Carnival ride Centripetal force question

In summary, the person would have to be standing very close to the center of the barrel to not feel the motion. If they are in a weightless environment, the centripetal force would be even stronger and they would not be able to move at all.
  • #1
AsaRand
8
0
First, please forgive my ignorance, as I do not really know enough to ask this question properly.

In the carnival ride, Barrel of Fun, the centripetal force holds you against the wall of the barrel while it spins. As anyone who has ridden this ride knows, you can feel the force holding you against the wall. While I never attempted it, I have thought about the idea of what might happen if someone decided to "stand up" on the wall while the barrel is moving. I have surmised the person doing it might become disoriented, maybe even dizzy, because of the speed the barrel is turning. This brings me to my question.

How far away from the center would the wall have to be for the centripetal force to exert 1g, and the person being held in place to not feel the motion?

A second question, would being in a weightless environment make a difference, since the direct affects of the gravitational pull of the Earth would be significantly reduced?
 
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  • #2
The distance from the centre is fairly irrelevant in this context. It is more down to the speed of the ride.

To ensure the riders are safe, you need to make sure they are pinned against the wall with more than 1g. That way if the wall tilts you, you will always overcome gravity.
 
  • #3
jarednjames said:
The distance from the centre is fairly irrelevant in this context. It is more down to the speed of the ride.

To ensure the riders are safe, you need to make sure they are pinned against the wall with more than 1g. That way if the wall tilts you, you will always overcome gravity.

The allegory to the ride was just a way to describe what it is I am trying to find out. I know that the speed of the rotation has an effect, but I don't know in what way, or to what extent. I am trying to figure out how large an object would have to be to allow someone to

* "stand" on the wall and not feel like they are being pushed into the wall
* walk around and not feel as though they are going to fall over,
be thrown about or in any other way negatively impacted
* basically feel the same as if they were standing on the surface of the planet

This is related to the concept of artificially simulated gravity.
 
  • #4
I did that barrel thing when I was a kid, and turned around and pushed myself away from the wall -- and started sliding down. So I quickly went back to the wall to avoid spoiling the ride.

I think that gravity pulls you downward constantly, but the force that pins you to the wall decreases as you, or part of you, moves away from the wall. I don't think you would stand a chance to stand on the wall.
 
  • #5
Oldfart said:
I did that barrel thing when I was a kid, and turned around and pushed myself away from the wall -- and started sliding down. So I quickly went back to the wall to avoid spoiling the ride.

I think that gravity pulls you downward constantly, but the force that pins you to the wall decreases as you, or part of you, moves away from the wall. I don't think you would stand a chance to stand on the wall.

That was the second part of my original question. I thought gravity on the planet's surface, or anywhere in the atmospheric envelope for that matter, would pull you down constantly, but I was not sure if it would matter in the ride because of the centripetal force pinning you to the wall. Your experiment as a child answers that. It obviously has an impact.

However, in the weightless environment of space, the pull of gravity is reduced. From what I've seen of the shuttle crews (and all other space flights in the past), it is negligible. Because of this, I believe you could stand on the wall, walk around, sit, even run and stay stuck to the wall. My thoughts have been that for the effect to simulate gravity, the centripetal force would have to equal 1g, though I am not certain that is correct.

Since the only part you really see of the barrel is the wall and the floor, you are unaware of the fact that the wall of the barrel is connected to the center by the same structure that spins the flloor (and raises it up and down). If you use the analogy of a ball attached to a string, the velocity of the string nearest the center is faster than the velocity at the ball.

What I am trying to figure out is, how fast would I have to spin the "wall" to achieve the correct centripetal force to allow someone at a distance from the center to not feel motion, and be able to move about freely without being thrown or tossed?

I think that maybe the speed at the center relates to the distance out (ie, faster spin = larger diameter wall, slower spin = smaller diameter wall) and achieve the same result. Not being a physicist, I really have no idea how to perform the calculations to get a specific answer, so I thought someone here might be able to throw out a few numbers and help me understand the dynamics involved.
 
  • #6
It's too late and tired to do numbers for you, but you can Google "Centripedal Force" and get more information than you'll ever need to get your numbers.

I'd speculate that standing on the wall in a rotating room that's only about ten feet in diameter might be pretty confusing in outer space, the balance organ in your ear would feel about zero g's and most of the rest of you would experience positive g forces. Though maybe you would get used to it, who knows? A 60 foot diameter spinning room would seem better, centripedal forces on your body would be more uniform. The important issues here are more related to biology than physics.
 
  • #7
Thank you for helping me. I am researching centripetal force more, and hopefully I can find the answers I seek.
 
  • #8
I did find an answer yesterday. It seems the minimum size necessary is approximately 1.8 miles radius, amounting to about 8 miles circumference, and to avoid the Coriolis effect the rate of spin would have to be between 1 and 2 rpm.
 

1. What is centripetal force?

Centripetal force is a force that acts on an object moving in a circular path, directed towards the center of the circle. It is responsible for keeping the object moving in a curved path instead of a straight line.

2. How does centripetal force work on a carnival ride?

In a carnival ride, centripetal force is created by the ride's mechanism, which keeps the riders moving in a circular path. The force is directed towards the center of the ride, and it is what makes the riders feel like they are being pushed outwards.

3. Is centripetal force the same as centrifugal force?

No, centripetal force is the force that keeps an object moving in a circular path, while centrifugal force is the apparent force that seems to push an object outwards from the center of rotation. Centripetal force is a real force, while centrifugal force is a fictitious force.

4. How is centripetal force calculated?

Centripetal force can be calculated using the formula Fc = mv²/r, where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

5. Can centripetal force be dangerous on a carnival ride?

Yes, if the centripetal force is too strong, it can cause the riders to experience discomfort or even injury. That's why ride designers carefully calculate and control the amount of centripetal force in carnival rides to ensure the safety of the riders.

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