Question #141807
An angular momentum of a fly wheel having a momentum of inertia of 0.125kgm^2 decreases form 3.0kgm^2/s to 2.0kgm^2/s in a period of 1.5secs.
(i) what is the average torque acting on the fly wheel during this period.
(ii) assuming a uniform angular acceleration, through how many revolution will the fly wheel have turned?
1
Expert's answer
2020-11-02T10:56:52-0500

Average torque=I×ΔwΔt=0.125×231.5=0.083=I\times\frac{\Delta w}{\Delta t}=0.125\times\frac{2-3}{1.5}=0.083


τ=Iα\tau=I\alpha

0.083=0.125×α0.083=0.125\times\alpha

α=0.67\alpha=0.67



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