{2}-\cos \theta )=\frac{1}{2}gR\cos \theta$

$\therefore \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

The distance travelled by the particle would be the length of the circular arc covered from ${30}^{\circ }$ to $\theta$:

$\Rightarrow \Delta \theta =\theta -\frac{\pi }{6}$, where all angular measurements are in radians.

Thus, the path-length will be $R\Delta \theta$, where $\Delta \theta$ is in radians.

Length travel $=R(\theta -\frac{\pi }{6})$