A pendulum is formed from a small ball of mass m on a string of length L. AsFIGURE $CP10.69$ shows, a peg is height$h=\frac{L}{3}$ above the pendulum’s lowest point. From what minimum angle u must the pendulum be released in order for the ball to go over the top of the peg without the string going slack?

Given a pendulum with a small ball of mass m.

Height of the peg,$h=\frac{L}{3}$

According to Newton's second law,

$F_{net}=ma$

$$\begin{gathered} F_{net}=F_G+T \\ \text{Acceleration}a=\frac{m{v}^2}{r} \\ \text{where}T\text{isthetensioninthestring} \end{gathered}$$

At the critical velocity $v_c$, the tension in the string will tend to $0$

Therefore the equation

$$\begin{gathered} mg=\frac{m{v}_c^2}{r} \\ {v}_c^2=gr \\ =\frac{gL}{3} \end{gathered}$$

According to the energy conservation equation, gravitational potential energy is transformed into the critical kinetic energy of the ball,

$$\begin{gathered} K_f+U_{gf}=K_i+U_{gi} \\ \frac{1}{2}m{v}_i^2+mg{y}_i=\frac{1}{2}m{v}_f^2+mg{y}_f \\ Since{v}_f=v_c \\ {v}_f=\frac{L}{3},y_f=\frac{L}{3}\text{andtheinitialvelocity}{v}_i=0 \\ mg{y}_i=\frac{1}{2}m{\left(\frac{L}{3}\right)}_{}^2+mg\left(\frac{L}{3}\right) \end{gathered}$$

$\frac{1}{2}m{v}_c^2=\frac{1}{2}mg\left(\frac{L}{3}\right)$

That is, the ball is a vertical distance$\frac{L}{2}$above the peg's location or a distance of$\frac{2L}{3}-\frac{L}{2}=\frac{L}{6}$

From the above figure, the minimum angle for the ball to go over the top of the peg:

$$\begin{gathered} \cos \left(\theta \right)=\frac{{\displaystyle \frac{L}{6}}}{L} \\ \cos \left(\theta \right)=\frac{1}{6} \\ \theta ={\cos }^{-1}\left(\frac{1}{6}\right) \\ =80.4° \end{gathered}$$