Circular Motion of a Ferris wheel

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SUMMARY

The discussion focuses on calculating the centripetal acceleration and effective weight of a person on a Ferris wheel with a radius of 16 meters and a rotation period of 20 seconds. The centripetal acceleration is determined using the formula a = ω² * r, where ω is derived from the wheel's rotation speed. The effective weight at the highest point is calculated as F = mg + ma, while at the lowest point, it is F = mg - ma. The correct values for these calculations yield an effective weight of 512.91 N at the top and 370.89 N at the bottom.

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  • Understanding of centripetal acceleration and its calculation
  • Familiarity with Newton's laws of motion
  • Knowledge of free body diagrams
  • Ability to calculate angular velocity (ω) from rotational period
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  • Learn how to derive angular velocity from rotational motion
  • Study free body diagrams to analyze forces in circular motion
  • Explore the relationship between centripetal force and gravitational force
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to explain concepts of centripetal force and effective weight in rotating systems.

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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 that's correct really.
and also strugling with b) and c)

But here's 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?
 
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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.
 
Oh yeah it was -2 not -1. Spelling error.


I Used for omega, ω = \frac{(2π*3)}{60}
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
 
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?
 
SHISHKABOB said:
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 = 370,89NOr is this as wrong as I can get? I'm new to this things so I might not be so good at it.
 
yeah that looks correct
 
Okey thanks for the help!
 
2Big2Bite said:
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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