Question: A uniform rod $AB$ of length $\ell$ and mass $M$ hangs from point $A$ at which it is freely hinged i...

Question

A uniform rod $AB$ of length $\ell$ and mass $M$ hangs from point $A$ at which it is freely hinged in a car moving with velocity $v_0$ . The rod can rotate in vertical plane about the axis at $A$ . If the car suddenly stops,

Answer 1

Solution

The problem describes a uniform rod of length $\ell$ and mass $M$ hinged at point A in a car moving with velocity $v_0$ . When the car suddenly stops, the hinge point A instantaneously comes to rest. We need to analyze the subsequent motion and energy changes.

1. Angular speed ( $\omega$ ) with which the rod starts rotating:

When the car suddenly stops, the hinge A's velocity changes from $v_0$ to $0$ . This is an impulsive event. During such an event, the angular momentum about the pivot point (A) is conserved because the impulsive force acts at A, and gravity is a non-impulsive force.

Initial state (just before stop): The rod is moving with a translational velocity $v_0$ . The velocity of its center of mass (CM), located at $\ell/2$ from A, is $v_0$ . The initial angular momentum about A is $L_i = M v_0 (\ell/2)$ .
Final state (just after stop): The rod starts rotating about A with an angular speed $\omega$ . The moment of inertia of a uniform rod about one end is $I_A = \frac{1}{3} M \ell^2$ . The final angular momentum about A is $L_f = I_A \omega = \frac{1}{3} M \ell^2 \omega$ .
Conservation of angular momentum: $L_i = L_f$

$M v_0 \frac{\ell}{2} = \frac{1}{3} M \ell^2 \omega$

$\frac{1}{2} v_0 = \frac{1}{3} \ell \omega$

$\omega = \frac{3 v_0}{2 \ell}$

Thus, statement (A) is correct.

2. Loss of energy during the process:

The "process" refers to the sudden stopping of the car, which causes the rod to transition from pure translation to pure rotation.

Initial kinetic energy (before stop): The rod has pure translational kinetic energy.

$K_i = \frac{1}{2} M v_0^2$ .
Final kinetic energy (just after stop): The rod has pure rotational kinetic energy about A.

$K_f = \frac{1}{2} I_A \omega^2 = \frac{1}{2} \left( \frac{1}{3} M \ell^2 \right) \left( \frac{3 v_0}{2 \ell} \right)^2$

$K_f = \frac{1}{6} M \ell^2 \frac{9 v_0^2}{4 \ell^2} = \frac{3}{8} M v_0^2$ .
Loss of energy: $\Delta E = K_i - K_f$

$\Delta E = \frac{1}{2} M v_0^2 - \frac{3}{8} M v_0^2 = \left( \frac{4}{8} - \frac{3}{8} \right) M v_0^2 = \frac{1}{8} M v_0^2$ .

Thus, statement (C) is correct, and statement (D) is incorrect.

3. Minimum value of $v_0$ for the rod to complete a full rotation:

After the car stops, the rod rotates about A under the influence of gravity. For the rod to complete a full rotation, its center of mass must reach the highest point (vertically above A) with at least zero angular velocity.

Initial state for this phase (just after car stops): The rod is vertically downwards. Its CM is at a depth $\ell/2$ below A. Let's set the potential energy $P=0$ at this position. The kinetic energy is $K_f = \frac{3}{8} M v_0^2$ . Total mechanical energy $E_1 = K_f + P_1 = \frac{3}{8} M v_0^2 + 0 = \frac{3}{8} M v_0^2$ .
Final state (rod at the highest point): The rod is vertically upwards. Its CM is at a height $\ell/2$ above A. The change in height of the CM from its initial position is $\Delta h = \ell$ . The potential energy at the highest point is $P_2 = Mg\ell$ . For minimum $v_0$ , the angular speed at the highest point $\omega_{top}$ can be $0$ . The kinetic energy at the highest point is $K_2 = \frac{1}{2} I_A \omega_{top}^2 = 0$ . Total mechanical energy $E_2 = K_2 + P_2 = 0 + Mg\ell = Mg\ell$ .
Conservation of mechanical energy: $E_1 = E_2$

$\frac{3}{8} M v_0^2 = Mg\ell$

$v_0^2 = \frac{8}{3} g\ell$

$v_0 = \sqrt{\frac{8}{3} g\ell}$ .

Thus, statement (B) is correct.

Since options (A), (B), and (C) are all correct based on our derivations, and the question format allows for multiple correct options, we select all of them.