CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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?


Open in App
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.


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Instantaneous Axis of Rotation
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon