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The square of the angular velcoity omega...

The square of the angular velcoity `omega` of a certain wheel increases linearly with the angular displacement during `100 rev` of the wheel's motion as shown. Compute the time `t` required for the increase.

Text Solution

Verified by Experts

The correct Answer is:
`(40pi)/(7)sec.`
We have `Delta theta =2pi xx (100 rev)=200pi rad`
So `omega ^(2)=omega _(0)^(2)+2 alpha Delta theta`
`rArr 1600=900+2alphaxx200pi`
`rArr alpha=(700)/(400pi)=(7)/(4pi)`
Also `omega=omega_(0)+alphat`
`rArr 40=30+(7)/(4pi)t`
`rArr t=(40pi)/(7)`
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