Two balls of mass \(2m\) and \(m\) collide with a rod of mass \(m\) and length \(L\) as shown. The balls stick to the rod after collision. Find \( \dfrac{v}{\omega} \) if the rod is hinged at its centre.
(Given: \( L = 8\,\text{m} \))
Show Hint
For collision problems involving rotation:
Choose the hinge or pivot point to apply angular momentum conservation
Linear momentum may not be conserved, but angular momentum can be
Concept:
Since the rod is hinged at its centre, external torque about the hinge during collision is zero.
Hence, angular momentum about the hinge is conserved.
\[
\text{Initial angular momentum} = \text{Final angular momentum}
\]
Step 1: Calculate initial angular momentum about the hinge.
From the diagram:
Ball of mass \(2m\) strikes at distance \( \dfrac{L}{4} \)
Ball of mass \(m\) strikes at distance \( \dfrac{L}{2} \)
Both contribute angular momentum in the same sense.
\[
L_i = 2m \cdot v \cdot \frac{L}{4} + m \cdot v \cdot \frac{L}{2}
\]
\[
L_i = \frac{m v L}{2} + \frac{m v L}{2} = m v L
\]
Step 2: Calculate total moment of inertia after collision.
Moment of inertia of rod about centre:
\[
I_{\text{rod}} = \frac{1}{12} m L^2
\]
Moment of inertia of ball \(2m\) at \( \frac{L}{4} \):
\[
I_1 = 2m\left(\frac{L}{4}\right)^2 = \frac{mL^2}{8}
\]
Moment of inertia of ball \(m\) at \( \frac{L}{2} \):
\[
I_2 = m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{4}
\]
Total moment of inertia:
\[
I = mL^2\left(\frac{1}{12} + \frac{1}{8} + \frac{1}{4}\right)
\]
\[
I = mL^2\left(\frac{11}{24}\right)
\]
Step 3: Apply conservation of angular momentum.
\[
m v L = I \omega = \frac{11}{24} mL^2 \omega
\]
\[
\omega = \frac{24}{11}\frac{v}{L}
\]
Step 4: Find \( \dfrac{v}{\omega} \).
\[
\frac{v}{\omega} = \frac{11}{24} L
\]
Given \( L = 8 \,\text{m} \):
\[
\frac{v}{\omega} = \frac{11}{24} \times 8 = \frac{11}{3}
\]
\[
\boxed{\dfrac{v}{\omega} = \dfrac{11}{3}}
\]
