# How Does Weight Distribution Affect the Balance of a Ferris Wheel?

• Jeremy74
In summary, the wheels with two, three, and four weights will stay in equilibrium by having the weights evenly spaced around the circumference.
Jeremy74
Homework Statement
Hello Everyone

This is my first question on this forum, I need help with something that is easy to observe in the real world, and on any real life model of a Ferris wheel, that is calculating which part of the Ferris wheel will be closest to the ground - 'pointing downwards' depending on the number of weights placed on its rim, and the angles between those weights. In other words, the simplest example: if we place one weight on the rim of the wheel - why thanks to gravity this weight will eventually be at the very bottom, and not X degrees from it?

The main issue is that most of the calculations relating to Ferris wheels that I could find talk about Ferris wheel that is already moving, and how many rotations it makes in a given time, which I don’t think would help me to find the answer to this question.
Relevant Equations
The closest thing to what I was trying to find was this website, which talks about calculating potential energy in a pendulum:
https://blogs.bu.edu/ggarber/interlace/pendulum/energy-in-a-pendulum/

I based my calculations on those equations for potential energy:
PE = mgh
PE = mgL(1 – COS θ)

With the following measurements:
Length of pendulum L=1 (meter)
Mass m=1 (kg)
Gravitational acceleration g=9.8 (m/s2)
I made a sample ‘test run’ with one, two, three and four weights of equal mass.

Wheel number 1: one weight placed on the rim.
Wheel number 2: two equal weights placed opposite to each other at 180 degrees angle, (‘diameter configuration’)
Wheel number 3: three equal weights placed at 120 degree angles, (‘equilateral triangle configuration’)
Wheel number 4: four equal weights placed at 90 degree angles, (’square configuration’)

For the purpose of those calculations the lowest point on the Ferris wheel is at 0 degrees, highest at 180 degrees, and two points halfway there are at 90 and 270 degrees.

Those pendulum equations seem to work great for wheel number 1 with only one weight:
If the weight is exactly at the bottom at 0 degrees, potential energy of the whole wheel is EP=0
If the weight is halfway there at 90 or 270 degrees, potential energy of the whole wheel is EP=9.80665 J
If the weight is at the very top at 180 degrees, potential energy of the whole wheel is EP=19.6133 J

I assumed that the configuration with the lowest potential energy of the whole wheel will be the one where the wheel won’t move, and thus the first configuration at 0 degrees was the one that would resemble real world situation with weight at the bottom.

Now, moving to wheel number 2, once I add the second weight to this picture things start to get tricky, because the potential energy of the whole wheel stays the same regardless of the position of those weights: it’s the same in both horizontal (90 and 270 degrees), and vertical (0 and 180 degrees) position, EP=19.6133 J

The same is true for wheels number 3 and 4, with three and four equal weights, as long as the angles between them are equal too. So the method of calculating the lowest potential energy of the whole wheel doesn’t work anymore.

I thought that perhaps I could calculate the potential energy from weights located in the upper half of the wheel, then subtract the potential energy from the weights located in the lower half of the wheel, and then choose the position of the wheel with the lowest difference in potential energy, which ‘kind of works’ for some configurations (wheels with one and three weights):

For example in wheel number 3 with three weights:
If the weights are positioned at 0, 120, and 240 degrees (one at the bottom, two in the upper half), the difference in the potential energy between upper and lower half is 29.41995 J

If in the same wheel the weights are positioned at 60, 180 and 300 degrees (two in the lower half, and one at the top), the difference in the potential energy between the upper and lower half is 9.80665 J
(I think that this position would resemble real world as well, because it would be ‘heavier’ in the lower half)But then new problem arrives especially in wheels with two and four weights, when I have weights at 90 and 270 degrees (do I count them to the upper or the lower half?).

For example in the wheel number 2 with two weights:
If the weights are positioned at 0 and 180 degrees (vertically - one at the bottom and one at the top), the difference in the potential energy between upper and lower half is 19.6133 J

If in the same wheel the weights are positioned at 90 and 270 degrees (horizontally) – should I count them to the upper or the lower half, or do something else?

Because if I count both of them as belonging to the lower half, then he difference in the potential energy between the upper and lower half is
–19.6133 J (that’s with minus – so I assume there’s something wrong).

And if I include both of those weights in the upper half of the wheel, the difference in the potential energy between the upper and lower half is 19.6133 J (this time with plus).

Overall I think that I’m missing something, or using the wrong equations to this Ferris wheel problem. I would really appreciate some advice on this matter. Thank you in advance.

Jeremy74 said:
once I add the second weight to this picture things start to get tricky, because the potential energy of the whole wheel stays the same regardless of the position of those weights:
I think you mean, regardless of the orientation of the wheel, keeping the weights equally spaced.
It's not tricky, it's just in dynamic equilibrium. It will stay wherever you put it.

The challenge is to consider arbitrary placements. Think about the mass centre of the system.

Jeremy74
Thank you for your response haruspex. You’re right - ‘regardless of the orientation of the wheel, keeping the weights equally spaced’ is what I had in mind.

I’ve read more about dynamic equilibrium, and the way I understand it is that those wheels with 2, 3, and 4 equal weights would ‘stay put’ in whatever orientation someone would place them, as long as no external force would be applied, and the overall total potential energy would stay the same (which would be true if the angles between the weights would be equal as well). Or am I interpreting it incorrectly?

Does it mean that I’m basically overthinking the calculations from my first post for wheels with 2, 3, and 4 equal weights, when I was dividing the wheel into upper and lower half and are they just wrong?

Also is calculating the total potential energy of the wheel the right way of calculating its orientation, or should I use different equations. I was using the potential energy ones because they looked the most promising at the beginning, but are there some other methods out there that I could use?

Thank you

Jeremy74 said:
I’ve read more about dynamic equilibrium, and the way I understand it is that those wheels with 2, 3, and 4 equal weights would ‘stay put’ in whatever orientation someone would place them, as long as no external force would be applied, and the overall total potential energy would stay the same
Yes.
Jeremy74 said:
overthinking the calculations from my first post for wheels with 2, 3, and 4 equal weights, when I was dividing the wheel into upper and lower half
Yes, separating into halves was not useful.
Jeremy74 said:
Also is calculating the total potential energy of the wheel the right way of calculating its orientation
The PE will not directly give you the orientation. You would have to find. Where the PE is minimised. As I posted, the easy way is to find the mass centre of the set of weights. Where will that be when it settles?

## What is a Ferris wheel gravity balance?

A Ferris wheel gravity balance is the force that keeps a Ferris wheel in equilibrium, preventing it from tipping over or falling.

## How does a Ferris wheel maintain its balance?

A Ferris wheel maintains its balance through a combination of its weight, the weight of the riders, and the centripetal force created by its rotation. The weight of the wheel and riders creates a downward force, while the centripetal force pushes outward, balancing the wheel.

## What factors affect the gravity balance of a Ferris wheel?

The gravity balance of a Ferris wheel can be affected by the weight and distribution of the riders, the speed of the rotation, and external forces such as wind or uneven ground.

## How is the gravity balance of a Ferris wheel calculated?

The gravity balance of a Ferris wheel can be calculated by considering the weight of the wheel, the weight and distribution of the riders, and the radius and speed of the rotation. This calculation is based on principles of Newton's laws of motion.

## Why is it important to maintain the gravity balance of a Ferris wheel?

Maintaining the gravity balance of a Ferris wheel is important for the safety and stability of the ride. If the balance is not maintained, the wheel could tip over or become unstable, posing a danger to the riders and those around the ride.

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