To make a pendulum, a 300 gball is attached to one end of a string that has a length of 1.4 gand negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

Mass of the pendulum =300 g

Length of the string =1.4 m

String makes an angle of$30.0°$ with the vertical

The connection between angle(measured from vertical) and height h(measured from the lowest point, which is our choice of reference position in computing the gravitational potential energy) is given by $\mathit{h}\mathbf{=}\mathit{L}\left(1-cos\theta \right)$where Lis the length of the pendulum.

We use energy conservation in the form of Eq. 8-17.

$$\begin{gathered} K_1+U_1=K_2+U_2 \\ 0+mgL\left(1-\cos {\theta }_1\right)=\frac{1}{2}m{v}_2^2+,gL\left(1-\cos {\theta }_2\right) \end{gathered}$$

With $L=1.4m,{\theta }_1={30}^o,and{\theta }_2={20}^o$we have

$v_2=\sqrt{2gL(\cos {\theta }_2-\cos {\theta }_1}=1.4m/s$

The maximum speed is at the lowest point. Our formula for h gives$h_3=0$when${\theta }_3=0$, as expected. From

$$\begin{gathered} K_1+U_1=K_3+U_3 \\ 0+mgL\left(1-\cos {\theta }_1\right)=\frac{1}{2}m{v}_3^2+0 \end{gathered}$$

we obtain$v_3=1.0m/s$

We look for an angle such that the speed there is $v_3=v_4/3$.To be as accurate as possible, we proceed algebraically (substituting$v_3^2=2gL(1-cos{\theta }_1)$) at the appropriate place) and plug numbers in at the end. Energy conservation leads to

$$\begin{gathered} K_1+U_1=K_4+U_4 \\ 0+mgL\left(1-\cos {\theta }_1\right)=\frac{1}{2}m{v}_4^2+mgL\left(1-\cos {\theta }_4\right) \\ mgL\left(1-\cos {\theta }_1\right)=\frac{1}{2}m\left(\frac{v_3^2}{9}\right)+mgL\left(1-\cos {\theta }_4\right) \\ =gL\cos {\theta }_1=\frac{1}{2}\frac{2gL\left(1-\cos {\theta }_1\right)}{9}-gL\cos {\theta }_4 \end{gathered}$$

where in the last step we have subtracted outand then divided bym. Thus, we obtain

${\theta }_4={\cos }^{-1}\left(\frac{1}{9}+\frac{8}{9}\cos {\theta }_1\right)=28.2°\approx 28°$