The angular frequency of a fan of moment of inertia `0.1kg 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