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A disc of moment of inertia 9.8//pi^(2)k...

A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done ?

A

1467J

B

1452J

C

1567J

D

1470 J

Text Solution

Verified by Experts

The correct Answer is:
D
`I= (9.8)/(pi^(2)) kg m^(2), f= (600)/(60)= 10 rps`
`omega_(0) = 2pi xx 10 = 20pi` rad/s
`omega= 2pi xx (300)/(60)= 10pi` rad/sec
Work done = Change is KE
`= (1)/(2) I (omega_(0)^(2)- omega^(2))`
`= (1)/(2) xx (9.8)/(pi^(2)) xx 4pi^(2) [10^(2)- 5^(2)]`
=1470 J
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