Ferris Wheel Forces: Exploring Normal Force and Centripetal Acceleration

In summary: So, now I go and repeat the correct calculations, using the equations for Newton's second law where Fnet = Fc.
  • #1
hagobarcos
34
0

Homework Statement



A student of weight 667 N rides a steadily rotating Ferris wheel. At the highest point, the magnitude of the normal force on the student from the seat is 556 N.

a) Does the student feel "light" or "heavy" there?

b) What is the magnitude of Fn at the lowest point?

c) If the wheel's speed is doubled, what is the Fn at the highest point,

d) and at the lowest point?

Homework Equations



Fc = m(v²/R)

Ac = (v²/R)

Fnet = m a



The Attempt at a Solution



Okay this is where I am a bit backwards:

At the top:

Fnet = may
Fn-Fc-mg = may = 0

^^ I think this part is wrong,

I also attempted it like this:

Fn -mg = Fc (Since rotating in a circle, some acceleration due to rotation)

Fn = mg + Fc

570 N = 667 N + Fc

Fc at top = -111 Newtons

So since the seat is being pulled downward, and the normal force is less than the full gravitational force, the student should feel lighter than normal.

at the bottom:

Okay here is where I really get messed up:

Fn -mg = Fc

Fc is now +111 N, still pointing towards the center of the circle.

Fn = mg + Fc = 667 N + 111 N = 778 N.

Now, when the wheel's speed is doubled, how do I calculate the new centripetal force?
 
Last edited:
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  • #2
***ay = acceleration in the y direction, sorry. ^^
 
  • #3
***ohhhh wait, hold on, also have equation for Ac wrong, centripetal acceleration should be Ac = v²/R
 
  • #4
[tex]F_{CP}=m\cdot\frac{v^2}{R}[/tex]
What happens with F when v doubles?
 
  • #5
Mmm... If (2v)^2 is placed in there, the resulting centripetal force will be four times as large, since the radius is the same for both.
 
  • #6
So, now I go and repeat the correct calculations, using the equations for Newton's second law where Fnet = Fc.

Thank you!
^.^
 
  • #7
hagobarcos said:
Mmm... If (2v)^2 is placed in there, the resulting centripetal force will be four times as large, since the radius is the same for both.

Exactly.
 

Related to Ferris Wheel Forces: Exploring Normal Force and Centripetal Acceleration

1. What is the purpose of "Ferris Wheel Problem Trial 2"?

The purpose of "Ferris Wheel Problem Trial 2" is to conduct a second trial of the experiment in order to gather more data and refine the results from the first trial. This will help to further understand the behavior and mechanics of the Ferris wheel and potentially improve its design and functionality.

2. What was the outcome of the first trial of the "Ferris Wheel Problem"?

The first trial of the "Ferris Wheel Problem" yielded data that showed inconsistencies and flaws in the design of the Ferris wheel, leading to a less than optimal ride experience. This prompted the need for a second trial to address and improve upon these issues.

3. How is the data collected in "Ferris Wheel Problem Trial 2"?

The data in "Ferris Wheel Problem Trial 2" is collected through various methods such as video recordings, sensors placed on the wheel, and surveys from riders. This data is then analyzed and compared to the data from the first trial to determine any improvements or changes needed.

4. What are some potential variables that may affect the results of "Ferris Wheel Problem Trial 2"?

Some potential variables that may affect the results of "Ferris Wheel Problem Trial 2" include weather conditions, weight distribution on the wheel, and speed of rotation. Other factors such as maintenance and technical malfunctions may also impact the results.

5. How will the results of "Ferris Wheel Problem Trial 2" be used?

The results of "Ferris Wheel Problem Trial 2" will be used to improve the design and functionality of the Ferris wheel. The data collected will be analyzed and compared to the first trial to determine any necessary changes or adjustments. These results may also be used for future research and studies on amusement park rides.

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