What is the centripetal acceleration of the Ferris wheel?

  • #1
A 39.0 kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 17.0 m.

(a) What is the centripetal acceleration of the child?


(b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride?

(c) What force does the seat exert on the child at the highest point of the ride?

(d) What force does the seat exert on the child when the child is halfway between the top and bottom?

Ok I need to know theory and a few questions about how to find Angular Velocity.

I understand V=RW. I also understand I can take the A=v2/r to get the centripetal acceleration.

I forgot how to take 4 revs per minute and convert it over to a angular velocity using a radius of 8.5.

but even when I finally get the angular velocity and centripetal acceleration, I am not sure how B,C, and D will be getting different answers. Since it is a radius shouldnt they all share the same distance and gravity values? Wether it is the highest point and lowest point they share the same distance and weight doesnt change??

Angular velocity...I take the distance of the circle, which would be 17*Pie=53.4.

So it revolutes 4 times per minute. This is where I am drawing a blank...this stuff use to be so easy for me 2 years ago but I am seriously drawing a blank here...ugh
 

Answers and Replies

  • #2
47
0
Well, ω is measured in rad/s, so you need to multiply revs by 2π to get the radian representation and then divide by 60 to turn minutes into seconds.
B,C,D give different answers because the weight of the child has a different relation to the centripetal force at each position. (For example, while the weight acts down towards the centre of the wheel at the highest point, it acts away from the centre at the lowest point.)
 

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