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As per the shown figure the central soli...


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

Text Solution

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`F=etaA(dv)/(dz)` where `(dv)/(dz)=(omegaR_(1)-0)/(R_(2)-R_(1))`
`F=eta(2piR_(1)lomegaR_(1))/(R_(2)-R_(1))` and `tau=FR_(1)=(2pietaR_(1)^(3)omegal)/(R_(2)-R_(1))`
`Ialpha=(2pietaR_(1)^(3)omegal)/(R_(2)-R_(1))implies(MR_(1)^(2))/(2)(-(domega)/(dt))=(2pietaR_(1)^(3)omegal)/(R_(2)-R_(1))`
or `-underset(omega_(0))overset(omega_(0)//2)(domega)/(omega)=(4pietaR_(1)l)/(m(R_(2)-R_(1)))underset(0)overset(t)intimpliest=(m(R_(2)-R_(1))ln2)/(4pietalR_(1))`
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