91影视

(a)

It is given that the mass of bird as $m=500\mathrm{g}$, the initial speed of the bird as $v_i=2.25\mathrm{rad}/\mathrm{s}$, the mass of the bar is $M=1.50\mathrm{kg}$, the length of the bar is $L=0.750\mathrm{m}$, the final speed of the bird as $v_f=0$ and the bird hit bar at 25.0cm below the top.

By angular momentum of the system, $L_i=L_f$ which implies $m{v}_i\left(L-0.250\right)=I{蝇}_f$ where $L-0.250$ is the distance between the bird and hinge.

Here, $m{v}_i\left(L-0.250\right)=I{蝇}_f$ can be written as, $m{v}_i\left(L-0.250\right)=\frac{1}{2}M{L}^2{蝇}_f$ then,

${蝇}_f=\frac{2m{v}_i\left(L-0.250\right)}{M{L}^2}$.

Substitute all the known values in ${蝇}_f$ and simplify.

$$\begin{gathered} {蝇}_f=\frac{3m{N}_j(L鈭�0.250)}{M{L}^2} \\ =\frac{3\left(0.50000\mathrm{kg}\right)(2.25\mathrm{m}/\mathrm{s})(0.750\mathrm{m}鈭�0.250\mathrm{m})}{\left(1.50\mathrm{kg}\right)(0.750\mathrm{m}{)}^2} \\ =\frac{\left(1.50000\mathrm{kg}\right)(2.25\mathrm{m}/\mathrm{s})\left(0.500\mathrm{m}\right)}{\left(1.50\mathrm{kg}\right)(0.750\mathrm{m}{)}^2} \\ =2.00\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the angular velocity after the hit by bird is ${蝇}_f=2.00\mathrm{rad}/\mathrm{s}$.

(b)

By conservation of energy principle $E_i+W=E_f$ then,

$Mg\left(\frac{L}{2}\right)+\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){蝇}_f^2=\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){蝇}_f^{\mathrm{鈥�}2}$.

Now, solve the above equation for $蝇{\text{'}}_f$.

$$\begin{gathered} {蝇}_f^{\mathrm{鈥�}2}=\frac{3\left(MgL+\frac{1}{3}M{L}^2{蝇}_f^2\right)}{M{L}^2} \\ {蝇}_f^{\mathrm{鈥�}}=\sqrt{\frac{3\left(MgL+\frac{1}{3}M{L}^2{蝇}_f^2\right)}{M{L}^2}} \\ =\sqrt{\frac{3ML\left(g+\frac{1}{3}L{蝇}_f^2\right)}{M{L}^2}} \\ =\sqrt{\frac{3g+L{蝇}_f^2}{L}} \end{gathered}$$

Substitute all the known values in $蝇{\text{'}}_f$ and simplify

$$\begin{gathered} 蝇{\text{'}}_f=\sqrt{\frac{3\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.750\mathrm{m}\right){\left(2.00\mathrm{rad}/\mathrm{s}\right)}^2}{0.750\mathrm{m}}} \\ =\sqrt{\frac{29.4\mathrm{m}/{\mathrm{s}}^2+\left(0.750\mathrm{m}\right){\left(2.00\mathrm{rad}/\mathrm{s}\right)}^2}{0.750\mathrm{m}}} \\ =6.57\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the angular velocity of the bar just as it reaches the ground is $蝇{\text{'}}_f=6.57\mathrm{rad}/\mathrm{s}$.