# Rotation of a system and conservation of angular momentum

Question:

Two small balls $A$ and $B$, each of mass $m$, are connected by a light rod of length $L$. The system is lying on a frictionless horizontal surface. A particle of mass $m$ collides with the ball $B$ horizontally with a velocity $v_o$ perpendicular to the rod and gets stuck to it. Find:

(a) the angular velocity of the system after collision.

(b) the velocities of $A$ and $B$ immediately after collision.

(c) the velocity of the center of the rod when the rod rotates through $90^\circ$ degree after the collision.

Doubt:

As the particle collides with $B$, the rod should rotate about its new center of mass that will be at a distance $L/3$ from $B$. The net torque of this system should be zero and therefore the angular momentum about the axis of rotation will be conserved. Therefore I got $\omega$ using this.

For the second part, the answer given is, $v_A=0\,,v_B=v_o/2$.

How is this possible unless the rod rotates about $A$?

Also, how do I find the third part?

Any help suggesting how this system moves will be appreciated.

## 1 Answer

For (b), obtain the velocity of the CM by momentum conservation. Then just use your answer in (a) to find the velocities relative to the CM, and then add the velocity of the CM, you get the answer of (b). Yes, it is instantaneously rotating about $A$.

For (c), it's similar to (b). The velocity is the vector sum of the CM velocity you used in (b) and the velocity relative to CM due to rotation about CM.

• Can you please tell why the rod will be instantaneously rotating about $A$? – Cheapstrike Jan 3 '17 at 20:00
• Because $A$ is instantaneously at rest. – velut luna Jan 3 '17 at 20:01
• The rod rotates about the new CM while the CM moves forward in a straight line. Relative to the CM the point A moves backwards (rotation). But the CM also moves forwards. These 2 motions just happen to cancel each other initially, so that A is instantaneously at rest relative to the ground. – sammy gerbil Jan 4 '17 at 0:43