# Uniform circular motion and Ferris wheel

1. Feb 21, 2006

### zolloz89

my teacher said my answer is wrong but i cannot figure out why

there is a ferris wheel. a rider in the ferris wheel is 35kg. the radius of the ferris wheel is 12m. it asks what is the magnitude of the force does the seat exert on the rider when he is halfway between the top and the bottom?

i got 343N by using the force summation equation, Fnet = Fn + Fg = 0 because there are no other forces acting on the rider in the vertical direction
Fg=35kg x 9.8m/s^2 = 343, Fn=-Fg, Fn=-(-343) = 343N

2. Feb 21, 2006

### topsquark

The Ferris wheel is moving the rider in a circle, presumably at a constant angular speed, so the net force on the rider is equal to the centripetal force. In the vertical direction, you are correct: Fnet(vert)=0 N. What about in the direction of the center of the Ferris wheel? Since the rider is not sliding in that direction, friction must be keeping him there. That's the origin for the centripetal force at this point in the motion.

-Dan

3. Feb 21, 2006

### gulsen

Between top and bottom, the centipetal force is pointing inwards, in horizontal, not vertical. And let's assume it wasn't, if you balance out centripedal force, how would he accelerate and draw a cirle (relative to the observe on the ground, of course).

4. Feb 21, 2006

### zolloz89

i think it is only asking about the vertical force on the rider exerted by the seat

5. Feb 21, 2006

### gulsen

...which equals to his weight.

6. Feb 21, 2006

### zolloz89

yeah so that would equal normal force which would be 35kg x 9.8m/s^2, right