91影视

It is given that pole length as L=6.00m , mass of the pole as M = 24.00kg, mass of the person as m =70.00kg and swing angle as $胃=72.0掳$.

By angular momentum of the system, $L_1=L_2$. Substitute $L_1=mvL$and $L_2=I蝇$then,

localid="1667828535645" $$\begin{gathered} mvl=I蝇 \\ m\sqrt{2ghL}=(I_{person}+I_{pole})蝇 \\ {m}^{2}2gh{L}^2={(I_{person}+I_{pole})}^2{蝇}^2 \\ h=\frac{{(I_{person}+I_{pole})}^2{蝇}^2}{2g{m}^2{L}^2} \end{gathered}$$

Thus, the height of the pole is localid="1667830093573" $h=\frac{{(I_{person}+I_{pole})}^2{蝇}^2}{2g{m}^2{L}^2}$鈥︹�� (1)

The final height of the person is given by $h^{\text{'}}$as,

$h^{\text{'}}=L-L\cos 胃$

$=L\left(1-\cos 胃\right)$

Then the final gravitational potential energy of the person is $mg{h}^{\text{'}}=mgL(1-cos胃)$and the pole is$Mg\frac{L}{2}-Mg\frac{L}{2}\cos 胃=Mg\frac{L}{2}(1-\cos 胃)$.

The final gravitational potential energy from kinetic energy can be written as follows:

$$\begin{gathered} \left(1-\cos 胃\right)+Mg\frac{L}{2}\left(1-\cos 胃\right)=\frac{1}{2}(I_{person}+I_{pole}){蝇}^2 \\ gL\left(1-\cos 胃\right)+\left(m+\frac{M}{2}\right)=\frac{1}{2}(I_{person}+I_{pole}){蝇}^2 \\ gL\left(1-\cos 胃\right)+\left(2m+M\right)=\frac{1}{2}(I_{person}+I_{pole}){蝇}^2 \\ {蝇}^2=\frac{gL\left(1-\cos 胃\right)+\left(2m+M\right)}{(I_{person}+I_{pole})} \end{gathered}$$

Substitute all the values in (1) and simplify.

$$\begin{gathered} h=\frac{(I_{person}+I_{pole})}{2g{m}^2{L}^2}\frac{gL\left(1-\cos 胃\right)\left(2m+M\right)}{(I_{person}+I_{pole})} \\ =\frac{(I_{person}+I_{pole})\left(1-\cos 胃\right)\left(2m+M\right)}{2m^{2}L} \end{gathered}$$

Replace $I_{person}=m{L}^{2}and{I}_{pole}=\frac{1}{3}M{L}^2$and simplify.

localid="1667829790484" $$\begin{gathered} h=\frac{\left(m{L}^2+\frac{1}{3}M{L}^2\right)(1-\cos 胃)(2m+M)}{2m^{2}L} \\ =\frac{(1-\cos 胃)(2m+M)(3m+M)}{6m^2} \end{gathered}$$

Substitute all the given values and find h.

$$\begin{gathered} h=\frac{\left(6\right)\left(1-\cos {72}^{掳}\right)\left(2\left(70\right)+24\right)\left(3\left(70\right)+24\right)}{6{\left(70\right)}^2} \\ =\left(6\right)\left(1+0.97\right)\left(164\right)\left(234\right) \\ =\frac{460512}{29400} \\ =5.41\mathrm{m} \end{gathered}$$

Therefore, the height of the pole to have maximum angle of swing $70.0^{掳}$after collision is 5.41m.