Question

A woman rides a carnival Ferris wheel at radius 15 m, completing five turns about its horizontal axis every minute. What are (a) the period of the motion, the (b) magnitude and (c) direction of her centripetal acceleration at the highest point, and the (d) magnitude and (e) direction of her centripetal acceleration at the lowest point?

Short answer

1. The period of motion is 12s.
2. The magnitude and direction centripetal acceleration at highest point$4.1\mathrm{m}/{\mathrm{s}}^2$
3. Direction of centripetal acceleration at highest point is downwards.
4. The magnitude of centripetal acceleration at lowest point$4.1\mathrm{m}/{\mathrm{s}}^2$
5. The direction centripetal acceleration at lowest point is upward towards the center of orbit.

Step-by-step solution

Given

The frequency of fan is 5 revolutions per minute.

Radius of wheel is 15.0m.

Understanding the concept time period and centripetal acceleration

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:

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

(a) Calculate the period of motion

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

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

Therefore, the time period is 12 s.

(b) Calculate the magnitude of her centripetal acceleration at the highest point

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),

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

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

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

(c) Calculate the direction of her centripetal acceleration at the highest point

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

(d) Calculate the magnitude of her centripetal acceleration at the lowest point

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

(e) Calculate the direction of her centripetal acceleration at the lowest point

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