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.

Accepted Answer

Given:
Mass of particle $= m$ ,
Radius of the sphere $= R$
The initial velocity of the particle $= v$
(a) Let the normal reaction at the topmost position $A$ be $N$ , as shown in the figure.

Applying Newton's second law in radial direction at point $B$ , we get,
$m g - N = \frac{m v^{2}}{R} ⟹ N = m g - \frac{m v^{2}}{R}$

(b) As the velocity increases normal reaction at point $A$ decreases. At a certain velocity, the normal reaction becomes zero and the particle ceases to slip on the sphere.
The minimum value of $v$ such that the particle jumps off from sphere at position $A$ ,
$N = 0 = m g - \frac{m v_{\min}^{2}}{R} \Rightarrow v_{\min} = \sqrt{g R}$
(c) Given:

$v = \frac{v_{\min}}{2} = \frac{\sqrt{g R}}{2}$
Let the velocity of the particle be $v_{B}$ and angle made by the particle with vertical be $θ$ when it leaves the contact with the sphere as shown in the figure.

At the time the particle loses contact, the normal reaction from this sphere will be zero.
Applying Newton's second law in radial direction at this point, we get,
$m g \cos θ - N = \frac{m v_{B}^{2}}{R} m v_{B}^{2} = m g R \cos θ \dots (1)$

Applying conservation of mechanical energy between $A and B$ , we get,
$Δ U + Δ K = 0 \Rightarrow - m g (R - R \cos θ) + \frac{1}{2} m v_{B}^{2} - \frac{1}{2} m {(\frac{\sqrt{g R}}{2})}^{2} = 0 \dots (2)$

From equations $(1)$ and $(2)$ ,
$- m g R + m g R \cos θ + \frac{m g R \cos θ}{2} - \frac{m g R}{8} = 0 \Rightarrow \cos θ = \frac{3}{4} \Rightarrow θ = \cos^{- 1} (\frac{3}{4})$

B M Sharma Solutions for Chapter: Circular Motion, Exercise 52: ILLUSTRATION 9.49

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 52: ILLUSTRATION 9.49

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 52: ILLUSTRATION 9.49 with Hints & Solutions

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