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 ν. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of ν for which the particle does not slip on the sphere? (c) Assuming the velocity ν 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

(a) Radius = R
Horizontal speed = \[v\] 

From the above diagram:
Normal force,

\[\text{ N = mg }- \frac{\text{ m }v^2}{\text{ R}}\]

(b) When the particle is given maximum velocity, so that the centrifugal force balances the weight, the particle does not slip on the sphere.

\[\text{So },\frac{\text{m}\nu^2}{R} = \text{mg}\]

\[ \Rightarrow \nu = \sqrt{\text{gR}}\]

(c) If the body is given velocity \[\nu_1\] at the top such that,

\[\nu_1 = \frac{\sqrt{gR}}{2}\]

\[ \nu_1^2 = \frac{gR}{4}\]

Let the velocity be \[\nu_2\]  when it loses contact with the surface, as shown below.

So,

\[\frac{\text{m}\nu_2^2}{R} = \text{mg} \cos \theta\]

\[ \Rightarrow \nu_2^2 = \text{Rg} \cos \theta . . . (\text{i})\]

\[\text{ Again,} \left( \frac{1}{2} \right) \text{m} \nu_2^2 - \left( \frac{1}{2} \right) m \nu_1^2 \]
\[ = \text{ mgR }\left( 1 - \cos \theta \right)\]
\[ \Rightarrow \nu_2^2 = \nu_1^2 + 2\text{ gR} \left( 1 - \cos \theta \right) . . . (ii)\]

From equations (i) and (ii),

\[Rg \cos \theta = \left( \frac{Rg}{4} \right) + 2gR \left( 1 - \cos \theta \right)\]
\[ \Rightarrow \cos \theta = \left( \frac{1}{4} + 2 - 2 \cos \theta \right)\]
\[ \Rightarrow 3 \cos \theta = \left( \frac{9}{4} \right)\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{3}{4} \right)\]