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 calculate in part,

(d) find the angle made by the radius.

Answer 1

(a )In the instant right after the impulse, the particle can be safely assumed to be in circular motion. Hence, it will require centripetal force, which will be provided by the resultant of $mg$ and $N$.

$\therefore m\frac{v^2}{R}=mg-N$

$\Rightarrow N=mg-m\frac{v^2}{R}$

(B) In the event that the particle doesn't slip on the surface of the sphere and instead flies off, the normal reaction $N$ is zero. Thus, the minimum velocity $v$ to prevent slipping would be given by:

$m\frac{v_{min}^2}{R}=mg$

$\Rightarrow v_{min}=\sqrt{Rg}$

(C) Let $v=\frac{v_{min}}{2}$, and final velocity in this case be $v_f$. Let the particle descend by an angle of $\theta$ before losing contact, and $h$ be the height of the initial point above the point at which it leaves contact.

Applying conservation of energy:

$\frac{1}{2}m{v}^2+mgh=\frac{1}{2}m{v}_f^2$

From the diagram, it can easily be deduced that $h=R(1-\cos \theta )$. Additionally by definition, $v=\frac{v_{min}}{2}=\frac{\sqrt{Rg}}{2}$.

$\Rightarrow \frac{1}{2}m\frac{Rg}{4}+mgR(1-\cos \theta )=\frac{1}{2}m{v}_f^2$

$\Rightarrow v_f^2=\frac{Rg}{4}(9-8\cos \theta )$

(D) And upon drawing the FBD of the particle at its final position and applying Newton's 2nd law along the radial direction, we get:

$mg\cos \theta -N=m\frac{v_f^2}{R}$

And since $N=0$ here, we get:

$v_f^2=Rg\cos \theta$

Substituting this value in equation obtained from conservation of energy:

$\frac{Rg(9-8\cos \theta )}{4}=Rg\cos \theta$

$\Rightarrow \cos \theta =\frac{3}{4}$

$\therefore \theta =co{s}^{-1}\left(\frac{3}{4}\right)$