(adsbygoogle = window.adsbygoogle || []).push({}); 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 speedw. If a chip of massmbreaks 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*w^{2}

I = 0.5*M*R^{2}

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, I_{new}= 0.5*M*R^{2}-m*R^{2}.

so based on the conservation of kinetic energy:

0.5*I*w^{2}= 0.5*I_{new}*w_{new}^{2}+ o.5*m*(R*w)^{2}

then I substituted in the new moment of inertia and simplified.

what i get forw_{new}is the same asw.

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.

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# Rotating Thin Disc

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