Question

When a ceiling fan rotating with an angular speed of 2.75 rad/s is turned off, a friction torque of 0.120 N.m slows it to a stop in 22.5s . What is the moment of inertia of the fan?

Answer 1

When the ceiling fan is turned off, the frictional torque brings it to stop. The frictional torque is given as

Torque \(\quad \tau=I \alpha \ldots \ldots \ldots \ldots \ldots \ldots\) (1)

The change in angular speed of the fan is given as

$$ \text { Angular speed } \quad \Delta \omega=\alpha t \ldots \ldots \ldots \ldots \ldots \text { (2) } $$

On substituting eq \((2)\) in eq (1), we get

$$ \tau=I\left(\frac{\Delta \omega}{t}\right) $$

Moment of inertia of the fan from the above equation is

$$ I=\frac{\pi t}{\Delta \omega} $$

Given that Torque \(\tau=(0.120 \mathrm{Nm})\)

Time \(t=(22.5 \mathrm{~s})\)

Angular speed is \(\Delta \omega=2.75 \mathrm{rad} / \mathrm{s}\)

$$ \begin{array}{l} =\frac{(0.120 \mathrm{~N} \cdot \mathrm{m})(22.5 \mathrm{~s})}{2.75 \mathrm{tad} / \mathrm{s}} \\ I=0.982 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{array} $$