Question

A ceiling fan with 80 cm-diameter blades is turning at 60 rpm. Suppose the fan coasts to a stop 25s after being turned off.

Short answer

The speed of the tip of the blade is 1.15 m/s

The number of revolutions is 12.5 rev

Step-by-step solution

Step 1. Given information

The diameter of the fan is 80 cm, its angular velocity is 60 rpm , and it needs to 25 seconds to stop.

The radius of the blade is $r=40\text{cm}$,

$d=(40\text{cm})\left(\frac{0.01\text{m}}{1\text{cm}}\right)=0.4\text{m}$

Initial angular speed of the fan is

${\omega }_i=\left(60\frac{\text{rev}}{\text{min}}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)=6.28\text{rad}/\text{s}$

The angular acceleration of the fan is,

$\alpha =\frac{{\omega }_{\text{f}}-{\omega }_{\text{i}}}{\Delta t}=\frac{0-6.28\text{rad}/\text{s}}{25\text{s}}=-0.251\text{rad}/{\text{s}}^2$

Part (a)

The angular speed of the tip of the blade is

${\omega }_{\text{f}}={\omega }_{\text{i}}+\alpha t=6.28\text{rad}/\text{s}+\left(-0.251\text{rad}/{\text{s}}^2\right)(10\text{s})=3.77\text{rad}/\text{s}$

The speed of the tip of the blade is,

$v=r{\omega }_{\text{f}}=(0.4\text{m})(3.77\text{rad}/\text{s})=1.51\text{m}/\text{s}$

Therefore, speed of the tip of the blade is $1.51\text{m}/\text{s}$.

Part (b)

The number of revolutions, in$\Delta t=25\text{s}$

$\begin{array}{l}\Delta \theta ={\omega }_{\text{i}}\Delta t+\frac{1}{2}\alpha {\left(\Delta t\right)}^2\\ =\left(6.28\text{rad}/\text{s}\right)\left(25\text{s}\right)+\frac{1}{2}\left(-0.251\text{rad}/{\text{s}}^2\right){\left(25\text{s}\right)}^2\\ =\left(78.6\text{rad}\right)\left(\frac{1\text{rev}}{2\pi \text{rad}}\right)\\ =12.5\text{rev}\end{array}$

Therefore, the number of revolutions is$12.5\text{rev}$