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Step-by-Step Solution
Step 1: Identify the Key Physical Principles
The rod is pivoted at its center, and an external mass sticks to one of its ends after a collision. Since the pivot exerts no external torque about the center (ideal pivot), the total angular momentum of the system about this pivot is conserved during the collision.
Step 2: Calculate the Initial Angular Momentum
• Let the mass of the rod be $M = 4m$ and its length be $l$.
• The point of collision is at the rod’s end, which is at a distance $\frac{l}{2}$ from the pivot.
• The mass $m$ approaches with speed $v$ at an angle $\theta = \frac{\pi}{4}$ to the rod’s axis.
• Because the rod’s axis is taken as horizontal, the radius vector (from pivot to collision point) is horizontal, and the incoming velocity makes a $45^\circ$ angle to this horizontal.
• The component of the velocity perpendicular to the rod is $v \,\sin\left(\frac{\pi}{4}\right) = \frac{v}{\sqrt{2}}$.
Therefore, the initial angular momentum of mass $m$ about the center is:
$$L_{\text{initial}} = m \left(\frac{v}{\sqrt{2}}\right)\!\left(\frac{l}{2}\right)
= \frac{m\,v\,l}{2\sqrt{2}}.$$
Step 3: Calculate the Total Moment of Inertia After Collision
1. Moment of inertia of a uniform rod of mass $M=4m$ about its center:
$$I_{\text{rod}} = \frac{1}{12} M\,l^2 = \frac{1}{12} \times 4m \times l^2 = \frac{1}{3}m\,l^2.$$
2. Moment of inertia of the attached mass $m$ at the rod’s end (distance $\frac{l}{2}$ from the pivot):
$$I_{\text{mass}} = m \left(\frac{l}{2}\right)^2 = \frac{m\,l^2}{4}.$$
Hence, the total moment of inertia of the rod-plus-mass system about the pivot is:
$$I_{\text{total}} = I_{\text{rod}} + I_{\text{mass}}
= \frac{1}{3}m\,l^2 + \frac{1}{4}m\,l^2
= \left(\frac{4}{12} + \frac{3}{12}\right)m\,l^2
= \frac{7}{12}\,m\,l^2.$$
Step 4: Apply Conservation of Angular Momentum
Before collision, the angular momentum is $L_{\text{initial}} = \frac{m\,v\,l}{2\sqrt{2}}$.
After collision, the rod and the mass move together with some common angular velocity $\omega$. Hence,
$$L_{\text{final}} = I_{\text{total}}\;\omega = \left(\frac{7}{12}m\,l^2\right)\omega.$$
By conservation of angular momentum:
$$L_{\text{initial}} = L_{\text{final}} \quad \Longrightarrow \quad
\frac{m\,v\,l}{2\sqrt{2}} = \left(\frac{7}{12}m\,l^2\right)\omega.$$
Step 5: Solve for the Angular Velocity
Cancel the common factors $m$ and $l$ from both sides:
$$\frac{v}{2\sqrt{2}} = \left(\frac{7}{12}l\right)\omega
\quad \Longrightarrow \quad
\omega = \frac{\frac{v}{2\sqrt{2}}}{\frac{7}{12}l}
= \frac{v}{2\sqrt{2}} \times \frac{12}{7\,l}
= \frac{12}{2\sqrt{2}}\times \frac{1}{7}\times \frac{v}{l}
= \frac{6}{\sqrt{2}}\times \frac{1}{7}\times \frac{v}{l}
= \frac{3\sqrt{2}}{7}\,\frac{v}{l}.$$
This matches the given correct answer.
Final Answer
$ \omega = \frac{3\sqrt{2}}{7} \, \frac{v}{l}.
$
