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As per the shown figure the centreal sol...

As per the shown figure the centreal solid cylinder starts with initial angular velocity `omega_(0)`. Find out the time after which the angular velocity becomes half. `(` Velocity gradien uniform)

Text Solution

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`F = etaA(dv)/(dz)`, where `(dv)/(dz) = (omegaR_(1) - 0)/(R_(2) - R_(1)`
`F = eta(2piR_(1)l_(omega)R_(1))/(R_(2) - R_(1))`
and `tau = FR_(1) = (2pietaR_(1)^(3)omegal)/(R_(2) - R_(1))`
`lalpha = (2pietaR_(1)^(3)omegal)/(R_(2) - R_(1))`
`rArr (mR_(1)^(2))/(2)(-(domega)/(dt)) = (2pietaR_(1)^(3)ometgal)/(R_(2) - R_(1))`
or `-overset(omega_(0)//2)underset(omega_(0))(int)(domega)/(omega)=(4pietaR_(1)l)/(m(R_(2)-R_(1)))overset(t)underset(0)(int)dt`
`rArr t = (m(R_(2)-R_(1))ln2)/(4pietalR_(1))`
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