Application of angular impulse to a problem

[Krushnaraj Pandya]

Homework Statement

two discs of radii r and 2r are pressed against each other. Initially disc with radius r is rotating with angular velocity w and other disc was stationary. Both discs are hinged at their respective centers and free to rotate about them Moment of inertia of 1st disc is I and bigger disc is 2I about their respective axis of rotation, find w of bigger disc after long time. Note*-I only have a problem with one step of the solution

Homework Equations

1) r1w1=r2w2 when slipping ceases
2) ∫(torque)dt=dL

The Attempt at a Solution

only force producing torque about center is friction. acting upwards on the smaller disc, from equation 1 w2=w1/2 where w1 and w2 are final angular velocities of disc 1 and 2. now total angular impulse on second disk is ∫f2Rdt=dL=2Iw2...(i) impulse on first however is written as ∫fRdt=dL=I(w-w1)...(ii) which is initial angular momentum minus final angular momentum, shouldn't it be written the other way round as final-initial. (i) and (ii) are later used to solve for w2 which comes out to be w/3, please explain why it is written this way and how it is correct

 

[CWatters]

Not sure I follow your description of the problem. You refer to w1 and w2 as the final angular velocities of the two discs and that w2= w1/2. But after a long time shouldn't they be rotating at the same angular velocity?

Also in the problem statement you say .. "Moment of inertia of first disc is l and [the second] bigger disc 2l. Later you say "(I)impulse on second however is written as... =l(w-w1)". That equation has "l" in it rather than "2l" so isn't it the loss of disc 1 rather than the gain of disc 2?

 

[Krushnaraj Pandya]

Not sure I follow your description of the problem. You refer to w1 and w2 as the final angular velocities of the two discs and that w2= w1/2. But after a long time shouldn't they be rotating at the same angular velocity?

Also in the problem statement you say .. "Moment of inertia of first disc is l and [the second] bigger disc 2l. Later you say "(I)impulse on second however is written as... =l(w-w1)". That equation has "l" in it rather than "2l" so isn't it the loss of disc 1 rather than the gain of disc 2?

the point of contact will have the same velocity, not w after a long time- therefore the equation w2=w1/2.

Sorry, I misspoke, I meant to say first disc- I've edited it to the correct statement now

 

[CWatters]

Ah ok. I was visualising them rotating on the same shaft. Sounds like they are more like two gears that are brought into mesh.

 

[Krushnaraj Pandya]

Ah ok. I was visualising them rotating on the same shaft. Sounds like they are more like two gears that are brought into mesh.

yes, exactly! now the only doubt I have is the initial-final part

 

[Nathanael]

impulse on first however is written as ∫fRdt=dL=I(w-w1)...(ii) which is initial angular momentum minus final angular momentum, shouldn't it be written the other way round as final-initial.

Yes you are technically correct. The impulse here is negative. However if you know that the impulse is negative, then you can find its magnitude by doing it backwards like this.

 

[Krushnaraj Pandya]

Yes you are technically correct. The impulse here is negative. However if you know that the impulse is negative, then you can find its magnitude by doing it backwards like this.

oh so since the direction of torque is the direction of impulse, they directly wrote it this way, how silly of me to overlook that

 

[Krushnaraj Pandya]

Yes you are technically correct. The impulse here is negative. However if you know that the impulse is negative, then you can find its magnitude by doing it backwards like this.

Thanks a lot!