Ferris Wheel acceleration Problem

In summary: The direction of the net force is determined by the direction of the torque, which is caused by the rotation of the wheel.Careful: While the net force has same magnitude at top and bottom, it does not have the same direction. The direction of the net force is determined by the direction of the torque, which is caused by the rotation of the wheel.
  • #1
JoshMP
38
0

Homework Statement



While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 15 m, and you use your watch to find that each loop around takes 25 s.

What is the magnitude of your acceleration?
What is the ratio of your weight at the top of the ride to your weight while standing on the ground?
What is the ratio of your weight at the bottom of the ride to your weight while standing on the ground?

Homework Equations



F=ma

The Attempt at a Solution



I started by finding the angular velocity. I don't know how to find the angular acceleration or the tangential acceleration.
 
Physics news on Phys.org
  • #2
JoshMP said:
I don't know how to find the angular acceleration or the tangential acceleration.
If we presume that the wheel turns at a uniform rate, those will be zero. What other acceleration is involved?
 
  • #3
Doc Al said:
If we presume that the wheel turns at a uniform rate, those will be zero. What other acceleration is involved?

Radial acceleration. Got it, thanks.

So what about the ratios of weight...would they just be 1 to 1?
 
  • #4
JoshMP said:
So what about the ratios of weight...would they just be 1 to 1?
No. (That would be too easy.) When they say "weight", they really mean apparent weight, not the force of gravity. You'll need to analyze forces using Newton's 2nd law.
 
  • #5
Doc Al said:
No. (That would be too easy.) When they say "weight", they really mean apparent weight, not the force of gravity. You'll need to analyze forces using Newton's 2nd law.

Ok tell me if I'm on the right track. I know that the radial acceleration is 0.95 m/s^2, which means that the net force is pointing down at the top of the wheel and up at the bottom. I use this net acceleration and solve for Fnet. Does this value of Fnet = the apparent weight?
 
  • #6
JoshMP said:
Does this value of Fnet = the apparent weight?
No, but you'll use it to determine the apparent weight. The apparent weight of an object is the normal force exerted by the surface supporting it. Find that normal force.
 
  • #7
Ok I got the ratio at the bottom correct. But I'm having trouble with the top. My FBD has gravity pointing down and normal pointing down (is that right? - it's a Ferris wheel...). When I solve for the normal force, I get a negative value, which is impossible!
 
  • #8
JoshMP said:
My FBD has gravity pointing down and normal pointing down (is that right? - it's a Ferris wheel...).
No. Think of yourself sitting in a seat which is always upright. The normal force of the seat on you is always up.
 
  • #9
Doc Al said:
No. Think of yourself sitting in a seat which is always upright. The normal force of the seat on you is always up.

Wouldn't that mean that the normal force at the top is equal to the normal force at the bottom? If the net force is the same at the top and bottom, and gravity is the same too, then the normal force must also be the same, and the ratios would be the same too.
 
  • #10
JoshMP said:
Wouldn't that mean that the normal force at the top is equal to the normal force at the bottom? If the net force is the same at the top and bottom, and gravity is the same too, then the normal force must also be the same, and the ratios would be the same too.
Careful: While the net force has same magnitude at top and bottom, it does not have the same direction.
 

Related to Ferris Wheel acceleration Problem

What is the Ferris Wheel acceleration problem?

The Ferris wheel acceleration problem refers to the calculation of the acceleration of a passenger on a Ferris wheel at any given point during the ride. It takes into account factors such as the radius of the wheel, the speed of rotation, and the position of the passenger on the wheel.

Why is the Ferris Wheel acceleration problem important?

The Ferris Wheel acceleration problem is important because it helps us understand the forces acting on passengers during a ride. This can be useful for designing safer and more comfortable Ferris wheels, as well as for predicting the experience of riders.

What is the formula for calculating acceleration in the Ferris Wheel acceleration problem?

The formula for calculating acceleration in the Ferris Wheel acceleration problem is a = v2/r, where a is the acceleration, v is the tangential velocity, and r is the radius of the wheel. This formula is derived from the centripetal acceleration equation, a = v2/r, where a is the centripetal acceleration, v is the linear velocity, and r is the radius.

How does acceleration change on a Ferris wheel?

Acceleration on a Ferris wheel constantly changes as the wheel rotates. At the bottom of the wheel, the acceleration is at its maximum, as the rider is moving in a circular motion with the greatest speed. At the top of the wheel, the acceleration is equal to the centripetal acceleration, which is directed towards the center of the wheel.

Are there any safety concerns related to the Ferris Wheel acceleration problem?

Yes, there are safety concerns related to the Ferris Wheel acceleration problem. If the acceleration is too high, it can cause discomfort or even injuries to riders. Therefore, it is important for engineers to carefully consider the acceleration when designing and operating Ferris wheels.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
6K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
31K
Back
Top