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Circular Motion of a Ferris wheel

  • Thread starter 2Big2Bite
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  • #1
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Homework Statement


An Ferris Wheel has a radius of 16m and rotates once every 20 seconds.

a) Find Centripedal acceleration.
b) Whats the effektive wheight of a 45 kg person at highest point?
c) Lowest point?

Homework Equations


a) I tried using the a=ω2*r in a) but not sure if its right
b)
c) We = mg - ma
Is this the correct equation for a and b?



The Attempt at a Solution


I got out 1,578m*s-2 in a) but not sure if thats correct really.
and also strugling with b) and c)

But heres c) atleast trying using that equation:
45kg*9,82m*s-2 - 45kg*1,578m*s-2= 370,89N Is that right?

and b) do i use same equation for upward motion?
 
Last edited:

Answers and Replies

  • #2
537
1
for a you should end up with an acceleration, not a velocity

a=ω^2*r works but you need the correct value of omega, what did you use?

b and c are just adding up the forces acting at those points. Just make sure you have a good understanding of which direction the centripetal force is directed.
 
  • #3
5
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Oh yeah it was -2 not -1. Spelling error.


I Used for omega, ω = [itex]\frac{(2π*3)}{60}[/itex]
The motions are up and down verticaly, but not sure if that formula is right or if i need to add other things to it
 
  • #4
537
1
well, basically the centripetal force is always pointing towards the center of the circle, and the force due to gravity is always pointed downwards.

So at the top, they are pointed in the same direction, at the bottom they are pointed in opposite directions. So how would you find the net force acting on someone at the top and then at the bottom?
 
  • #5
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well, basically the centripetal force is always pointing towards the center of the circle, and the force due to gravity is always pointed downwards.

So at the top, they are pointed in the same direction, at the bottom they are pointed in opposite directions. So how would you find the net force acting on someone at the top and then at the bottom?
Well when going upward you add weight since your pushed "into the seat", and down you take weight when the seat accelerate away from you.

So it would be F = mg+ma at the top since its "added weight"
45kg*9,82m/s^2+45*1,578m/s^2= 512,91N

And when you go down it will be the same but with (-):

45kg*9,82m/s^2-45*1,578m/s^2 = 370,89N


Or is this as wrong as I can get? I'm new to this things so I might not be so good at it.
 
  • #6
537
1
yeah that looks correct
 
  • #7
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Okey thanks for the help!
 
  • #8
PhanthomJay
Science Advisor
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Well when going upward you add weight since your pushed "into the seat", and down you take weight when the seat accelerate away from you.

So it would be F = mg+ma at the top since its "added weight"
45kg*9,82m/s^2+45*1,578m/s^2= 512,91N

And when you go down it will be the same but with (-):

45kg*9,82m/s^2-45*1,578m/s^2 = 370,89N


Or is this as wrong as I can get? I'm new to this things so I might not be so good at it.
This is not correct. The person's effective or 'apparent' weight is the magnitude of the normal force of the seat that pushes up on him or her. The person feels lighter at the top than at the bottom. Use good free body diagrams and newton's 2nd law to find the normal force at the top, and then at the bottom.
 

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