Ok here is the question that is being asked. Note that I know how to solve it one way, but when I go about trying to solve this problem another way that I think should give the the same result, I don't end up with the same result.(adsbygoogle = window.adsbygoogle || []).push({});

Question:

A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotating disk, in its center, causing both disks to spin at a new angular velocity ω. Assuming a negligible loss of energy to friction what is ω?

Ok so this problem is pretty easy just set initial angular momentum to final angular momentum

I(large disk)(ω0)=I(large disk)(ω)+I(small disk)(ω) and just solve for ω.

which gives me an answer of ω=((R**2)ω0)/(R**2+r**2)

So I know that is the correct answer. But then I also thought that I should be able to solve this using conservation of energy. I set this up like so

(1/2)I(large disk)(ω0**2)=(1/2)I(large disk)(ω**2)+(1/2)I(small disk)(w**2)

however when I solve I get

ω=R(ω0)/sqrt(R**2+r**2)

I checked my work multiple times and cannot find an error. I just don't know where my logic is going wrong. I mean if there is conservation of angular momentum doesn't that imply conservation of energy? Or since you are adding mass to the system can I not set up my energy conservation equation that way because I am not accounting for the rest energy of the second disk? I guess it is an inelastic collision when you drop the second disk on the top of the first so maybe that is why as well. But they say assuming a negligible loss to friction which throws me off. I mean isn't there significant loss of friction in inelastic collisions or do they just mean a small loss compared to the total rotational energy.

Thanks for reading

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# Disk rotation probelm

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