The angular frequency of a fan of moment of inertia 0.1 kg `m^(2)` is increased from 30 rpm to 60 rpm when a torque of 0.03 Nm acts on it. Find the number revolutions made by the fan while the angular frequency is increased from 30 rpm to 60 rpm.

Explanation

Work done `=(1)/(2)I(omega_(f)^(2)-omega_(i)^(2)) " " tau theta = (1)/(2)I(omega_(f)^(2)-omega_(i)^(2))`
`theta=(1)/(2)((I)/(tau))(omega_(f)^(2)-omega_(i)^(2))=(0.1)/(2xx0.03)((2pi)^(2)-pi^(2))=5pi^(2)`
The number of revolutions, `(theta)/(2pi)=(5pi^(2))/(2pi)=(5pi)/(2)=7.855` rev.