Conserving Angular Momentum (Rigid Body Dynamics)

[AdityaDev]

Lets say i have a rod (length = L) hinged at one end (A).It is initially at rest.Now if an impulse (J) acts on the other end (B),can i conserve the angular momentum about A(the hinge)? that is can i write: JL=Iw?(I=moment of inertia,w=angular velocity)
this is what i saw in the book.
My Doubt: But due to the impulse, there is a torque.Then how can i conserve angular momentum if net torque is not equal to zero?(i have thinked but I am still confused :()
CASE 2:
Now if change the situation, instead of impulse if a particle collides at B,can i conserve the angular momentum now?(Now the torque applied will be internal)

 

[Simon Bridge]

Welcome to PF;

Case 1. By definition: impulse = change in momentum. $J_L=I\Delta\omega$ would be the expression for the angular impulse delivered to an object whose moment of inertia does not change. If we write $\Delta\omega = \omega-\omega_0$ - then, in your example, $\omega_0=0$ and the expression turns into the one you have written.

Note: This is not a conservation of angular momentum equation.

Case 2. It is only total momentum that gets conserved.
If you find it is not conserved for some system, then you are not looking at the complete system.

In case 1 - whatever produces the impulse presumably loses whatever angular momentum is gained by the rod.
In case 2 - collisions can get complicated. For instance, sound and heat of the collision could carry off some momentum.

 

[AdityaDev]

thanks, i understood case1.
for case2:
(From halliday resnick)"we will consider rod and particle as our system.This system will experience external impulse due to hinge. But angular momentum about hinge can be conserved since impulse due to hinge is zero."(expression they have written: mux=mvx+Iw where u=initial velocity of particle and v= velocity of particle just after collision)
the rod experiences a torque about the hinge.Then how can you conserve angular momentum about hinge?Without torque how will the rod start rotating?

 

[Simon Bridge]

H&R were trying to second guess possible students objections there... they'd say "wait a minute: what about the reaction forces at the pivot?"

Notice that the pivot (the hinge) does not accelerate? What does that tell you about the net torque on the pivot?
Consider that the impulse at the pivot acts at zero distance from the pivot - so what is the torque due to the impulse?
Therefore, what is the change in momentum due to the impulse?

The impulse at the pivot is not what starts the rod rotating - the rod rotates due to an impulse delivered at the other end due to the collision.

In the linear case, a mass sliding along a frictionless level surface conserves momentum in collisions with another mass despite the fact that there is a (normal) force acting on the mass from the ground. How does this work?

During the collision, there will be forces acting on the objects involved. These forces will be action-reaction pairs so the momentum gained by one object gets lost by another. The exact details of how this happens can be arbitrarily complicated ... but, since we have the complete system, we can ignore the details and just talk a bout the final outcome in terms of the initial conditions, knowing that in the final analysis the total energy and momentum must be conserved. The trick is to identify the smallest amount of information you need to know in order to do this.