Question

A simple pendulum of length $L$ having a bob of mass $m$ is deflected from its rest position by an angle $\theta$ and released (figure). The string hits a peg which is fixed at a distance $x$ below the point of suspension and the bob starts going in a circle centred at the peg.
(a) Assuming that initially the bob has height less than the peg, show that the maximum height reached by the bob equals its initial height
(b) If the pendulum is released with $\theta ={90}^{\circ }$ and $x=L/2$ find the maximum height reached by the bob above its lowest position before the string becomes slack
(c) Find the minimum value of $X/L$ for which the bob goes in a complete circle about the peg when the pendulum is released from $\theta ={90}^{\circ }$.

Answer 1

1. As LCM, the peg will not disturb the oscillation of bob. So, from energy conservation, $△KE=0$

There must not be change in potential energy.

2. Velocity when bob reaches the bottom point $v=\sqrt{2gl}$

Velocity required at bottom point to complete circle with radius $x$ is

$\sqrt{5gx}$ which is greater than, $\sqrt { 2gl }$ so it will not complete circle.

Let H be the height at which tension in the string will be zero,

$\frac{m{v}^2}{x}=mgsin\theta v=\sqrt{\frac{glsin\theta }{2}}$

Applying energy conservation,

$\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(\sqrt{2gl}\right)}^2=mg(x+xsin\theta )$

$sin\theta =\frac{2}{3}$

The value of height$x=x(1+sin\theta )$

$h=\frac{5L}{6}$

3. Velocity required at the lowest point to complete circle$=\sqrt { 5gx }$.

It must be equal to $\sqrt{2gl}$.

$\sqrt{5gx}=\sqrt{2gl}$

$\frac{x}{l}=\frac{2}{5}$