Question

A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v (a) Find the normal force between the sphere and the particle just after the impulse? (b) What should be the minimum value of v for which the particle does not slip on the sphere (c) Assuming the velocity v to be half the minimum calculated in part (b) find the angle made by the radius through the particle with the vertical when it leaves the sphere ?

Answer 1

Here in the question ,
Assume Radius = R
Horizontal speed = h

Now for (a)

Particle is on the top = mg (on downward)Normal force towards particle = N (on Upward)The centripetal force on the particle  = mg -NAlso, Centripetal acceleration on thew particle  = mv²/RTherefore ,
 $mg-N = \frac{mv^2}{R}$
Normal force, N = $mg - \frac{mv^2}{R}$ .

Hence the normal force b/w the sphere is $mg - \frac{mv^2}{R}$ .

For- (b)
when a particle is at maximum velocity for a limiting state , then N= 0 .
Therefore,
mg - mv²/R = 0
$mg = \frac{mv^2}{R}$ -----(i)
$v = \sqrt{gR}$ .


For -(c).
Given that velocty v₁ = 1/2 √gR
Now v₁² = 1/4 gR
$v^2_{1} - \frac{1}{4} *gR$

When the particle leaves the surface assume that the velocity would be v₂.
Now from equation (i) we get,

$\frac{mv^2_{2}}{R} = mgcos\theta$ 
Hence
$v^2_{2} = Rgcos\theta$   ----(ii)

Now equating the initial and final points.

$\frac{1}{2} mv^2_{2} - \frac{1}{2}mv^2_{1} = mgR (1-cos\theta)$ 
from this equation we get
v²₂= v²₁ + 2gR (1-cosθ) ----(iii)

Now from equation (ii) & (iii) we get

$Rgcos\theta = \frac{Rg}{4} +2gR (1-cos\theta)$ 
3cosθ = 9/4
$\boxed{\theta = cos^-(3/4)}$

Hope it Helps :-)