In Fig. 8-38, the string is L=120 cmlong, has a ball attached to one end, and is fixed at its other end. A fixed peg is at pointP. Released from rest, the ball swings down until the string catches on the peg; then the ball swings up, around the peg. If the ball is to swing completely around the peg, what value must distancedexceed? (Hint: The ball must still be moving at the top of its swing. Do you see why?)

Length of the string is, L =120 cm=125 m .

The problem is based on the principle of conservation of mechanical energy. According to this principle the energy can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature. By finding the radius of the circle and applying mechanical energy conservation we can find the velocityvof the ball. Then we have to find initial and final potential energy. Finally, applying conservation energy, we can find the value of exceeded distanced.

Formula:

The mechanical energy conservation,

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

The potential energy is,$U=mgL$

Note that the radius of circle is r=L-d.

Applying mechanical energy conservation, we get

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

Where,vis speed,mis the mass of the ball and r=L-d .

$T+mg=\frac{m{v}^2}{\left(L-d\right)}$

For the ball passes the highest point with the least speed, the tension is zero. Then we get,

$$\begin{gathered} mg=\frac{m{v}^2}{\left(L-d\right)} \\ \Rightarrow v^2=\frac{mg\left(L-d\right)}{m} \\ \Rightarrow v=\sqrt{g\left(L-d\right)} \end{gathered}$$

Here, initial potential energy is given by,

$U_i=mgL$

And initial kinetic energy is zero, since the ball starts from rest.

The final potential energy at the top oftheswing is given by

$U_f=mg(L-d)$

And final kinetic energy is given by,

$\frac{1}{2}m{v}^2=\frac{1}{2}mg(L-d)$

Therefore, using equation for, and conservation of energy, we get,

$mgL=2mg(L-d)+\frac{1}{2}/mg(L-d)$

$$\begin{gathered} \Rightarrow L=\frac{5}{2}\left(L-d\right) \\ \Rightarrow d=\frac{6}{10}L \\ \Rightarrow d=\frac{3}{5}L \end{gathered}$$

With, L=1.20 m solve for the distance as:

$$\begin{gathered} \Rightarrow d=\frac{3}{5}\times 1.20 \\ \Rightarrow d=0.72\mathrm{m} \end{gathered}$$