Tricky problem about angular momentum / rotational mechanics.

  • #1

Homework Statement


The figure shows two cylinders of radius r1 and r2 rotating about their axes (along the length) with speeds [tex]\omega_1[/tex] and [tex]\omega_2[/tex]
The two cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the point of contact but slipping finally ceases due to friction between them.

Find the angular speeds of the cylinders after the slipping ceases.
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Homework Equations



  1. After slipping ceases, if [tex]\omega_{1}'[/tex] and [tex]\omega_{2}'[/tex] are new angular speeds then...
    w1' * r1 = w2' * r2 .................(1)
  2. Angular momentum = Iw.

The Attempt at a Solution



I have succeeding in solving the problem by equating the angular impulse due to friction to the change in angular momentums of both cylinders and eliminating friction force from the two equations.

But I'd like to solve this using conservation of angular momentum as well since it should be pretty straightforward and easy.

Since no external torque acts on the system, conserving angular momentum, I get:

[tex]I_{1}\omega_{1}+I_{2}\omega_{2}=I_{1}\omega_{1}' + I_{2}\omega_{2}'[/tex]


Using this with w1' * r1 = w2' * r2 (the equation 1 above), I get

[tex]\omega_{1}' = \frac{I_{1}\omega_{1}+I_{2}\omega_{2}}{I_{1}r_{2}+I_{2}r_{1}} r_{2}[/tex]
and a similar expression for w2'.


The answer, however, is [tex]\omega_{1}' = \frac{I_{1}\omega_{1}r_{2}+I_{2}\omega_{2}r_{1}}{I_{1}r_{2}^{2}+I_{2}r_{1}^{2}} r_{2}[/tex]
 

Answers and Replies

  • #2
799
0
Not so tricky if you realize that you're applying the law of angular conservation wrongly :biggrin:

The law says, if there is no external torque, with respect to one fixed axis, the angular momentum is conserved (and also, under certain interactions, too; but that's out of the context of Newtonian mechanics). Here you are not even considering one axis, let alone whether it is fixed or not. You are considering TWO axes, each of which is the central axis of each cylinder.

Choosing a fixed axis here is quite tough (remember, the axis must be fixed throughout the process, from before the collision to after the collision). The situation after the collision is even more complicated: the frictional forces force the cylinders to move, or to gain translational motion, and that makes thing more complicated to predict the final phenomenon in order to choose an appropriate fixed axis to calculate angular momentum conveniently.

Then let's try dynamics - force / torque analysis. Call the impulse that cylinder (1) exerts on cylinder (2) throughout the process X, then cylinder (2) exerts -X on cylinder (1). We then have:
[tex]Xr_1 = I_1(w_1-w_1')[/tex]
[tex]-Xr_2=I_2(w_2-w_2')[/tex]
The next step is yours :wink:
 
  • #3
Then let's try dynamics - force / torque analysis. Call the impulse that cylinder (1) exerts on cylinder (2) throughout the process X, then cylinder (2) exerts -X on cylinder (1). We then have:
[tex]Xr_1 = I_1(w_1-w_1')[/tex]
[tex]-Xr_2=I_2(w_2-w_2')[/tex]
The next step is yours :wink:
Thanks for clearing that.
As I said, I have successfully solved the problem before using the method you just suggested.

I was also aware of the requirement of one axis throughout the problem, but I neglected/couldn't figure out in a hundred years that any translational motion would also take place.

That said, is it impossible to apply the conservation of angular momentum here to solve this problem?
I am ready/would love to to discuss the 'very complicated' method.
 
  • #4
799
0
Oops, sorry, I didn't notice that you solved the problem :biggrin:

I think I overlooked one problem: In order to obtain the equation [tex]w_1r_1=w_2r_2[/tex] , we have to assume that the translational speed is negligible; otherwise, the equation should look like this: [tex]w_1r_1+v_1=w_2r_2+v_2[/tex] . However, we have: [tex]X = m_1v_1[/tex] (assume that the cylinders have zero initial translational speed). So if the speed goes to zero, that also means [tex]w_1=w_1'[/tex] as [tex]Xr_1=I_1(w_1-w_1')[/tex] . If there is no change in angular speed, there is no friction, or initially there is no relative motion between the 2 points of contact. That means, initially we must have [tex]r_1=r_2[/tex] and [tex]w_1=w_2[/tex] , as we assume that the initial translational speed is zero. So plug them into both results (the answer provided and the result got from applying angular momentum conservation law in your post), they yield the same result.

In the general case, I suppose that the angular momentum equation must take into account the contribution of translational motion, and thus, the masses [tex]m_1[/tex] and [tex]m_2[/tex] must appear in the final result. That means, I think, the 2 results we have obtained so far are meaningless.
 
Last edited:

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