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I have a question regarding what I think should be a simple dynamics problem, but somehow I’m struggling with it.

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Imagine an object with mass $m$ coming from the left with velocity $v$. An identical object is hanging on a massless rope. I chose a frame of reference in which the axis around which the second object can rotate is at the origin. Object 1 hits object 2, and since they and identical, it transfers all its momentum to the object 2. To find out how fast object 2 starts spinning, I can approach this in two ways:

  1. Conservation of linear momentum. Initially, object 1 has a linear momentum of $p=mv$. After the collision, object 2 must have the same linear momentum, so it starts rotating with velocity $v$. The angular velocity is therefore $\omega=v/R$.

  2. Conservation of angular momentum. Initially, object 1 has an angular momentum of $L=Rp$. Again, it transfers it to object 2. On the other hand, an angular momentum of an object rotating around the origin of the frame of reference is $L=mR^2\omega,$ so again, the angular velocity is $\omega=p/(mR)=v/R$

Now, just to make my life more complicated, I change the frame of reference so that the origin is where object 2 is initially at rest. Now, object 1 carries no angular momentum, but the way I see it, object 2 will still end up rotating, therefore it must has some angular momentum. How does this agree with the conservation of angular momentum?

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My first thought was that maybe, somehow, for this particular choice of the frame of reference, the angular momentum of object 2 will be 0. So I decided to calculate it.

If $\pmb{r}$ is the vector from the origin to object 2 at any time, then $\pmb{L}=\pmb{r}\times \pmb{p}=m\pmb{r}\times \pmb{v}$. We can write $\pmb{r}=(R\sin(\omega t),R(1-\cos(\omega t)),0)$, where $\omega$ is the angular velocity at which object 2 rotates. The velocity is the time derivative of position, so $\pmb{v}=R\omega(\cos(\omega t),\sin(\omega t),0)$. The cross-product $\pmb{r}\times \pmb{v}=(0,0,R^2\omega(1-\cos(\omega t)))$ and the amplitude of the angular momentum is $L=mR^2\omega(1-\cos(\omega t))$. So not only it is not 0, but it is also not time independent, as my intuition tells me it should be, since there are no external torques acting on object 2.

So the two questions I have are:

  1. Why is angular momentum not conserved?
  2. Why is angular momentum time dependent?

Or, where in my calculation/reasoning did I make a mistake?

basketas
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  • This might help - Clarification about what makes a system isolated – mmesser314 Jan 06 '24 at 00:48
  • @mmesser314 Thank you, I think I get it now. I didn't take into account the tension of the rope, which in the second case results in non-zero torque such that $T=dL/dt$. And for the conservation, I need to consider the angular momentum of whatever object 2 is attached to. However, this got me thinking about a slightly different case, where object 2 is replaced with a free rod that object 1 hits at one end. Now the rod is not attached to anything, so there is nothing else to take into account to conserve the angular momentum. I'm now struggling to understand such a case. – basketas Jan 06 '24 at 02:18
  • In the case of the rod the angular momentum is indeed conserved, and it is zero. The rotating angular momentum is canceled by the angular momentum of the CM. – Pato Galmarini Jan 06 '24 at 03:43
  • @ basketas If you introduce a rigid rod, you do not succeed in removing external forces. Instead of the rope tension you now have a force of constraint at the pivot point of the rod, which constrains the body to rotate. – Albertus Magnus Jan 07 '24 at 21:48
  • @Pato Galmarini For a rod rotating about a pivot at the end, in the coordinates chosen according to the OP, the angular momentum of the COM does not cancel the angular momentum of the rigid body. – Albertus Magnus Jan 07 '24 at 21:59
  • @basketas You can get rid of forces of constraint by envisioning a rod in free space unattached to anything, if the incoming mass strikes precisely at the bottom, then the rod will rotate at a constant angular frequency about the center of mass. The angular momentum in the chosen coordinates will be non-zero. – Albertus Magnus Jan 07 '24 at 22:16
  • @AlbertusMagnus I thought he meant a free rod. – Pato Galmarini Jan 08 '24 at 00:12
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    I see there was some confusion regarding the rod. What I meant was a free rod, not attached to anything, so that after collision it starts rotating and moving away. And indeed, as @PatoGalmarini explained, the angular momentum due to the rotation is canceled by the angular momentum of the moving center of mass. Thank you for pointing this out. – basketas Jan 08 '24 at 19:08

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