Thapar University - Sem 2 - Mechanics

A particle of mass m is at rest A when it is slightly displaced and allowed to slide down the cylindrical surface of radius r (Fig.4). Neglecting friction, find the value of $\theta$ when the particle leaves the surface.

Two positions of the particle are shown. For convenience, the datum, which is horizontally fixed, passes through position 1.

When the particle is in position 1

$\hspace{0.5cm}$ Potential energy of particle $V_1=0$,

And

$\hspace{0.5cm}$ Kinetic energy of particle $T_1=0.$

When the particle is in position 2

$\hspace{0.5cm}$ potential energy of particle $V_2=-mg(r-r \cos \theta)$

and

$\hspace{0.5cm}$ Kinetic energy of particle $T_2=\bigg(\dfrac12\bigg)mv^2,$

When $v$ is the velocity of particle at angle $\theta$ Applying the principle of conversion of energy, $$T_1+V_1=T_2+V_2$$

$0+0=\dfrac12mv^2-mg(r-r \cos \theta)$

$\hspace{0.5cm} V^2=2gr (1-\cos \theta)$

$\boxed{v=\sqrt{2gr(1-\cos\theta)}}$

Fig shows the FBD of the particle. Resolving forces along the normal to the surface

$\sum F_n=0\rightarrow R+\dfrac{mv^2}{r}-mg \cos\theta=0 \\ R=mg \cos \theta -\dfrac{mv^2}{r}$

Particle leaves the surfaces if R=0, therefore

$0=mg \cos \theta- \dfrac{mv^2}{r} \\ v^2=gr \cos \theta$

From Equations (i) & (ii) , we have

$\hspace{11cm}2gr(1- \cos \theta)=gr \cos \theta$

OR $\hspace{10.5cm}2g-2g \cos \theta=g \cos \theta$

OR $\hspace{12cm}$ $\theta=\cos^{-1}\dfrac23 $

$\hspace{13cm}$ $\boxed{\theta=48.18^0}$