Question

A nail is located at a certain distance vertically below the point of suspension of a simple pendulum. The pendulum bob is released from the position where the string makes an angle of ${60}^{\circ }$ from the vertical. The distance of the nail from a point of suspension is $\frac{x}{10}$ such that the bob will just perform revolution with the nail as center. Assume the length of the pendulum to be $1m$. Find the value of $x$.

Answer 1

$\text{Step 1: Minimum Speed at point P}$

Suppose distance of nail from point of suspension is y

After reaching point P the bob with undergo in a circle of radius $R=(l-y)$

For just performing complete revolution

Speed at lowest point P, $v_p=\sqrt{5gR}$ (Result to be remembered)

So, $v_p=\sqrt{5g(l-y)}$ $....\left(1\right)$

$\text{Step 2:Work energy theorem from point A to P}$

Work done by gravity from point $A$ to $P=mgh$

From figure $h=l-lcos\theta$

So, $W_{gravity}=mgl(1-\cos {60}^o)$ $=\frac{mgl}{2}$ $....\left(2\right)$

By work energy theorem:

$W_{ext}+W_{int}=\Delta K.E.$

Here, Work done by Tension force is zero, as Tension is always perpendicular to speed. Only gravity is doing work.

Therefore, $\Delta KE=W_{gravity}$

$\Rightarrow \frac{1}{2}m(v_p{)}^2-\frac{1}{2}m(0{)}^2=\frac{mgl}{2}$ (Using Equation $2$)

$\Rightarrow 5g(l-y)=gl$ (putting value of $v_p$ from Equation $1$)

$\Rightarrow y=\frac{4l}{5}$

and $l=1m$

So, $y=\frac{4}{5}=\frac{8}{10}$

Hence, $x=8$