A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses $m$

Question

A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses $m$ and $2 m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in 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 :

Professor PC Thomas Classes · Answer

1. Locate the Center of Mass (COM)
First, we find the new center of mass of the system after the masses stick to the bar. Let the center of the bar be the origin $(x=0)$.
- Mass $2 m$ is at $x=-2 \mathrm{~cm}$
- Mass $m$ is at $x=+4 \mathrm{~cm}$
$
X_{c o m}=\frac{20 m(0)+2 m(-2)+m(4)}{20 m+2 m+m}=\frac{-4 m+4 m}{23 m}=0
$
The COM of the entire system remains exactly at the center of the bar.
2. Calculate the Moment of Inertia (I)
Next, we find the total moment of inertia of the system about this central axis.
- $I_{\mathrm{bar}}=\frac{M L^2}{12}=\frac{20 m \cdot(12)^2}{12}=240 m$
- $I_{2 m}=2 m \cdot(2)^2=8 m$
- $I_m=m \cdot(4)^2=16 m$
- Total $I=240 m+8 m+16 m=\mathbf{2 6 4 m}$
3. Conserve Angular Momentum
Since no external torque acts on the system, the initial angular momentum about the COM equals the final angular momentum ( $L_i=L_f$ ).
- Initial Angular Momentum ( $L_i$ ): Both masses create a torque in the same rotational direction.
$
L_i=(2 m \cdot v \cdot 2)+(m \cdot v \cdot 4)=4 m v+4 m v=\mathbf{8 m v}
$
- Final Angular Momentum ( $L_f$ ):
$
L_f=I \omega=264 \mathrm{~m} \omega
$
Equating the two:
$
\begin{aligned}
& 8 m v=264 m \omega \
& \frac{v}{\omega}=\frac{264}{8}=33
\end{aligned}
$