A 500.0-g bird is flying horizontally at 2.25 m/s, not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 cm below the top (Fig. P10.85). The bar is uniform, 0.750 m long, has a mass of 1.50 kg, and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterward (but soon recovers to fly happily away). What is the angular velocity of the bar (a) just after it is hit by the bird and (b) just as it reaches the ground?

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

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

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

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

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

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

(b)

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

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

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

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

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

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

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