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A horizontal disc rotating freely about a vertical axis through its center makes 90 revolutions per minute. A small piece of wax of massm falls vertically on the disc and sticks to it at a distancer 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

Step 1. Given data

  1. A horizontal disc rotating freely about a vertical axis through its center makes 90 revolutions per minute, i.e. ω1=90rpm
  2. The mass of the wax =m
  3. Wax sticks to the disk at a distance of r from the axis
  4. After wax fell on the disc the new angular velocity is ω2=60rpm
  5. Let, the moment of inertia of the disc is I.

Step 2. Formula used

In this case, the total angular momentum of the system must be conserved.

L=Iωis conserved.

Where L is the angular momentum, I moment of inertia, ωis angular velocity

Step 3. Find the moment of inertia

The initial angular momentum is L1=I1ω1=90I1

The wax is attached to the disc at a distance of r from axis, so the new moment of inertia of the disc with the wax piece is I+mr2.

The final angular momentum L2=ω2I+mr2=60I+mr2

From Law of conservation of angular momentum we have

L1=L2Iω1=(I+mr2)ω290I=60I+mr230I=60mr2I=2mr2

Hence, the moment of inertia of the disc is 2mr2.


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