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 $\nu$. (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.

Figure

Answer 1

(a) Radius = R
Horizontal speed = $\nu$



From the above diagram:
Normal force,
$N=mg-\frac{m{v}^2}{R}$

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

$$\begin{gathered} \text{So,}\frac{m{\nu }^2}{\mathrm{R}}=mg \\ \Rightarrow \nu =\sqrt{g\mathrm{R}} \end{gathered}$$

(c) If the body is given velocity ${\nu }_1$ at the top such that,

$$\begin{gathered} {\nu }_1=\frac{\sqrt{g\mathrm{R}}}{2} \\ {\nu }_1^2=\frac{g\mathrm{R}}{4} \end{gathered}$$

Let the velocity be ${\nu }_2$ when it loses contact with the surface, as shown below.



So, $$\begin{gathered} \frac{m{\nu }_2^2}{\mathrm{R}}=mg\cos \mathrm{\theta } \\ \Rightarrow {\mathrm{\nu }}_2^2=\mathrm{Rg}\cos \mathrm{\theta }...\left(\mathrm{i}\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Again,}\left(\frac{1}{2}\right)m{\nu }_2^2-\left(\frac{1}{2}\right)m{\nu }_1^2 \\ =mg\mathrm{R}\left(1-\cos \mathrm{\theta }\right) \\ \Rightarrow {\nu }_2^2={\nu }_1^2+2g\mathrm{R}\left(1-\cos \mathrm{\theta }\right)...\left(ii\right) \end{gathered}$$

From equations (i) and (ii),

$$\begin{gathered} \mathrm{R}g\cos \mathrm{\theta }=\left(\frac{\mathrm{R}g}{4}\right)+2g\mathrm{R}\left(1-\cos \mathrm{\theta }\right) \\ \Rightarrow \cos \mathrm{\theta }=\left(\frac{1}{4}+2-2\cos \mathrm{\theta }\right) \\ \Rightarrow 3\cos \mathrm{\theta }=\left(\frac{9}{4}\right) \\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$