# Inelastic collision of ball and rod - rotation problem

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## Homework Statement

A uniform rod of mass M and length l is hinged at the center. a particle of mass m and speed u sticks after hitting the end of the rod. find the angular velocity of the rod just after collision

## Homework Equations

Energy conservation-0.5mu^2=0.5(m+M)v^2
CoM=m1x1+m2x2/m1+m2

## The Attempt at a Solution

Can we apply energy conservation since the hinge isn't specified to be frictionless?
I'm sure we can conserve angular momentum about hinge though, taking hinge as origin the CoM comes out to be at a distance mL/2(M+m) and moment of Inertia about CoM can be calculated using parallel axis theorem but this gives a lot of complicated terms. Is there a simpler way to solve this question then?

PeroK
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Why do you need to calculate the COM?

Orodruin
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The friction at the hinge is not your main problem
sticks after hitting the end of the rod
What does this tell you about the particle-rod collision?

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The friction at the hinge is not your main problem

What does this tell you about the particle-rod collision?
Its purely inelastic, so we can't conserve total kinetic energy... but we can still write m1v1+m2v2=(m1+m2)v3. so mu=(M+m)v. but using v=Lω/2 gives an incorrect answer

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Why do you need to calculate the COM?
to apply initial angular momentum=Iω. I is about CoM

PeroK
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to apply initial angular momentum=Iω. I is about CoM

Is there a simpler way to solve this question then?

Let me rephrase my question: Do you really need to calculate the COM?

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Let me rephrase my question: Do you really need to calculate the COM?
could I write Iω about hinge for the sytem, then add Mv(com)*distance of com (for rod)+ mv*distance of com(for particle) instead? I'd still have to find the CoM though, I don't see any other way

PeroK
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could I write Iω about hinge for the sytem, then add Mv(com)*distance of com (for rod)+ mv*distance of com(for particle) instead? I'd still have to find the CoM though, I don't see any other way

True or false:

The rod has an angular momentum.
The mass has an angular momentum.
The angular momentum of the rod-mass system is the sum of the two, regardless of whether they are stuck together or not?

Orodruin
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Also a warning: The CoM is a very bad choice for a reference point here. Due to the forces at the hinge, the angular momentum about the CoM will not be conserved.

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True or false:

The rod has an angular momentum.
The mass has an angular momentum.
The angular momentum of the rod-mass system is the sum of the two, regardless of whether they are stuck together or not?
True! Oh, how silly of me. That was amazingly simple, I was just overthinking it. I got the correct answer as 6mu/(3m+M)L, Thank you very much :D

• PeroK
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Also a warning: The CoM is a very bad choice for a reference point here. Due to the forces at the hinge, the angular momentum about the CoM will not be conserved.
Ohh, I was under the impression that for a combined system we always need to write I about CoM while using initial AM=Iω. but adding them separately makes a lot of sense as I now discovered

Orodruin
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