Ferris wheel problem - vertical circles and centripetal acceleration

In summary, the problem involves finding the centripetal acceleration and forces experienced by a rider on a Ferris wheel that rotates 5 times per minute with a diameter of 19m. The answer to part a is 2.6018 m/s^2, while the answers to parts b, c, and d are 518.2676 N, 518.2676 N, and 72 kg, respectively. The problem also involves considering the rider in a non-inertial frame and taking into account pseudo forces such as the centrifugal force. The force due to gravity does not need to be considered as the Ferris wheel is powered by a motor.
  • #1
therest
25
0

Homework Statement


Given a Ferris wheel that rotates 5 times each minute and has a diameter of 19 m, with the acceleration of gravity as 9.8 m/s^2, what is the centripetal acceleration of a rider? Answer in units of m/s^2.

(There's a diagram that shows a ferris wheel with radius 9.5 m spinning at ω radians per second.)

Homework Equations


centripetal acceleration = v2 / r = ω2r

The Attempt at a Solution


one rotation was equal to 2pi * r, so
5 rotations = 5(2pi * r) radians in one minute
or 10pi * r / 60 radians/second = ω

I used centripetal acceleration = ω2r = (10pi * r / 60)2 * 9.5. I got 4.97 for centripetal acceleration. But this was wrong! I don't know where I made a mistake.

Oh, also, how does the force due to gravity figure into this problem? Shouldn't it be ignored since the ferris wheel is being powered by a motor?

----------
While I'm at it, I might as well ask about the other three parts of the problem.
Part b

Homework Statement


What force does the seat exert on a 72 kg rider at the lowest point of the ride? Answer in units of N.

Homework Equations


Fnet = Fcentripetal, so since net force must be upwards to maintain a centripetal force, normal force will be stronger than net force.
So: Fnormal - Fg = Fnet = Fc = m * v2/r
Fnormal = Fg + Fc = m(g + ac)

The Attempt at a Solution


Isn't the force exerted by the seat the normal force?
I think it would just be the mass times the sum of acceleration due to gravity and the centripetal acceleration, or m(g + ac). I can't do this though as I have not found centripetal acceleration yet. That was part one.

Part c

Homework Statement


What force does the seat exert on a 72 kg rider at the highest point of the ride? Answer in units of N.

Homework Equations


At the highest point, Fnet = Fc = Fg - Fnormal
Fnormal = mg - mac

The Attempt at a Solution


Operation seems wasy with the information from part a.

Part d

Homework Statement


What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom? Answer in units of N.

Homework Equations


Fg = mg

The Attempt at a Solution


Wouldn't this just be the weight of the rider since there would be no centripetal force pulling him up or down? Or would it be the equilibrant to the force due to gravity nd the centripetal force?

Thanks for help in advance!
Bonus information: I have only two hours to submit the answer to this problem electronically. Because I am a procrastinator. :(
 
Last edited:
Physics news on Phys.org
  • #2
The way I solve it, I consider that the Ferris wheel seat is a part of the wheel, and since the seat is at the rim of the wheel, we can say that the centripetal acceleration of the seat is (w^2)r, where w=(2pi x no. of rotations per min)/ 60.

we get a = 2.6018 m/s^2.
This gives us part (a).

Now for (b) and (c) :
We consider, the person to be in a non inertial frame that is accelerating at 2.6018 m/s^2 towards the centre. As a result, i consider, he is facing a pseudo force of the same, in a dir'n opposite to the dir'n of the centripetal force.

At the highest point, we use the equation of equilibrium.
We know Normal force, N is in the upward direction, and the centrifugal force F is in the upper direction too.
The weight of the person, W, is in the downward direction now.

This gives us an equation : N=W-F.

Now substitute values, and the answer is N=518.2676 N.

(d) halfway between top and bottom: the seat is again vertical, but the centripetal force is in the direction of the centre. Hence the centrifugal force (non inertial frame again) is in the horizontal outward direction ! As a result the net force experienced by the man is N=W. This solves it.

Now, we are considering the man sits still, and does not cause the seat to oscillate. :)

part (b) again : see, we consider the centripetal force as the net force.
hence : the net force is inward.
this means that the weight is greater than the normal reaction by the seat.
as a result : centripetal force=W-N.

therefore : N=W-ma ( considering that we subsequently note centripetal acc- centrifugal acc, only dir'ns are opp. as the frame of references are not the same.)
Also, centripetal force is real, while centrifugal force is pseudo, a concept :)

Somak.
 
  • #3


I would like to first commend you for taking the time to understand and solve this problem. It shows a great level of dedication and curiosity towards science. Now, let's break down the problem and address each part individually.

Part a:
Your approach to finding the centripetal acceleration is correct. However, the mistake you made was in the units. In your calculation, you used meters for the radius, but in the given answer, the units should be m/s^2. This is because the final answer should be in terms of acceleration, which is measured in m/s^2. So, the correct answer for the centripetal acceleration would be 4.97 m/s^2.

Part b:
You are correct in thinking that the force exerted by the seat is the normal force. And you are also correct in using the equation Fcentripetal = m * v^2/r. However, in this case, the rider is at the bottom of the ride, where the net force is equal to the centripetal force, so the equation becomes Fnet = Fcentripetal = Fnormal + Fg. Plugging in the values, we get Fnormal = 72 kg * (9.8 m/s^2 + 4.97 m/s^2) = 1,044.24 N.

Part c:
At the highest point of the ride, the net force is equal to the centripetal force, which is equal to the weight of the rider. So the equation becomes Fnet = Fcentripetal = Fg - Fnormal. Plugging in the values, we get Fnormal = 72 kg * (9.8 m/s^2 - 4.97 m/s^2) = 356.16 N.

Part d:
At the halfway point, the rider is experiencing both the force due to gravity and the centripetal force. So the equation becomes Fnet = Fg + Fcentripetal = Fnormal. Plugging in the values, we get Fnormal = 72 kg * (9.8 m/s^2 + 2.485 m/s^2) = 846.72 N.

Bonus information:
I would also like to remind you that procrastination is not a good habit and it can lead to mistakes and stress. It's always better to start early and give yourself enough time to understand and solve problems. Good
 

Related to Ferris wheel problem - vertical circles and centripetal acceleration

1. What is the Ferris wheel problem and why is it important?

The Ferris wheel problem is a classic physics problem that involves a rotating wheel and the calculation of centripetal acceleration. It is important because it allows us to understand and analyze the forces at play in circular motion, which is a fundamental aspect of many real-world situations and technologies.

2. How does the height of the Ferris wheel affect the centripetal acceleration?

The height of the Ferris wheel does not directly affect the centripetal acceleration. The centripetal acceleration is determined by the speed of rotation and the radius of the wheel, not the height. However, a taller Ferris wheel may require a faster speed to maintain a comfortable ride, which would result in a higher centripetal acceleration.

3. What is the relationship between the speed of the Ferris wheel and the centripetal acceleration?

The speed of the Ferris wheel and the centripetal acceleration are directly proportional. This means that as the speed of the wheel increases, the centripetal acceleration also increases. This relationship is described by the equation a = v^2/r, where a is the centripetal acceleration, v is the speed, and r is the radius of the wheel.

4. Can a person experience weightlessness on a Ferris wheel?

No, a person cannot experience true weightlessness on a Ferris wheel. This is because the centripetal force exerted by the wheel is always acting on the person, keeping them in circular motion and preventing them from falling. However, a person may feel a temporary sensation of weightlessness at the very top of the wheel when the centripetal force is momentarily equal to their weight.

5. How does the mass of the riders affect the centripetal acceleration on a Ferris wheel?

The mass of the riders does not directly affect the centripetal acceleration. The centripetal acceleration is determined by the speed and radius of the wheel, not the mass of the riders. However, a heavier load on the wheel may require a faster speed to maintain a comfortable ride, resulting in a higher centripetal acceleration.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
221
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
28
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Back
Top