# Conservation of angular momentum ## Homework Statement

A uniform thin rod AB is equipped at both ends with the hooks as shown in the figure and is supported by a frictionless horizontal table. Initially the rod is hooked at A to a fixed pin C about which it rotates with a constant angular velocity w1 . Suddenly end B of the rod hits pin D and gets hooked to pin D, causing end A to be released. Determine the magnitude of the angular velocity w2 of the rod in its subsequent rotation about D. (Assume length and mass of the hook is negligible. Pin C & D are lying on a same horizontal line)

## Homework Equations

Angular momentum of the rod = moment of inertia about axis of rotation X angular velocity

## The Attempt at a Solution

While I understand that I need to apply Conservation of angular momentum to solve this. But the problem here is that I have to compute the angular momentum about an axis other than the axis of rotation just before the other end hits pin D. I am unable to compute this. Any help will be highly appreciated.

## Answers and Replies

Angular to linear to angular

Angular to linear to angular
I am sorry but I don't get it! What is the angular momentum of the rod about pin D just before it hits it (i.e., just before it hits the pin D)? I have tried taking the rod as a large number of very small masses and computing their momentum about pin D. But this doesn't give the correct answer.

I am sorry but I don't get it! What is the angular momentum of the rod about pin D just before it hits it (i.e., just before it hits the pin D)?

At the moment of collision the angular velocity about C constitutes a linear velocity down which is an angular velocity about D.

Orodruin
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At the moment of collision the angular velocity about C constitutes a linear velocity down
I do not believe that this is correct. Since the rod is rotating just before the collision, different parts of it will have different velocities. However, it is still possible to compute the angular momentum relative to D using the instantaneous velocities just before the collision.

• haruspex
I do not believe that this is correct. Since the rod is rotating just before the collision, different parts of it will have different velocities. However, it is still possible to compute the angular momentum relative to D using the instantaneous velocities just before the collision.

I believe that what you are suggesting does not contradict what I said.

vela
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Are you sure angular momentum is conserved?

haruspex
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I believe that what you are suggesting does not contradict what I said.
It does.
If you could treat it as a linear motion then you should get the same answer if, when the collision occurs, the rod is not rotating. We can test that by asking what would happen if there were no linear motion, only a rotation about the rod's centre. The end of the rod near C would have a greater angular momentum about D than the end that strikes D, so there would be subsequent rotation about D (clockwise in the picture).

haruspex
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Are you sure angular momentum is conserved?
It depends what axis you choose. The impulse is at D only, so angular momentum about D is conserved.
There is a force at C, but that can be discounted for two reasons:
- it is a steady force, so imparts no appreciable momentum or angular momentum during the infinitesimal period of the impulse
- at the instant of collision it acts on a line through D, so has no torque about D.

haruspex
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I have tried taking the rod as a large number of very small masses and computing their momentum about pin D. But this doesn't give the correct answer.
That ought to work. Please post your attempt.

It does.

It does not. What Orodruin is suggesting is to take each point along the rod, from the angular velocity about C determine the linear velocity down, multiply by the distance from D to get the angular velocity about D and integrate to get the angular momentum about D. How is that in any way inconsistent with what I said?

“At the moment of collision the angular velocity about C constitutes a linear velocity down which is an angular velocity about D.”

I wasn’t going to do the whole problem for the OP, but that is a pretty solid hint at what to do, and it is an absolutely true statement.

• roam
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How is that in any way inconsistent with what I said?
It constitutes a linear velocity down plus a rotation about its mass centre.
Rightly or wrongly, @Orodruin and I both interpreted your statement as saying that the bar can be treated as moving purely linearly, i.e. ignoring its rotation about its centre. If that is not what you meant then at the least it was somewhat misleading.

Last edited:
That ought to work. Please post your attempt.

Here is my attempt:
Angular momentum of the rod about point D just before hitting the point D shall be: Σdm.x×v, where x is the distance from D and v is the velocity of that small mass.[Also taking mass of rod as M, length as L, and remember ω1 is the angular velocity of the rod about point C]

Now, dm = (M/L).dx, and, v = ω1 × (L-x)

So, angular momentum about point D just before collision = ∫(M/L).dx.x.ω1.(L-x) = (M/L)ω1.[Lx2/2 - x3/3]
This gives angular momentum about point D just before collision as (ML2/6)ω1

And final momentum about D just after collision shall be = Iω = (ML2/12)ω2

Equating initial and final we get ω2 = 2.ω1

which is fundamentally wrong!!! how can angular velocity increase? I am sure I am missing something here.

haruspex
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(ML2/12)
D is not the mass centre.

• studentofphysics83
D is not the mass centre.
Oh, got it! Thank you!

Pls someone solve this question I'm not getting it.

haruspex
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Pls someone solve this question I'm not getting it.
Per forum rules, you must show some attempt; or at the least, your thoughts on the matter.

Iw1=Iw2-ml^2w2

haruspex
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Iw1=Iw2-ml^2w2
That's not so much an attempt as a wild guess. You will need to explain your reasoning.

I found the angular momentum about c and then about d . The angular momentum remains conserved

Okay I got it. Thanks.

Error in right hand thumb rule!

haruspex
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I found the angular momentum about c and then about d . The angular momentum remains conserved
Angular momentum is always in respect of a chosen axis. About which axis is it conserved?