91Ó°ÊÓ

The frequency of fan is 5 revolutions per minute.

Radius of wheel is 15.0m.

If the object travels along a circle or circular arc at a constant speed then it is said to be in a uniform circular motion and has an acceleration of constant magnitude. If we know the time period and radius of the circular motion, we can find the acceleration using the formula for the acceleration. From the given situation we can find the time period of motion by using the given frequency and from the given radius, it is easy to find the magnitude of acceleration.

Formulae:

$$\begin{gathered} \mathrm{Circumferenceof}\mathrm{circle}\left(\mathrm{C}\right)=2\mathrm{Ï„Ï„°ù}...\left(1\right) \\ \mathrm{velocity}=\frac{\mathrm{distance}}{\mathrm{time}}....\left(2\right) \\ \mathrm{Acceleration}\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}}...\left(3\right) \\ \mathrm{frequency}\left(\mathrm{f}\right)=\frac{1}{\mathrm{time}\left(\mathrm{T}\right)}....\left(4\right) \end{gathered}$$

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

Using equation (iv) and the given the frequency of object i.e. 5 turns per minute gives

$$\begin{gathered} \mathrm{f}=\frac{1}{\mathrm{T}} \\ \mathrm{T}=\frac{1}{\mathrm{f}} \\ =\frac{1}{5\mathrm{rpm}} \\ =0.2×60\mathrm{s} \\ =12\mathrm{s} \end{gathered}$$

Therefore, the time period is 12 s.

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

$a=\frac{v^2}{r}$

But for that, we need the velocity of revolution which can be obtained from the distance covered by the object in one revolution. It is equal to the circumference of the circle. Therefore, from equation (i),

$$\begin{gathered} C=2\mathrm{Ï„}\mathrm{Ï„}r \\ =2(3.14)\left(15\mathrm{m}\right) \\ =94.2\mathrm{m} \end{gathered}$$

Now using equation (ii),

$$\begin{gathered} \mathrm{Velocity}=\frac{\mathrm{distance}}{\mathrm{time}} \\ =\frac{94.4\mathrm{m}}{12\mathrm{s}} \\ =7.85\mathrm{m}/\mathrm{s} \end{gathered}$$

So, if we plug all obtained values in the equation (iii), we can find acceleration

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}} \\ =\frac{{(7.85\mathrm{m}/\mathrm{s})}^2}{15\mathrm{m}} \\ =4.1\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

When the passenger is at the top, the acceleration is directed downward (at center of orbit) with the same magnitude.

When passenger is at the lowest point, acceleration is same as part (b).

The direction would be upward towards the center of the orbit.