Question

A uniform smooth rod (mass m and length l) placed on a smooth horizontal floor is hit by a particle (mass m) moving on the floor, at a distance $l/4$ from one end elastically (e =1). The distance travelled by the centre of the rod after collision when it has completed three revolutions will be

Answer 1

2πl

Answer 2

This problem involves an elastic collision between a particle and a uniform rod, followed by the motion of the rod. We need to apply the principles of conservation of linear momentum, conservation of angular momentum, and the definition of the coefficient of restitution to determine the velocities of the rod after the collision.

Let:

• Mass of the particle = $m$
• Initial velocity of the particle = $u$ (let's assume it moves along the positive x-axis)
• Mass of the rod = $m$
• Length of the rod = $l$
• Initial velocity of the rod = $0$
• Coefficient of restitution $e = 1$ (elastic collision)

After the collision:

• Velocity of the particle = $v_p$
• Translational velocity of the center of mass (CM) of the rod = $V_{CM}$
• Angular velocity of the rod = $\omega$

The collision occurs at a distance $l/4$ from one end. This means the distance from the center of mass of the rod to the point of impact is $r = l/2 - l/4 = l/4$. Let's assume the particle hits the rod on the side that causes a positive (counter-clockwise) angular velocity, so we take $r = l/4$.

1. Conservation of Linear Momentum:

The total linear momentum of the system (particle + rod) is conserved since there are no external horizontal forces.

Initial momentum = $mu$ Final momentum = $mv_p + mV_{CM}$

$mu = mv_p + mV_{CM}$

$u = v_p + V_{CM}$ --- (1)

2. Conservation of Angular Momentum about the CM of the rod:

The net external torque about the CM of the rod is zero during the collision.

Initial angular momentum = $m u r = m u (l/4)$ Final angular momentum = $m v_p r + I_{CM} \omega$

The moment of inertia of a uniform rod about its center of mass is $I_{CM} = \frac{ml^2}{12}$.

$m u (l/4) = m v_p (l/4) + \frac{ml^2}{12} \omega$

Dividing by $m$ and multiplying by $12/l$:

$3u = 3v_p + l\omega$ --- (2)

3. Coefficient of Restitution (e=1):

The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact.

Velocity of approach = $u - 0 = u$ Velocity of the point of contact on the rod just after collision = $V_{contact} = V_{CM} + r\omega = V_{CM} + (l/4)\omega$ Velocity of separation = $V_{contact} - v_p = (V_{CM} + (l/4)\omega) - v_p$

Since $e=1$:

$1 = \frac{V_{CM} + (l/4)\omega - v_p}{u}$

$u = V_{CM} + (l/4)\omega - v_p$ --- (3)

Now we solve the system of equations (1), (2), and (3) for $V_{CM}$ and $\omega$.

From (1), we can express $v_p$ as $v_p = u - V_{CM}$.

Substitute $v_p$ into (2):

$3u = 3(u - V_{CM}) + l\omega$

$3u = 3u - 3V_{CM} + l\omega$

$0 = -3V_{CM} + l\omega \implies l\omega = 3V_{CM}$ --- (4)

Substitute $v_p$ into (3):

$u = V_{CM} + (l/4)\omega - (u - V_{CM})$

$u = V_{CM} + (l/4)\omega - u + V_{CM}$

$2u = 2V_{CM} + (l/4)\omega$ --- (5)

Now we have two equations (4) and (5) with $V_{CM}$ and $\omega$.

From (4), $\omega = \frac{3V_{CM}}{l}$.

Substitute this expression for $\omega$ into (5):

$2u = 2V_{CM} + \left(\frac{l}{4}\right) \left(\frac{3V_{CM}}{l}\right)$

$2u = 2V_{CM} + \frac{3V_{CM}}{4}$

$2u = \frac{8V_{CM} + 3V_{CM}}{4}$

$2u = \frac{11V_{CM}}{4}$

$V_{CM} = \frac{8u}{11}$

Now, calculate $\omega$:

$\omega = \frac{3V_{CM}}{l} = \frac{3}{l} \left(\frac{8u}{11}\right) = \frac{24u}{11l}$

So, after the collision, the center of mass of the rod moves with a constant velocity $V_{CM} = \frac{8u}{11}$ and the rod rotates with a constant angular velocity $\omega = \frac{24u}{11l}$.

4. Distance traveled by the center of the rod:

The problem asks for the distance traveled by the center of the rod when it has completed three revolutions.

The total angular displacement for three revolutions is $\theta = 3 \times 2\pi = 6\pi$ radians.

The time taken to complete three revolutions is $T = \frac{\theta}{\omega}$.

$T = \frac{6\pi}{\frac{24u}{11l}} = \frac{6\pi \times 11l}{24u} = \frac{11\pi l}{4u}$

Since the floor is smooth, there are no horizontal forces acting on the rod after the collision, so its center of mass moves with a constant velocity $V_{CM}$.

The distance traveled by the center of the rod is $D = V_{CM} \times T$.

$D = \left(\frac{8u}{11}\right) \times \left(\frac{11\pi l}{4u}\right)$

$D = \frac{8 \times 11 \times \pi l \times u}{11 \times 4 \times u}$

$D = \frac{8\pi l}{4} = 2\pi l$

The distance traveled by the center of the rod after collision when it has completed three revolutions will be $2\pi l$.