Question

A particle of mass m is kept 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 travelled by the particle before it leaves contact with the sphere.

Answer 1

(a) When the particle is released from rest

(Fig.1),the centrifugal force is zero.

N force =$mgcos\theta =mgcos{30}^{\circ }$

$=\frac{\sqrt{3}}{2}mg$

(b) When the particle leaves contact with the surface (Fig.2) N =$\theta$

So, $\frac{m{v}^2}{r}=mgcos\theta$

$\Rightarrow v^2=Rgcos\theta$ ...(i)

Again,$\left(\frac{1}{2}\right)m{v}^2=mgR(cos{30}^{\circ }=cos\theta )$

$\Rightarrow v^2=2Rg(\frac{\sqrt{3}}{2}-cos\theta )$. ...(ii)

From equation (i) and equation (ii),

$Rgcos\theta =2Rg[\frac{\sqrt{3}}{2}-cos\theta ]$

$\Rightarrow 3cos\theta =\sqrt{3}$

$\Rightarrow cos\theta =\frac{1}{\sqrt{3}}$

or $\theta =co{s}^{-}1\frac{1}{\sqrt{3}}$

So, the distance travelled by the particle before leaving contact,

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

[because ${30}^{\circ }=\left(\frac{\pi }{6}\right)$]

Putting the value of $\theta ,$we get

$L=0.43R$