A thin uniform rod of mass m and length l is free to rotate about its upper end.When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately after impact

a-the angular momentum of the rod is Jl

b-the angular velocit of the rod is 3J/ml

c-the kinetic energy of the rod is 3J^2/2m

d-the linear velocity of the mid-point of the rod is 3J/2m

To analyze the situation of a thin uniform rod receiving an impulse at its lowest point, we need to apply the principles of angular momentum, rotational motion, and kinetic energy. Let's break down the problem step by step to determine the correct statements regarding the angular momentum, angular velocity, kinetic energy, and linear velocity of the rod after the impulse is applied.

Understanding Angular Momentum

When the impulse \( J \) is applied at the lowest point of the rod, it generates an angular momentum about the pivot point (the upper end of the rod). The angular momentum \( L \) can be calculated using the formula:

• Angular Momentum: \( L = r \times p \)

Here, \( r \) is the distance from the pivot to the point where the impulse is applied (which is the length \( l \) of the rod), and \( p \) is the linear momentum imparted by the impulse. The impulse \( J \) is equal to the change in momentum, so:

• Linear Momentum: \( p = J \)

Thus, the angular momentum immediately after the impulse is:

• Angular Momentum: \( L = l \cdot J \)

This confirms that statement (a) is correct: the angular momentum of the rod is \( Jl \).

Calculating Angular Velocity

Next, we need to find the angular velocity \( \omega \) of the rod after the impulse. The moment of inertia \( I \) of a thin rod rotating about one end is given by:

• Moment of Inertia: \( I = \frac{1}{3}ml^2 \)

Using the relationship between angular momentum and angular velocity, we have:

• Angular Momentum: \( L = I \cdot \omega \)

Substituting the expressions we have:

• Angular Momentum: \( Jl = \frac{1}{3}ml^2 \cdot \omega \)

Solving for \( \omega \), we get:

• Angular Velocity: \( \omega = \frac{3J}{ml} \)

This means statement (b) is also correct: the angular velocity of the rod is \( \frac{3J}{ml} \).

Determining Kinetic Energy

The kinetic energy \( K \) of a rotating object can be calculated using the formula:

• Kinetic Energy: \( K = \frac{1}{2} I \omega^2 \)

Substituting the moment of inertia and the angular velocity we found earlier:

• Kinetic Energy: \( K = \frac{1}{2} \cdot \frac{1}{3}ml^2 \cdot \left(\frac{3J}{ml}\right)^2 \)

Calculating this gives:

• Kinetic Energy: \( K = \frac{1}{2} \cdot \frac{1}{3}ml^2 \cdot \frac{9J^2}{m^2l^2} = \frac{3J^2}{2m} \)

This confirms that statement (c) is correct: the kinetic energy of the rod is \( \frac{3J^2}{2m} \).

Finding Linear Velocity of the Mid-Point

Finally, to find the linear velocity \( v \) of the mid-point of the rod, we can use the relationship between linear velocity and angular velocity:

• Linear Velocity: \( v = r \cdot \omega \)

For the mid-point of the rod, \( r \) is \( \frac{l}{2} \), so:

• Linear Velocity: \( v = \frac{l}{2} \cdot \frac{3J}{ml} = \frac{3J}{2m} \)

This confirms that statement (d) is also correct: the linear velocity of the mid-point of the rod is \( \frac{3J}{2m} \).

Summary of Results

To summarize, all four statements regarding the rod after receiving the impulse are correct:

• Angular momentum: \( Jl \)
• Angular velocity: \( \frac{3J}{ml} \)
• Kinetic energy: \( \frac{3J^2}{2m} \)
• Linear velocity of the mid-point: \( \frac{3J}{2m} \)

This analysis illustrates the interconnectedness of linear and angular quantities in rotational dynamics, showcasing how an impulse can affect a rigid body in motion.