Conservation of angular momentum

AI Thread Summary
The discussion centers on the application of the conservation of angular momentum in a collision scenario involving a rod and a ball. A participant expresses confusion about the correct formulation of angular momentum conservation, specifically regarding the inclusion of the rod's center of mass and the velocities involved post-collision. It is clarified that angular momentum conservation must be applied about a fixed point, and using the center of mass of an accelerating object complicates the analysis due to the need for pseudo forces. The conversation emphasizes that if there is net torque acting on the system, the conservation of angular momentum may not hold. Ultimately, understanding the reference point and the system's dynamics is crucial for correctly applying the conservation laws.
RingNebula57
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Hello everyone!
I have a problem , to which I do not understand the law of conservation of angular momentum... I searched this problem on the web and it is obvious that I am making the mistake.
So we have a rod of length ##L## and mass ##m## that is lying on a horizontal frictionless table. We hit the rod at one end, perpendicular to the rod, with a ball of mass ##m## that is rolling on the table with initial speed ##v_0##. After they collide elastically , the rod begins rotational and translational motion with angular velocity ##\omega##(about its CM) and velocity ##v_1##(of the CM) , while the ball continues its translation with velocity ##v_2##, in the same direction as the initial velocity ##v_0##.
Now , if we consider the moment of inertia of the rod about its center of mass to be ## I ## , I say that the conservation of angular momentum about the center of mass of the rod is:

## m \cdot v_0 \cdot \frac{L}{2} = m \cdot (v_2 - v_1) \cdot \frac{L}{2} + I \cdot \omega ##
But the solution says:

## m \cdot v_0 \cdot \frac{L}{2} = m \cdot v_2 \cdot \frac{L}{2} + I \cdot \omega ##

##( I = m \cdot \frac {L^2}{12} )##

why ?
Isn't the conservation of angular momentum always relative to the center of mass of the rod?
 
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Hi,
RingNebula57 said:
Isn't the conservation of angular momentum always relative to the center of mass of the rod?
It's always. (period). And here you choose the center of mass for the calculation. ##v_1## can't appear in there. (but it does of course appear in the conservation of linear momentum)
 
Conservation of momentum applies when angular momentum is calculated about a fixed point which may or may not be the center of momentum of some particular object within your closed system and which may or may not be at the center of mass of the entire system. The key is that it is a fixed point. Anchoring it to an object which undergoes acceleration is a good way to have conservation of [angular] momentum not apply.

In some treatments, the reference point is taken as the origin of a coordinate system in your chosen inertial frame. But one can use any point that is in uniform motion by simply translating to an [inertial] reference frame where that point is the origin.
 
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Likes vanhees71 and Andreas C
Before applying angular momentum conservation think about 2 things.
First:
point about which you are applying angular momentum conservation. Whether point is stationary or moving or accelerated.
If accelerated you have to apply pseudo force to CENTER OF MASS OF THE SYSTEM ON WHICH YOU APPLY ANGULAR MOMENTUM CONSERVATION.
Second: system on which you apply it. If there is net tourqe acting (including psuedo force torque) angular momentum conservation invalid.

So here COM of rod is accelerated frame of reference, psuedo force has to be applied on system's COM i.e. on COM of rod plus bullet system. And definitely COM of rod does not match COM of system so you can't apply angular momentum conservation (psuedo force torque)
 
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