Asked by hriday
A horizontal disc rotating freely about a vertical axis makes 90 revolutions per minute. A small piece of wax of mass “m” falls vertically on the disk & sticks to it at a distance “r” from the axis. If the number of revolutions reduces to 60, then find the moment of inertia of the disc?
1. mr2
2.3/2mr2
3.2mr2
4.3mr2
(the last ‘2’in each option should be considered as square, please)
1. mr2
2.3/2mr2
3.2mr2
4.3mr2
(the last ‘2’in each option should be considered as square, please)
Answers
Answered by
bobpursley
Conservation of momentum is applied.
Momentum before=momentum after
I disk*(90*2PI/60)=Idisk(60*2PI/60)+mr^2*60(2PI/60)
solve for I disk.
Momentum before=momentum after
I disk*(90*2PI/60)=Idisk(60*2PI/60)+mr^2*60(2PI/60)
solve for I disk.
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