Uniform circular motion and Ferris wheel

AI Thread Summary
The discussion revolves around calculating the force exerted by the seat on a rider in a Ferris wheel. The rider's weight is correctly calculated as 343N, but the net force must also account for the centripetal force required for circular motion. At the halfway point, the seat must exert an additional upward force to provide the necessary centripetal acceleration, which is not considered in the initial calculation. Therefore, the correct force exerted by the seat is greater than just the weight of the rider. Understanding the distinction between vertical forces and the additional centripetal force is crucial for accurate calculations.
zolloz89
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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

please help me understand why my methods are wrong
thanks in advance
 
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zolloz89 said:
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

please help me understand why my methods are wrong
thanks in advance

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
 
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).
 
i think it is only asking about the vertical force on the rider exerted by the seat
 
zolloz89 said:
i think it is only asking about the vertical force on the rider exerted by the seat
...which equals to his weight.
 
yeah so that would equal normal force which would be 35kg x 9.8m/s^2, right
 
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