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 30° with the vertical. The particle is released from the 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

Given in the question :-
The particle is released on the smooth sphere . 

For (a)
Hence the Normal force on the surface = N
weight of the particle = mgcosθ

Now equating the conditions 
N = mgcos30°
N = √3/2 (mg) .
Hence, Force exerted by the sphere is √3/2(mg)

For (b)
The particle makes the angle θ when it leaves the surface, so N= 0
Hence the velocity v ,
$\frac{mv^2}{R} = mgcos \theta$ 
$v^2 = Rgcos( \theta + 30^0)$ ----(i) .

Now change in Potential energy
= $mg{Rcos30^0-Rcos(\theta+30^0)}$  ----(ii)

Change in Kinetic Energy
= $\frac{1}{2} mv^{2}$ -----(iii)

Now equating equation (ii) & (iii) .

$1/2 mv^2 = mg{Rcos30^0-Rcos(\theta+30^0)}$
$v^2 = 2g{Rcos30^0-Rcos(\theta+30^0)}$

Now from here , put the value of v² in equation (i)

$gRcos(\theta+30^0)=2g{Rcos30^0-Rcos(\theta+30^0)}$

$3cos(\theta+30^0) = \sqrt{3}$
cos(θ+30°) = √3/3 = 1/√3
1/√3 ≈cos55°

Now
θ = 55°-30°
θ = 25°
or 0.43 radian

Hence the distance covered by the particle 0.43 radian.


Hope it Helps :-)