91Ó°ÊÓ

Momentum of inertia of the uniform bar about the center

$I=\frac{m{\mathrm{ι}}^2}{12}$

Here, mass of the bar $=m$

Length of the bar= $l$

EXECUTIVE:

Length of the bar,

localid="1667799587809" $$\begin{gathered} \mathrm{ι}=80.0\mathrm{cm} \\ =(80.0 \mathrm{cm})\frac{1×{10}^2}{1\mathrm{cm}} \\ =0.8\mathrm{m} \end{gathered}$$

Mass of the bar,localid="1667799591543" $M=3.80\mathrm{kg}$

Mass of each ball, localid="1667799596757" $\mathrm{m}=2.50\mathrm{kg}$

Moment of inertia of the bar with a small ball glues at one end about the axis passing through centre of the bar and perpendicular is given by

$I\text{'}=\frac{M{\mathrm{ι}}^2}{12}+m{\left(\frac{1}{2}\right)}^2............\left(1\right)$

Now, torque at one end of the bar after ball fallen from other end is given by

$I\text{'}\mathrm{Î\pm }=mg\frac{1}{2}..................\left(2\right)$

By using the equations (1) and (2),

We get angular acceleration after fall of the mass as follows.

Angular acceleration is,

$$\begin{gathered} \mathrm{Î\pm }=\frac{6mg}{(M\mathrm{ι}+3ml)} \\ =\frac{6(2.50kg)(9.80 m/s^2)}{\left[\right(3.80kg\left)\right(0.8m)+3(2.50kg\left)\right(0.8\left)\right]} \\ =\boxed{16.3 rad/s^2} \end{gathered}$$

Hence, the angular acceleration is $16.3rad/s^2$.

No, the angular acceleration cannot remain constant.

The angular acceleration with decreases as the continue to swing, this is because after some swing, the gravitational force due to small ball will not stay perpendicular on the end of the bar. So, the torque on the bar will decreases continuously.

Using the conservation of energy law gravitational energy change of the small ball at vertical position of the bar will converts to rotational kinetic energy of the system.

$$\begin{gathered} \frac{1}{2}I\text{'}{\mathrm{Ó\neg }}^2=mg\left(\frac{1}{2}\right) \\ \left[M{\mathrm{ι}}^2+m{\left(\frac{1}{2}\right)}^2\right]{\mathrm{Ó\neg }}^2=mg\mathrm{ι} \\ {\mathrm{Ó\neg }}^2=\frac{12mg}{(M\mathrm{ι}+3m\mathrm{ι})} \\ \mathrm{Ó\neg }=\sqrt{\frac{12mg}{(M\mathrm{ι}+3m\mathrm{ι})}} \end{gathered}$$

By substituting the values in the above equation, we get

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt{\frac{12(2.50kg)(9.80m/s^2)}{\left[\right(3.80kg\left)\right(0.8m)+3(2.50kg\left)\right(0.8m\left)\right]}} \\ =\boxed{5.7 rad/s} \end{gathered}$$

Hence, the angular velocity of the bar is$5.7rad/s$.