Question

A pendulum, comprising a light string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. P8.68). (a) Show that if the sphere is released from a height below that of the peg. It will return to this height after the string strikes the peg. (b) Show that if the pendulum is released from rest at the horizontal position$\mathbf{(}\mathbf{\theta }\mathbf{=}{\mathbf{90}}^{\mathbf{o}}\mathbf{)}$and is to swing in a complete circle centered on the peg, the minimum value of d must be $\frac{\mathbf{3}\mathbf{L}}{\mathbf{5}}$.

Short answer

(a) If the sphere is released from a height below that of the peg. it will reach the same point from where it began.

(b) If the pendulum is released from rest at the horizontal position$(\mathrm{\theta }={90}^{\mathrm{o}})$ and is to swing in a complete circle centered on the peg, the minimum value of d must be $\mathrm{d}=\frac{3\mathrm{L}}{5}$.

Step-by-step solution

Step 1:

For an isolated system, the total mechanical energy of the system remains conserved.

${\mathbf{\Delta E}}_{\mathrm{mech}}\mathbf{=}\mathbf{0}$

This can be written as:

$\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}\mathbf{0}$

Here:

role="math" localid="1663596774612" $\mathbf{\Delta U}\mathbf{=}$Change in Potential energy

$\mathbf{\Delta K}\mathbf{=}$Change in kinetic energy

${\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}$Change in internal energy

Step 2

Part (a):

As no work will be done by the stationary peg, the total mechanical energy of the system must remain constant. So, there is no effect of the peg on the velocity of the ball, and it will reach the same point from where it started.

Part (b):

The ball will swing in a circle of radius$\mathrm{R}=(\mathrm{L}-\mathrm{d})$about the peg. For the ball to travel along the circular path, the minimum centripetal acceleration, at the top of the circle, should be equal to the acceleration due to gravity.

$\begin{array}{c}\frac{{\mathrm{mv}}^2}{\mathrm{R}}=\mathrm{g}\\ {\mathrm{v}}^2=\mathrm{g}(\mathrm{L}-\mathrm{d}).\end{array}$

When the ball is released from rest, ${\mathrm{U}}_{\mathrm{i}}=\mathrm{mgL}$, and when it is at the top of the circle,${\mathrm{U}}_{\mathrm{i}}=2\mathrm{mg}(\mathrm{L}-\mathrm{d})$, where the height is measured from the bottom of the swing, by energy conservation, we have:

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ {(\mathrm{K}+\mathrm{U})}_{\mathrm{i}}={(\mathrm{K}+\mathrm{U})}_{\mathrm{f}}\\ 0+\mathrm{mgL}=\mathrm{mg}2(\mathrm{L}-\mathrm{d})+\frac{1}{2}{\mathrm{mv}}^2\end{array}$

For ${\mathrm{v}}^2=\mathrm{g}(\mathrm{L}-\mathrm{d})$, the above equation becomes as shown below.

$\begin{array}{c}\mathrm{mgL}=\mathrm{mg}2(\mathrm{L}-\mathrm{d})+\frac{1}{2}\mathrm{mg}(\mathrm{L}-\mathrm{d})\\ \mathrm{mgL}=\left(2+\frac{1}{2}\right)\mathrm{mg}(\mathrm{L}-\mathrm{d})\\ \mathrm{L}=\frac{5}{2}\mathrm{L}-\frac{5}{2}\mathrm{d}\\ \frac{5}{2}\mathrm{d}=\frac{3}{2}\mathrm{L}\end{array}$

By solving further, we get:

$\mathrm{d}=\frac{3\mathrm{L}}{5}$

Hence, proved.