# Conservation of Angular Momentum about the point of collision

A uniform rod AC of length l and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity v strikes the rod at point B making an angle 37 degree with the rod. The collision is elastic. I am trying to solve the problem by conserving angular momentum of rod, linear momentum of rod and particle, and energy conservation.

My question is

1. Can the angular momentum of Rod be conserved about point B although the point B is accelerated during collision?
2. Does elastic collision mean that the Velocity of approach = velocity of seperation about Common Normal through point B?
• Why do you want to look at the angular momentum about point B? Elastic collision usually means that the kinetic energy is conserved. May 7, 2017 at 22:58
• I want to look at angular momentun about point B as it will give relation with 2 variables ( Vel. Of COM, angular velocity) only 2 unknown, Can the angular momentum about B be conserved for rod? May 7, 2017 at 23:00

Can the angular momentum of Rod be conserved about point B although the point B is accelerated during collision?

If you mean B as a point in an inertial frame, then yes. If you mean B as a point attached to the rod, then no. You're correct that if the axis under consideration is accelerating then it makes things difficult. In general, you don't want to do this unless the accelerating axis contains the center of mass of the object you are considering to rotate.

Does elastic collision mean that the Velocity of approach = velocity of seperation about Common Normal through point B?

It means that the total mechanical energy of the system (all kinetic in this case) is the same before and after the collision. In a frame where the center of mass is at rest, then the velocities before and after will be the same. In other frames, it may not be. A golf ball and a golf club may have a nearly elastic collision, but the ball will have a very different velocity before and after being struck on a tee.

What is conserved in the linear momentum about the line of action of the impact (path of ball here). There is an exchange of momentum ${\bf p}$ along this line which results in a change in angular momenum about the respective centers of mass.

If the velocities at the center of mass are ${\bf v}_1$ and ${\bf v}_2$ and the vectors from the contact point to each center of mass ${\bf c}_1$ and ${\bf c}_2$ then the conservation of momentum changes the velocities by

\begin{align} \Delta \mathbf{v}_1 &= -\frac{\mathbf{p}}{m_1} & \Delta \mathbf{v}_2 &= +\frac{\mathbf{p}}{m_2} \\ \Delta \boldsymbol{\omega}_1 & =- \mathrm{I}_1^{-1} (\mathbf{p} \times \mathbf{c}_1) & \Delta \boldsymbol{\omega}_2 & =+ \mathrm{I}_2^{-1} (\mathbf{p} \times \mathbf{c}_2) \end{align}

Total angular momentum is not conseved in general, unless it is measured on the contact point (with zero ang. momentum there).

• So we can conserve angular momentum of rod about contact point? Can it be considered inertial frame ( accelerated due to collision?)? May 8, 2017 at 5:29
• Is the equation of coefficient of restitution valid for non central collision ( Vel. Approach = Vel. Of Seperation) May 8, 2017 at 5:31
• Yes. if ${\bf n}$ is the direction of the contact force then impact velocity is $v_{imp} = \mathbf{n}\cdot ( \mathbf{v}_2^B - \mathbf{v}_1^B )$ The law of collision is valid and it is expressed as $$\mathbf{n}\cdot ( \Delta \mathbf{v}_2^B - \Delta \mathbf{v}_1^B ) = -(1+\epsilon) v_{imp}$$ May 8, 2017 at 5:40
• No, the angular momentum about the contact point will not change, because the contact line of action is through B. May 8, 2017 at 5:43