A wheel is rotating at 60 rotations per minute. If 480 J of energy is required to double its rotational speed, calculate the moment of inertia of the wheel.

Explanation

Data :' ` f_(1) = 60 rpm = 1Hz, f_(2) =2f_(1) = 2 Hz, W = 480 J`
` E_("rot") = 1/2 Iomega^(2) " " therefore W= 1/2 I (omega_(2)^(2) - omega_(1)^(2))`
` W = 1/2 I (2 pi f _(2))^(2) -(2pif_(1))^(2) = 2 pi^(2) I (f_(2)^(2) -f_(1)^(2))`
` I = W/(2pi^(2) (f_(2)^(2) -f_(1)^(2))) = 480 /(2pi^(2) (4-1)) = 80/ pi^(2) = 80/((4.142)^(2))= 8.104 kg.m^(2)`
This is the moment of inertia of the wheel.