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Question

A horizontal disc rotating freely about a vertical axis through its center makes 90 revolutions per minute. A small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the number of revolutions per minute reduce to 60, then the moment of inertia of the disc is?

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Solution

Let moment of inertia of the disc is I.
Initial angular momentum is L1=I1ω1=90I

When the mass m sticks at distance r, new moment of inertia is
I2=I+mr2

Now, angular moment of inertia is L2=(I+mr2)ω2=60(I+mr2)

From conservation of angular momentum-

L1=L2

90I=60I+60mr2

30I=60mr2

I=2mr2

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