Question

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be $15m$, and you use your watch to find that each loop around takes $25s$.

a. What are your speed and the magnitude of your acceleration?

b. What is the ratio of your weight at the top of the ride to your weight while standing on the ground?

c. What is the ratio of your weight at the bottom of the ride to your weight while standing on the ground?

Short answer

a). My speed is $3.77\mathrm{m}/\mathrm{s}$and the magnitude of the acceleration is $0.947\mathrm{m}/{\mathrm{s}}^2$.

b). The ratio of the weight at the top of the ferries wheel to the weight while standing at the ground is $0.903$.

c). The ratio of your weight at the bottom of the ride to your weight while standing on the ground is $1.09$.

Step-by-step solution

Given Information (Part a)

We know that the radius of the ferries wheel is $r=15\mathrm{m}$, and the period of one rotation is $T=25\mathrm{s}$. We have to calculate the speed and the magnitude of acceleration.

Explanation (Part a)

The speed of an object in circular motion is given by

$v=\omega r$

where $\omega$is the angular velocity and is given by $\omega =\frac{2\pi }{T}$. Hence, the speed is given by

$v=\frac{2\pi }{T}r$

Substituting the values we have

$v=\frac{2\pi }{(25\mathrm{s})}(15\mathrm{m})$

$=3.77m/s$

And the acceleration here will be centripetal acceleration, which is given by

$a={\omega }^{2}r=\frac{4{\pi }^2}{T^2}r$

Substituting the values we have

$a=\frac{4{\pi }^2}{(25\mathrm{s}{)}^2}(15\mathrm{m})$

$=0.947m/s^2$

Final Answer (Part a)

My speed is $3.77\mathrm{m}/\mathrm{s}$and the magnitude of acceleration is$0.947\mathrm{m}/{\mathrm{s}}^2$.

Given Information (Part b) 

We know that the acceleration of the gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Explanation (Part b)

Consider my mass is $m$. Hence, my weight standing at the ground is

$w=mg$

While at the top, the centrifugal force acts vertically upward and the weight acts vertically downward. Hence, net weight at the top is given by

$w_1=m(g-a)$

Hence, the ratio is

$\frac{w_1}{w}=\frac{(g-a)}{g}$

Substituting the values we have

$\frac{w_1}{w}=\frac{\left[\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)-\left(0.947\mathrm{m}/{\mathrm{s}}^2\right)\right]}{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}$

$=0.903$

Final Answer (Part b)

The ratio of the weight at the top of the ferries wheels to the weight while standing at the ground is $0.903$.

Given Information (Part c)

We know that the acceleration of the gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$

Explanation (Part c)

At the bottom, the centrifugal force and the weight, both are directed vertically downwards. Hence, the effect will add up. Hence, the net weight of me will be

$w_2=m(g+a)$

Hence, the ratio is

$\frac{w_2}{w}=\frac{g+a}{g}$

Substituting the values we have

$\frac{w_2}{w}=\frac{\left[\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.947\mathrm{m}/{\mathrm{s}}^2\right)\right]}{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}$

$=1.09$

Final Answer (Part c) 

The ratio of your weight at the bottom of the ride to your weight while standing on the ground is$1.09$.