A uniform rod of length L and mass 2m rests on a smooth horizontal table. A point mass m moving horizontally at right angle to the rod with an initial velocity v collides with one end of the rod and sticks to it. Determine

(a) the rotational inertia I about the axis passing through the center of mass after the collision,

(b) the angular velocity of the system after the collision,

(c) the position of the point on the rod which remains stationary immediately after the collision, and

(d) the change in kinetic energy of the system as a whole as a result of the collision

Sol90:

Before the collision, the rod is at rest, and the point mass is moving with a velocity v.

(a) The rotational inertia of a uniform rod about its center of mass is I = (1/12)ML^2. Since the rod is at rest initially, its center of mass is not moving, so the axis of rotation passes through the center of mass. The point mass sticks to one end of the rod, so the new system consists of a rod-point mass combination rotating about the center of mass of the rod. The total mass of the system is 3m, and the distance of the point mass from the center of mass of the rod is L/2. Therefore, the rotational inertia of the system is:

I = (1/12)ML^2 + mL^2/4 = (1/12)(2m)L^2 + (m/4)L^2 = (5/24)mL^2

(b) The angular momentum of the system is conserved during the collision, since there are no external torques acting on the system. Initially, the point mass has an angular momentum of mvL/2 about the center of mass of the rod, and the rod has zero angular momentum about its center of mass. After the collision, the system rotates about the center of mass of the rod, and the angular momentum is:

L = I? = (5/24)mL^2?

Conservation of angular momentum gives:

mvL/2 = (5/24)mL^2?

? = mvL/(12mL) = v/12

Therefore, the angular velocity of the system after the collision is ? = v/12.

(c) The point on the rod which remains stationary immediately after the collision is the one closest to the center of mass of the rod. Let x be the distance of this point from the center of mass. Since the rod is uniform, the center of mass is at the midpoint, so x = L/4. Conservation of momentum in the horizontal direction gives:

mv = (2m + m)V

where V is the velocity of the rod-point mass combination immediately after the collision. Therefore,

V = v/3

The velocity of the center of mass of the rod-point mass combination is also V. At the instant when the point on the rod closest to the center of mass is stationary, its velocity relative to the center of mass is zero. Therefore, the velocity of the center of mass is entirely due to the motion of the rest of the system, which is rotating about the center of mass of the rod. The speed of a point on the rod at a distance r from the center of mass is given by:

v(r) = r?

At the point closest to the center of mass, r = L/4, so:

v(L/4) = (L/4)(v/12) = v/48

Therefore, the velocity of the center of mass is:

V = v/3 + v/48 = 17v/48

(d) The kinetic energy of the system before the collision is (1/2)(2m)v^2 = 2mv^2. After the collision, the kinetic energy is (1/2)I?^2, where I is the rotational inertia of the system and ? is the angular velocity. Substituting the values found in parts (a) and (b), we get:

K = (1/2)I?^2 = (1/2)(5/24)mL^2(v/12)^2 = mv^2/288

Therefore, the change in kinetic energy is:

?K = K - 2mv