Question

On a smooth table two particles of mass $m$ each, travelling with a velocity $v_0$ in opposite directions, strike the ends of a rigid massless rod of length $l$, kept perpendicular to their velocity. The particles stick to the rod after the collision. Find the tension in rod during subsequent motion.

Answer 1

$\frac{2mv_0^2}{l}$

Answer 2

The problem involves a collision followed by rotational motion. We need to apply conservation laws during the collision and then analyze the forces during the subsequent motion.

1. Analysis of the Collision:

• System: Two particles (mass $m$ each) and a massless rigid rod (length $l$).
• Initial State: The two particles are moving with velocity $v_0$ in opposite directions, perpendicular to the rod. Let's assume the rod is along the x-axis, centered at the origin.
  • Particle 1 (at $x=l/2$): Velocity $\vec{v}_1 = -v_0 \hat{j}$ (downwards).
  • Particle 2 (at $x=-l/2$): Velocity $\vec{v}_2 = v_0 \hat{j}$ (upwards).
• Final State: The particles stick to the rod, forming a rigid body that rotates.

2. Conservation of Linear Momentum: The total initial linear momentum of the system is: $\vec{P}_{initial} = m\vec{v}_1 + m\vec{v}_2 = m(-v_0 \hat{j}) + m(v_0 \hat{j}) = \vec{0}$ Since there are no external forces acting on the system in the horizontal plane (smooth table), the total linear momentum is conserved. Thus, the final linear momentum must also be zero. This implies that the center of mass (CM) of the system remains at rest. The CM of the system (two particles + massless rod) is at the midpoint of the rod. Therefore, the system will only undergo rotational motion about its center.

3. Conservation of Angular Momentum: We conserve angular momentum about the center of mass (the origin).

• Initial Angular Momentum ($L_{initial}$): For particle 1 (at $\vec{r}_1 = (l/2)\hat{i}$): $\vec{L}_1 = \vec{r}_1 \times m\vec{v}_1 = \left(\frac{l}{2}\hat{i}\right) \times (m(-v_0)\hat{j}) = -\frac{mlv_0}{2}(\hat{i} \times \hat{j}) = -\frac{mlv_0}{2}\hat{k}$ For particle 2 (at $\vec{r}_2 = (-l/2)\hat{i}$): $\vec{L}_2 = \vec{r}_2 \times m\vec{v}_2 = \left(-\frac{l}{2}\hat{i}\right) \times (m(v_0)\hat{j}) = -\frac{mlv_0}{2}(\hat{i} \times \hat{j}) = -\frac{mlv_0}{2}\hat{k}$ The total initial angular momentum is: $\vec{L}_{initial} = \vec{L}_1 + \vec{L}_2 = -\frac{mlv_0}{2}\hat{k} - \frac{mlv_0}{2}\hat{k} = -mlv_0\hat{k}$ The magnitude is $L_{initial} = mlv_0$.

• Final Angular Momentum ($L_{final}$): After the collision, the two particles stick to the ends of the massless rod and rotate together. The moment of inertia of this system about its center of mass (the midpoint of the rod) is: $I = m\left(\frac{l}{2}\right)^2 + m\left(\frac{l}{2}\right)^2 = m\frac{l^2}{4} + m\frac{l^2}{4} = \frac{ml^2}{2}$ If the system rotates with an angular velocity $\omega$, the final angular momentum is: $L_{final} = I\omega = \left(\frac{ml^2}{2}\right)\omega$

• Conservation: Equating initial and final angular momenta: $mlv_0 = \frac{ml^2}{2}\omega$ Solving for $\omega$: $\omega = \frac{mlv_0}{ml^2/2} = \frac{2v_0}{l}$

4. Tension in the Rod: During the subsequent motion, each particle of mass $m$ is moving in a circular path of radius $r = l/2$ about the center of the rod with angular velocity $\omega$. The rigid rod provides the necessary centripetal force to keep the particles in circular motion. This force is the tension ($T$) in the rod. For one particle: $T = m r \omega^2$ Substitute $r = l/2$ and $\omega = \frac{2v_0}{l}$: $T = m \left(\frac{l}{2}\right) \left(\frac{2v_0}{l}\right)^2$ $T = m \left(\frac{l}{2}\right) \left(\frac{4v_0^2}{l^2}\right)$ $T = m \frac{4v_0^2 l}{2l^2}$ $T = \frac{2mv_0^2}{l}$

The tension in the rod during subsequent motion is $\frac{2mv_0^2}{l}$.

Explanation of the solution:

1. Linear Momentum Conservation: Initial linear momentum of the two particles is zero ($m\vec{v}_0 + m(-\vec{v}_0) = 0$). Since there are no external forces, the center of mass remains at rest.
2. Angular Momentum Conservation: Calculate the initial angular momentum of the two particles about the center of the rod: $L_{initial} = m v_0 (l/2) + m v_0 (l/2) = mlv_0$.
3. Moment of Inertia: After sticking, the moment of inertia of the system (two particles at ends of massless rod) about its center is $I = m(l/2)^2 + m(l/2)^2 = ml^2/2$.
4. Final Angular Velocity: By conservation of angular momentum, $L_{initial} = I\omega$, so $mlv_0 = (ml^2/2)\omega$. This gives $\omega = \frac{2v_0}{l}$.
5. Tension: Each particle undergoes circular motion with radius $l/2$. The tension in the rod provides the centripetal force: $T = m (l/2) \omega^2$. Substitute $\omega$ to get $T = m (l/2) (\frac{2v_0}{l})^2 = \frac{2mv_0^2}{l}$.