# Moment of inertia in a perfectly inelastic collision

Imagine a rod rotating about one of its ends colliding in a perfectly inellastic way with a ball hanging in the pivot (rotational axis). The rod goes towards the ball (initially stopped). Angular momentum is conserved due to the fact that the torque exerted on the ball by the rod is equal and opposite to the torque exerted on rod by the ball.

Before the collision we all agree with the idea that moment of inertia of the system (ball+rod) is equal to the one of rod + the one of the ball.

However, I am having difficulty understanding why after the perfectly inellastic collision the moment of inertia of the system is just the one of the rod.

The only hypothesis I have is that the mass of the ball is converted into heat, and to be honest I am quite unsure about it.

• I have added a diagram that I think shows the situation you describe. If my diagram is wrong please roll back my edit or ping me and I will remove it. However I do not understand your question since the moment of inertia after the collision should be sum of the rod and the ball. – John Rennie Apr 14 '18 at 7:20
• If you say inelastic collision , then the ball must cling to the rod . Then, the moment if inertia will have the form of that of rod , provided that the mass changes .... – Nehal Samee Apr 14 '18 at 9:20
• @JohnRennie I described the angular momentum of the system as follows: $\Delta L = L_f - L_i$ = 0; therefore $(\frac{ML_1^2}{3} + mL_2^2)\omega_i - (\frac{ML_1^2}{3} + mL_2^2)\omega_f$ = 0 Then we both agree with your statement: 'the moment of inertia after the collision should be sum of the rod and the ball'. – JD_PM Apr 14 '18 at 11:37
• Note $L_1$ is the length of the rod, $L_2$ is the length of the string, $\omega_f$ is the angular velocity of the system after the inelastic collision and $\omega_i$ is the angular velocity of the system before the inelastic collision – JD_PM Apr 14 '18 at 11:47

## 1 Answer

Before the collision the rod has angular momentum $I \omega_i$ where $I$ is the moment of inertia of the rod about the pivot and $\omega_i$ is the angular velocity of the rod immediately before the collision.

Whether the ball can be considered to have any moment of inertia before the collision is irrelevant, because it has no angular velocity so it has no angular momentum. If the ball did not stick to the rod after the collision, but bounced off in a straight line, then it would be even more perplexing to work out what its moment of inertia and angular velocity are, because both are constantly changing as the ball moves off in a straight line. However the ball's angular momentum is constant and easily calculated as linear momentum x perpendicular distance from the pivot.

After the inelastic collision the ball sticks to the rod, and both rotate about the pivot with the same angular velocity $\omega_f$. Both the rod and the ball now have angular momentum, the total of which is $(I+mL^2) \omega_f$. Whether you consider the ball and rod to be a single object in which there is no internal motion (ie a rigid body), or two separate objects moving such as to maintain their relative positions, does not make any difference.

It ought to be obvious from the Conservation of Angular Momentum, which you accept does apply in this case, that $\omega_i \ne \omega_f$ if $m \ne 0$.

It is not clear to me what you mean when you suggest that the mass of the ball is converted to heat. Some kinetic energy is lost in the collision, which is why it is called inelastic, and some of this loss becomes heat. But the mass of the ball (or the rod) does not change to any significant degree.

• I realized where I got confused. The fact is I should have noticed the moment of inertia of the ball before the collision is irrelevant, as it does not have angular velocity (therefore it does not have angular momentum) – JD_PM Apr 18 '18 at 17:14