Rotational Motion

Step 1: The angular momentum of a particle about the origin is given by the cross product:

where is the position vector and is the velocity vector.

Step 2: The magnitude of the angular momentum is given by:

where is the angle between and .

Step 3: Since is constant and the particle moves in a straight line, the angular momentum varies with , and the correct expression is:

1. Step 1: The ball is moving in a horizontal circle, which means that the forces acting on it in the vertical direction (the gravitational force mg) are balanced by the vertical component of the tension in the string. The ball is subject to two forces:
Gravitational force: mg and Tension in the string: T
2. Step 2: Since the ball is in circular motion, the horizontal component of the tension T sin θ provides the centripetal force required for the circular motion. The vertical component T cos θ balances the weight of the ball.
3. Step 3: The torque about the center O of the horizontal circle is given by the cross product of the force and the radius vector. The torque τ due to the forces acting on the ball is calculated as:
τ = r × F
where r is the radius vector (the length L) and F is the force (in this case, the tension). However, since the tension is acting along the string, there is no torque about the center of the horizontal circle.
4. Step 4: Therefore, the total torque about the center is zero, because the force creating the circular motion (tension in the string) does not create any rotational effect about the center of the horizontal circle.

1. Step 1: To find the velocity of the centre of mass, we use the formula:

   Where and are the velocities of the two ends of the rod, and and are their masses. Since the mass of the rod is uniform, .

2. Step 2: Given that the velocities of the ends of the rod are and , we can substitute these into the formula:

   Therefore, the velocity of the centre of mass is .

3. Step 3: The rotational kinetic energy and angular velocity can be found using the rotational dynamics of the rod. However, the key result from the question is the velocity of the centre of mass.

Conclusion: The velocity of the centre of mass is .