Question

A thin rod of uniform mass m and length l is free to rotate at its upper end. When it is at rest, it receives an impulse J at the lowest point, normal to its length immediately after the impact. Then:
(A) Angular momentum of the rod is $ Jl $
(B) Angular velocity of the rod is $ \dfrac{{3J}}{{ml}} $
(C) Kinetic energy of the rod is $ \dfrac{{3{J^2}}}{{2m}} $
(D) Linear velocity of the midpoint of the rod is $ \dfrac{{3J}}{{2m}} $

Answer 1

Hint : A rod is a rigid body. In the question, a rod is free to rotate about its upper end. Hence, the upper end will be taken to be hinged or the point of rotation. Then we will simply apply the known formulas for rotational dynamics to get the answer. We should also keep in mind that the conservation principles have to be applied when an impulse acts on a rigid body.
The moment of inertia of a rod along the end of the rod is $ I = \dfrac{{m{l^2}}}{3} $
 $ K = \dfrac{1}{2}I{\omega ^2} $ where K is the kinetic energy, I is the moment of inertia and $ \omega $ is the angular velocity.

Complete Step By Step Answer:
 In the case of rigid bodies, the angular momentum is the product of the impulse and the length along which it acts.
Hence the angular momentum of the rod is $ Jl $
After the impact, the impulse will be equal to the change in the momentum.
 $ Jl = I\omega - 0 $
For a rod the moment of inertia is given as $ I = \dfrac{{m{l^2}}}{3} $
Further solving we get,
 $ Jl = \dfrac{{m{l^2}}}{3} \times \omega $
 $ \Rightarrow \omega = \dfrac{{3J}}{{ml}} $
Hence the angular velocity of the rod is $ \dfrac{{3J}}{{ml}} $
The kinetic energy is given by $ K = \dfrac{1}{2}I{\omega ^2} $
Substituting the values we get,
 $ K = \dfrac{1}{2} \times \dfrac{{m{l^2}}}{3} \times {(\dfrac{{3J}}{{ml}})^2} $
 $ \Rightarrow K = \dfrac{{3{J^2}}}{{2m}} $
The linear velocity is related to the angular velocity as $ v = \omega r $ where v is the linear velocity and r is the effective distance between the point and the axis.
Since it is the midpoint, $ r = \dfrac{l}{2} $ .
So, the linear velocity becomes $ v = \dfrac{{3J}}{{ml}} \times \dfrac{l}{2} $
 $ \Rightarrow v = \dfrac{{3J}}{{2m}} $
Hence, all four options are correct.

Note :
Inertia is the analog of mass in rotational dynamics. The moment of inertia is defined for a specific position from the center of mass and the axis of rotation. If the axis of rotation changes, the moment of inertia also changes, unlike mass which does not change. In every type of motion, the conservation principles always hold. So, we can use the conservation of energy and momentum to solve such problems.