Circular Motion: Ferris Wheel Dynamics

In summary: Your understanding of "apparent weight" is correct. All the forces in the problem are proportional to mass. It will cancel out.
  • #1
Mehta29
17
0
I'm on the right track, but I'm stuck here...

The radius of a Ferris wheel is 5 m and it makes one rev in 10 sec

a Find the difference b/w the apparent weight of a passenger at the highest and lowest points, expressed as a fraction of his weight, W

b What would the time for one rev be if teh apparent weight at the top were zero?
c What would be the apparent weight at the low point??

I have at the top that

mv^2/r = mg - Fn

and at the bottom

mv^2/r = Fn -mg

I really don't undestand what a is asking...(Fnbottom - Fntop)/ W ??
 
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  • #2
Mehta29 said:
I really don't undestand what a is asking...(Fnbottom - Fntop)/ W ??
Yes you do. It's exactly what you said. Write your answer as Fnb - Fnt = _______W
 
  • #3
but I am confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
Fnbottom - Fntop = xW

b would just be v^2/r = g? and then t = 2pir/v

and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
 
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  • #4
Mehta29 said:
but I am confused as if anything else would be needed...liek that blank befor e the W...would i need to expand any further or would i just keep it
Fnbottom - Fntop = xW

b would just be v^2/r = g? and then t = 2pir/v

and c would be Fnb = mv^2/r + mg...but how would i eliminate m?
Solve your earlier top and bottom equations for Fn. Take the difference between the two. Your result will be of the form
Fnb - Fnt = mA where A is a number that can be computed from the given information. You can do that. Once you have that form, multiply and divide by g
mA = mgA/g = WA/g
You can calculate A/g to express your answer as
Fnb - Fnt = (A/g)W with A/g replaced by a number.
 
  • #5
i got A and B...but how would i manage part c?

im not seein anyway to cancel out the mass...
 
  • #6
Mehta29 said:
i got A and B...but how would i manage part c?

im not seein anyway to cancel out the mass...
What is your understanding of "apparent weight". All the forces in the problem are proportional to mass. It will cancel out.
 

FAQ: Circular Motion: Ferris Wheel Dynamics

What is circular motion?

Circular motion refers to the movement of an object along a circular path. This type of motion is characterized by a constant speed and a change in direction, as the object continuously moves around the center of the circle.

How does a ferris wheel use circular motion?

A ferris wheel is designed to rotate around a central axis, using a motor to provide the necessary circular motion. As the wheel turns, the riders move along a circular path, experiencing both a change in direction and a constant speed.

Why do ferris wheels rotate in a circle?

Ferris wheels are designed to rotate in a circle to provide an enjoyable and thrilling experience for riders. The circular motion allows for a smooth and continuous ride, while also providing a sense of height and movement.

What is the centripetal force in a ferris wheel?

The centripetal force in a ferris wheel is the force that keeps the riders moving in a circular path. This force is provided by the structure of the wheel and the motor that drives its rotation.

How does the speed of a ferris wheel affect the riders?

The speed of a ferris wheel can affect the riders in different ways. A slower speed may provide a more relaxed and leisurely experience, while a faster speed can increase the thrill and excitement of the ride. The speed also affects the centripetal force, which can impact the feeling of weightlessness or g-forces experienced by the riders.

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