91Ó°ÊÓ

(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 $m{v}_i\left(L-0.250\right)=I{\mathrm{Ó\neg }}_f$ where $L-0.250$ is the distance between the bird and hinge.

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

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

Substitute all the known values in ${\mathrm{Ó\neg }}_f$ and simplify.

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

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

(b)

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

$Mg\left(\frac{L}{2}\right)+\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){\mathrm{Ó\neg }}_f^2=\frac{1}{2}\left(\frac{1}{3}M{L}^2\right){\mathrm{Ó\neg }}_f^{\mathrm{′}2}$.

Now, solve the above equation for $\mathrm{Ó\neg }{\text{'}}_f$.

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

Substitute all the known values in $\mathrm{Ó\neg }{\text{'}}_f$ and simplify

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

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