# Ferris Wheel Problem Trial 2

1. May 4, 2014

### hagobarcos

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Fc = m(v²/R)

Ac = (v²/R)

Fnet = m a

3. 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: May 4, 2014
2. May 4, 2014

### hagobarcos

***ay = acceleration in the y direction, sorry. ^^

3. May 4, 2014

### hagobarcos

***ohhhh wait, hold on, also have equation for Ac wrong, centripetal acceleration should be Ac = v²/R

4. May 4, 2014

### mafagafo

$$F_{CP}=m\cdot\frac{v^2}{R}$$
What happens with F when v doubles?

5. May 5, 2014

### hagobarcos

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. May 5, 2014

### hagobarcos

So, now I go and repeat the correct calculations, using the equations for Newton's second law where Fnet = Fc.

Thank you!!
^.^

7. May 5, 2014

Exactly.