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.

The angular frequency of a fan of moment of inertia 0.1 kg m^(2) is in

1.	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.
Answer» SOLUTION :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.