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. If the pendulum is released with $\theta ={90}^{\circ }$ and x = $\frac{L}{2}$ find the maximum height reached by the bob above its lowest position before the string becomes slack.

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

When the pendulum is released with $\theta ={90}^{\circ }$ and x = $\frac{L}{2}$, (figure) the path of the particle is shown in the figure.

At point C, the string will become slack and so the particle will start making projectile motion.

$\frac{1}{2}m{v}_c^2-0=mg\left(\frac{L}{2}\right)(1-cos\theta )$

Because, distance between A and C in the vertical direction is $\frac{L}{2}(1-cos\theta )$

$\Rightarrow v_c^2=gL(1-cos\theta )$------------(1)

Again, from the free body diagram (fig)

$\frac{m{v}^2}{\frac{L}{2}}mgcos\theta (because{T}_c=0)$

So, $v_c^2=\frac{gL}{2}cos\theta$ ---------------------(2)

From Eqn.(1) and eqn.(2),

$gL(1-cos\theta )=\frac{gL}{2}cos\theta$

$\Rightarrow 1-cos\theta =\frac{1}{2}cos\theta$

$\Rightarrow \frac{3}{2}cos\theta =1\Rightarrow cos\theta =\frac{2}{3}$ ----------------(3)

To find highest position C,before the string becomes slack

$BF=\frac{L}{2}+\frac{L}{2}cos\theta =\frac{L}{2}+\frac{L}{2}\times \frac{2}{3}=L(\frac{1}{2}+\frac{1}{3})$

So,BF = $\left(\frac{5L}{6}\right)$