Just a confusion -- A mass falling onto a pivoting rod

[Satvik Pandey]

Homework Statement

A rod of mass $M$ and length $l$ is pivoted at the center($O$) in horizontal position. An object of equal mass(having velocity $v$) falls vertically on the rod at distance $\frac{l}{4}$ from $O$ and sticks to it. Find the angular velocity of the rod just after that object falls on the rod.

Homework Equations

The Attempt at a Solution

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I know this question can be solved easily by conserving angular momentum about point O. But I have a confusion. If we consider that rod and object as a system then we can conserve angular momentum about $O$ only if no (net) torque act on the system about point $O$. right?
Now, torque on this system about $O$=torque on rod about $O$ +torque on that object about $O$. right?
When the object is just going to collide with rod. At that moment only $Mg$ force acts on the rod. But as that passes through point$O$ so torque due to that force will be 0.
But when the object is just going to collide with the rod, at that moment $Mg$ force acts on the object. But it is at distance $\frac{l}{4}$ from point $O$. And as the it falls vertically. So the torque due to that force is $Mg\frac{l}{4}$.
So net torque on the system about point $O$ is not zero then how can we use conservation of angular momentum?

 

[BvU]

That torque is not coming from outside the system under consideration. So it's part of the angular momentum conservation equation.

Actually, the force on the rod is a lot bigger: the object has to decelerate very rapidly.

 

[Orodruin]

I believe the problem maker intends for you to diseregard gravity. (Why else would the rod be at rest?)

Otherwise you might be able to consider the impact time small such that gravity does not impair any angular momentum on the rod in the short time of the collision.

 

[Satvik Pandey]

That torque is not coming from outside the system under consideration..

I don't get you. That torque it due to external force on the system. So, isn't it coming from outside the system under consideration?
Sorry if I am missing something very obvious.

 

[Orodruin]

I should read better. Still, the impact time is small and gravity does not give any angular momentum in the short impact time.

 

[BvU]

Satvik: You're right. See Oro's comment: the collision time is apparently considered short enough to ignore this mg.

 

[NascentOxygen]

Your torque of MgL/4 would be correct if the situation was that the object was delicately sat on the rod with zero speed. He actual scenario is that it is dropped from some height, so it carries significant momentum into the collision.

 

[BvU]

@O₂: I think Satvik is correct, no matter what.

And, for the record:

That torque is not coming from outside the system under consideration. So it's part of the angular momentum conservation equation.

is completely wrong!

 

[Satvik Pandey]

I should read better. Still, the impact time is small and gravity does not give any angular momentum in the short impact time.

Ok. So, that collision last for very short period of time(dt). And that torque $\frac{Mgl}{4}$ does not bring considerable change in the angular momentum of the system because it acted for very short period of time. right guys?

 

[BvU]

Excellent. Who knows you score extra by mentioning this. Distinguishes you from the crowd that doesn't even make the distinction ! (I walked straight into it ...)

 

[Orodruin]

Ok. So, that collision last for very short period of time(dt). And that torque $\frac{Mgl}{4}$ does not bring considerable change in the angular momentum of the system because it acted for very short period of time. right guys?

Right. In the idealised case, dt goes to zero andthe torque imparts no angular momentum at all during the collision. Hence conservation during the (idealised) collision. After the collision, you will have to take it into account to know the evolution of the system.

Similar considerations could be made for the linear momentum of objects colliding completely inelastically in a gravitational field.

 

[Satvik Pandey]

Right. In the idealised case, dt goes to zero andthe torque imparts no angular momentum at all during the collision. Hence conservation during the (idealised) collision. After the collision, you will have to take it into account to know the evolution of the system.

I have one more confusion. How we define the time period over which collision occurs? Do we define it as the time period over which impulsive force acts between the bodies?
In elastic collision the bodies simply collide with coefficient of restitution($e$)=1. This collision acts for very small interval of time.
In completely inelastic collision the bodies stick with each other after collision. How we define it's time period? In this case the impulse acting on the object reduces its speed to such limit so that after it the object moves along with the rod. right?

Similar considerations could be made for the linear momentum of objects colliding completely inelastically in a gravitational field.

Can this consideration be used for elastic and inelastic collision?

 

[Orodruin]

This would depend on the actual collision. For the purposes of most problems that are reasonable to give at this level, it will suffice to say that the collision time is small (and in many actual cases this might be a good approximation).

Yes, it should be possible also for elastic collisions - or any collision with known coefficient of restitution.

If you assume some collision time dt and constant force, you will see that the value of dt simply will not matter as long as it is small (and that constant external forces are irrelevant).

 

[Satvik Pandey]

Thank you Orodruin for clearing my confusion.:)