# Rotating Thin Disc

by weesiang_loke
Tags: disc, rotating
 P: 33 1. The problem statement, all variables and given/known data A horizontal thin disc of mass M and radius R rotates about its horizontal axis through its centre with angular speed w. If a chip of mass m breaks off at the edge of the disc, what is the final angular speed of the disc? 2. Relevant equations Initial rotational kinetic energy = 0.5*I*w2 I = 0.5*M*R2 3. The attempt at a solution I assume that the chip broke off can be considered as some point mass. So the new moment of inertia of the thin disc, Inew = 0.5*M*R2 - m*R2. so based on the conservation of kinetic energy: 0.5*I*w2 = 0.5*Inew*wnew2 + o.5*m*(R*w)2 then I substituted in the new moment of inertia and simplified. what i get for wnew is the same as w. That means the speed is unaffected. I am not sure whether this is the correct way to do it? and if so why the speed is not affected by the process? Thanks for any help give.
 P: 431 I am not sure why conservation of energy works here but a safer option would have been conservation of angular momentum. If you use that you would get the same result and that can be explained as the disc's angular momentum decreases corresponding to it's moment of inertia. Will find out if conservation of energy here is really applicable.
 P: 33 Hi aim1732, but i still dont get it why the speed still remained the same even when some parts fell off.
P: 431

## Rotating Thin Disc

That is because the angular momentum decreased but L=Iw---since there is corresponding decrease in I,w happens to remain same.
 P: 33 ok. i get it now. thanks...
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Thanks
PF Gold
P: 26,127
hi weesiang_loke!

(have an omega: ω )
 Quote by weesiang_loke Hi aim1732, but i still dont get it why the speed still remained the same even when some parts fell off.
it's because there was no interaction

it's exactly the same as if the chip was never part of the disc, but was attached to the same axle by a string, and just happened to be rotating at the same angular speed …

when the string is cut, of course it has no effect on the disc!

only if you were told that the chip "pushed off", or that some energy was lost, would there be an interaction and therefore an effect on the disc
 P: 33 hi tiny-tim, i think i fully understand it now. thanks for the explanation. It's really useful.

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