Question

Discuss the motion in vertical circle. Find an expression for the minimum velocities at the lowest point and top point. Also find tension at these points?

Answer 1

The motion of a Particle in Vertical Circle Whenever a body is released from a height, it travels vertically downward towards the surface of the earth.

This is due to the force of gravitational attraction exerted on the body by the earth.

The acceleration produced by this force is called acceleration due to gravity and is denoted by $‘g^{\prime }$. Value of $‘g^{\prime }$ on the surface of the earth is taken to the 9.8 $m/s^2$ and it is same for all the bodies.

It means all bodies (whether an iron ball or a piece of paper), when dropped (u = 0) from same height should fall with the same rapidity and should take the same time to reach the earth, it is the minimum velocity given to the particle at the lowest point to complete the circle.

The tendency of the string to become slack is maximum when the particle is at the topmost point of the circle.

At the top, tension is given by T = $\frac{m{v}_T^2}{R}$ - mg,

where $v_T$ = speed of the particle at the top.

$\frac{m{v}_T^2}{R}$ = $T+mg$

For $v_T$ to be minimum, $T=0$

$v_T=\sqrt{gR}$

If $V_B$ be the critical velocity of the particle at the bottom,

then from conservation of energy:

$Mg\left(2R\right)$ + $\frac{1}{2}m{v}_T^2$ = 0 + $\frac{1}{2}m{v}_B^2$

As $v_T=\sqrt{gR}$ > $2mgR$ + $\frac{1}{2}mgR$

=$\frac{1}{2}m{v}_B^2$

so $v_B=\sqrt{5gR}$

Highest point $H(h=2r)$

We have, $v$ = $\sqrt{u^2-2gh}$

$v$ = $\sqrt{(\sqrt{5gr}{)}^2-2g(2r)}$

$v$ = $\sqrt{5gr-4gr}$

= $\sqrt{gr}$

Tension at the lowest point:

$T-mg$ = $\frac{m{v}^2}{R}$.........(since,v = $\sqrt{5gr}$)

$T=6mg$

Tension at highest point:

$T+mg$ = $\frac{m{v}^2}{R}$.......(since, v = $\sqrt{gr}$)

$T=0$