91影视

Mass of the pendulum =300 g

Length of the string =1.4 m

String makes an angle of$30.0掳$ 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 $\mathit{h}\mathbf{=}\mathit{L}\left(1-cos胃\right)$where Lis the length of the pendulum.

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

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

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

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

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

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

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$v_3^2=2gL(1-cos{胃}_1)$) at the appropriate place) and plug numbers in at the end. Energy conservation leads to

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

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

${胃}_4={\cos }^{-1}\left(\frac{1}{9}+\frac{8}{9}\cos {胃}_1\right)=28.2掳鈮�28掳$