Question

A uniform rod of length \(L\) and mass \(3m\) is held vertically hinged at its base. A mass \(m\) moving horizontally with velocity \(v\) strikes the rod at the top and sticks to it. The angular velocity with which the rod hits the ground is:

• A. \(\sqrt{\dfrac{5g}{L} + \dfrac{v^2}{L^2}}\)
• B. \(\sqrt{\dfrac{5g}{2L} + \dfrac{v^2}{4L^2}}\)
• C. \(\sqrt{\dfrac{g}{2L} + \dfrac{v^2}{L^2}}\)
• D. \(\sqrt{\dfrac{g}{5L} + \dfrac{4v^2}{L^2}}\)

Hint:

For collision + rotation problems:
Use angular momentum conservation during collision
Use energy conservation after collision
Always choose hinge/contact point as reference for angular momentum

Answer 1

The Correct Option is A

Step 1: Angular momentum conservation at the instant of collision. Take moments about the hinge (no external impulsive torque about hinge). Initial angular momentum due to particle: \[ L_i = m v \cdot L \] Moment of inertia about the hinge after collision: \[ I = I_{\text{rod}} + I_{\text{particle}} \] For the rod: \[ I_{\text{rod}} = \frac{1}{3}(3m)L^2 = mL^2 \] For the particle stuck at the top: \[ I_{\text{particle}} = mL^2 \] \[ \Rightarrow I = 2mL^2 \] Applying angular momentum conservation: \[ m v L = (2mL^2)\omega_0 \] \[ \Rightarrow \omega_0 = \frac{v}{2L} \]
Step 2: Use energy conservation after collision. After collision, the system rotates and falls until it hits the ground. Initial rotational kinetic energy: \[ K = \frac{1}{2}I\omega_0^2 = \frac{1}{2}(2mL^2)\left(\frac{v}{2L}\right)^2 = \frac{mv^2}{4} \] Loss of gravitational potential energy:
Rod (mass \(3m\)), centre at \(L/2\): \[ \Delta U_{\text{rod}} = 3m g \cdot \frac{L}{2} = \frac{3}{2}mgL \]
Particle (mass \(m\)) at height \(L\): \[ \Delta U_{\text{particle}} = mgL \] Total loss: \[ \Delta U = \frac{3}{2}mgL + mgL = \frac{5}{2}mgL \]
Step 3: Final angular speed when rod hits the ground. Let final angular speed be \(\omega\). Total energy just before hitting ground: \[ \frac{1}{2}I\omega^2 = \frac{mv^2}{4} + \frac{5}{2}mgL \] Substitute \(I = 2mL^2\): \[ \frac{1}{2}(2mL^2)\omega^2 = \frac{mv^2}{4} + \frac{5}{2}mgL \] \[ mL^2\omega^2 = \frac{mv^2}{4} + \frac{5}{2}mgL \] Divide by \(mL^2\): \[ \omega^2 = \frac{v^2}{4L^2} + \frac{5g}{2L} \] \[ \omega = \sqrt{\frac{5g}{L} + \frac{v^2}{L^2}} \] Final Answer: \[ \boxed{\omega = \sqrt{\dfrac{5g}{L} + \dfrac{v^2}{L^2}}} \]