Posted by **Jon** on Wednesday, December 15, 2010 at 12:35pm.

A ferris wheel, radius 5.0m, rotates once every 8.4 seconds.

a) What is the centripetal acceleration of a 60kg person at the circumference?

b) calculate the force of the seat on the person at the highest point.

c) calculate the force of the seat on the person at the lowest point.

d) Assume the seat is horizontal, that it does not have a back and that the seat surface is rough. If the person on the seat is midway between the highest and lowest point on the ferris wheel, calculate the minimum coefficient of static friction between the seat and the person in order that the person does not fall off the ferris wheel.

So.

a) 2.8m/s^2

b) 420N [upwards]

c) 756 N [upwards]

d) I'm not sure how to approach this question. The answer in the back of the book says 0.286, but can someone describe how to get to the solution please

- Physics -
**drwls**, Wednesday, December 15, 2010 at 1:39pm
The outer rim speed is

V = 2 pi*R/t = 3.74 m/s

a) the centripetal acceleration is

V^2/R = 2.80 m/s^2

(b) M (g-a) = 420 N

(c) M (g+a) = 756 N

d) M a*mu_s = M g

mu_s = g/a = 3.5

You'd need something like Velcro for that high a static friction coefficient

- Physics -
**Jon**, Wednesday, December 15, 2010 at 2:09pm
I got 3.5 as well!

But the solutions manual says 0.286?

- Physics -
**Sabrina**, Thursday, December 13, 2012 at 12:58pm
I know this is an old question, but I was just working through it myself and found the answer. So I will post it here for future people trying to solve this question.

Nu_s = ma

mg*u_s = ma

Therefore, u_s = a/g = 2.80/9.8

u_s = 0.286

:)

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