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The rotor's velocity of a helicopter eng...

The rotor's velocity of a helicopter engine changes from 330 rev/min to 110 rev/min in 2 minutes. How long does the rotor blades take to stop

A

2min

B

4min

C

5min

D

6min

Text Solution

Verified by Experts

The correct Answer is:
A
`omega_(0)= 2pi xx (330)/(60)= 11pi` rad/sec
`omega= 2pi xx (110)/(60)= (11pi)/(3)` rad/s
`alpha= ((11pi)/(3)- 11pi)/(2 xx 60) = (-22pi)/(3 xx 2 xx 60)= (-11pi)/(180)` rad/s
`omega= omega_(0) + alpha t_(1) rArr t_(1)= 120= 2` min
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