A uniform bar of length \(12\,\text{cm}\) and mass \(20m\) lies on a smooth horizontal table. Two point masses \(m\) and \(2m\) are moving in opposite directions with the same speed \(v\) and in the same plane as the bar, as shown in the figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency \(\omega\). The ratio of \(v\) and \(\omega\) is
Show Hint
In collision problems on a smooth surface, linear momentum may not be conserved, but angular momentum is conserved if external torque is zero.
Since the table is smooth, there is no external torque acting on the system about the vertical axis.
Hence, angular momentum is conserved.
Step 1: Choose the axis.
We take the axis perpendicular to the plane of motion and passing through the center of mass of the system.
Step 2: Initial angular momentum.
From the figure, the distances of the point masses from the center of the bar are:
\[
r_1 = 2\,\text{cm}, \quad r_2 = 4\,\text{cm}.
\]
The initial angular momentum due to the two particles is:
\[
L_i = (2m)v(2) + (m)v(4) = 4mv + 4mv = 8mv.
\]
Step 3: Moment of inertia after collision.
Moment of inertia of the uniform bar about its center:
\[
I_{\text{bar}} = \frac{1}{12}(20m)(12)^2 = 240m.
\]
Moment of inertia of the point masses:
\[
I_1 = 2m(2)^2 = 8m, \quad I_2 = m(4)^2 = 16m.
\]
Total moment of inertia:
\[
I = 240m + 8m + 16m = 264m.
\]
Step 4: Apply conservation of angular momentum.
\[
L_i = I\omega
\]
\[
8mv = 264m\omega
\]
\[
\frac{v}{\omega} = \frac{264}{8} = 33.
\]
Final Answer:
\[
\boxed{33}
\]
