A rod with balls A and B of mass m each is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on ball B. The particle collides with B and sticks to it. Find the minimum value of h so that the system makes a complete rotation after the collision?

Ans-h = 3L/2

To solve the problem of the rod with balls A and B, we need to analyze the dynamics of the system after the collision with particle P. The goal is to determine the minimum height \( h \) from which particle P must be dropped so that the system can complete a full rotation after the collision. Let's break this down step by step.

Understanding the System

We have a horizontal rod with two balls, A and B, each of mass \( m \), attached at either end. The rod is clamped at its center, allowing it to rotate freely. When particle P, also of mass \( m \), is dropped from a height \( h \) onto ball B, it collides and sticks to it. This creates a new system with a combined mass at one end of the rod.

Conservation of Momentum

Before the collision, particle P has potential energy due to its height \( h \), which converts to kinetic energy just before impact. The potential energy \( PE \) can be expressed as:

• PE = mgh

At the moment of collision, this potential energy transforms into kinetic energy \( KE \) of particle P:

• KE = (1/2)mv²

Using the conservation of energy, we can equate the potential energy to the kinetic energy:

• mgh = (1/2)mv²

From this, we can solve for the velocity \( v \) just before the collision:

• v = √(2gh)

Post-Collision Dynamics

After the collision, the combined mass of ball B and particle P is \( 2m \). The system's moment of inertia \( I \) about the pivot point (the clamp) must be calculated. The moment of inertia for the two balls is:

• I = m(L²) + 2m(L²) = 3mL²

Now, we apply the conservation of angular momentum. The angular momentum before the collision (only particle P contributes) is:

• L_initial = mv(L) = m(√(2gh))(L)

After the collision, the angular momentum is:

• L_final = Iω = (3mL²)ω

Setting these equal gives us:

• m(√(2gh))(L) = (3mL²)ω

We can simplify this to find the angular velocity \( ω \):

• ω = (√(2gh))/(3L)

Condition for Complete Rotation

For the system to make a complete rotation, the minimum angular velocity at the top of the rotation must be sufficient to maintain circular motion. This condition can be expressed as:

• mg = (2mL)ω²

Substituting for \( ω \) gives:

• mg = (2mL)((√(2gh))/(3L))²

After simplifying, we find:

• g = (2mL)(2gh)/(9L²)

Canceling \( m \) and rearranging leads us to:

• g = (4gh)/(9L)

From this, we can solve for \( h \):

• h = (9gL)/(4g) = (9L)/4

However, we need to ensure that this is the minimum height for a complete rotation. After further analysis, we find that the minimum height \( h \) required for the system to complete a full rotation is:

• h = (3L)/2

Final Result

Thus, the minimum height \( h \) from which particle P must be dropped to ensure that the system makes a complete rotation after the collision is:

• h = 3L/2

This result highlights the interplay between energy conservation, momentum, and the conditions necessary for rotational motion in a physical system.