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Vitek1552 [10]
2 years ago
11

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass MMM and radius RRR about an axis perpendicular to th

e hoop's plane at an edge.
Physics
1 answer:
slega [8]2 years ago
3 0

Answer:

I = sum m *r^2    where m represents the (small) individual masses and r is the distance of that mass from center of rotation

Note:  sum m = M

For the hoop given all masses are at a distance RRR from the center of rotation

I = MMM * RRR^2

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
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<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

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b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

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