A rotating fan completes 1200revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a)Through what distance does the tip move in one revolution? What are (b) the tip’s speed and (c) the magnitude of its acceleration? (d) What is the period of the motion?

The frequency of fan is 1200 revolutions per minute.

Distance of center of axis and tip’s blade is r = 0.15 m.

It is given that the fan completes certain revolutions in a particular time with the given radius of the blade. So, we have to find the distance covered by the blade in one revolution. The speed and acceleration of tip can be found by using simple formulae.

Formulae:

$$\begin{gathered} Circumference\mathit{}of\mathit{}circle,C=2\tau \tau r \\ \mathrm{velocity}=\frac{\mathrm{distance}}{\mathrm{time}} \\ A\mathrm{cceleration},a=\frac{v^2}{r} \\ \mathrm{frequency}=\frac{1}{\mathrm{time}} \end{gathered}$$

Let’s take the radius of the circle for calculating the distance covered by the tip of fan in one revolution.

Tip of the fan covers the distance equal to its circumference in its revolution.

So, the circumference of the circle is

$$\begin{gathered} C=2\tau \tau r \\ =2(3.14)(0.15\mathrm{m}) \\ =0.94\mathrm{m} \end{gathered}$$

Therefore, the circumference of the circle is 0.94m.

We are giventhefrequency ofthefan, so we can find out its time. By using this value, we find its speed

$$\begin{gathered} \mathrm{Frequency}=\frac{1}{\mathrm{time}} \\ \mathrm{time}=\frac{1}{\mathrm{frequency}} \\ =\frac{1}{1200\mathrm{rev}/\min }\times 60\mathrm{s} \\ =0.05\mathrm{s} \end{gathered}$$

Now we can use$\mathrm{velocity}=\frac{\mathrm{distance}}{\mathrm{time}}$to find velocity.

$$\begin{gathered} \mathrm{velocity}=\frac{\mathrm{distance}}{\mathrm{time}} \\ =\frac{0.94\mathrm{m}}{0.05\mathrm{s}} \\ =18.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocity is 18.8 m/s.

As the motion is circular, theobject has centripetal acceleration and is given by

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}} \\ =\frac{{(18.8\mathrm{m}/\mathrm{s})}^2}{0.15\mathrm{m}} \\ =2.356\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \\ \approx 2.4\times {10}^3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the centripetal acceleration of the fan is equal to $2.4\times {10}^3\mathrm{m}/{\mathrm{s}}^2$.

We are giventhefrequency ofthefan, so we can find out its time. By using this value, we find its speed

$$\begin{gathered} \mathrm{Frequency}=\frac{1}{\mathrm{time}} \\ \mathrm{time}=\frac{1}{\mathrm{frequency}} \\ =\frac{1}{1200}\times 60\mathrm{s} \\ =0.05\mathrm{s} \end{gathered}$$

Therefore, the time period of motion is 0.05 s.