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?

Let the moment of inertia of the disc is I. Initial angular momentum is L1=I1ω1=90I When the mass m sticks at distance r, a new moment of inertia is given by I2=I+mr2 Now, the angular moment of inertia is L2=(I+mr22 =60(I+mr2) From the conservation of angular momentum- L1=L2 90I=60I+60mr2 30I=60mr2 I=2mr2

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