Question regarding centripetal force

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Centripetal force on rides like the Gravitron pushes riders toward the center, while their momentum is tangential to the circular path. If the walls of the ride suddenly disappear, the force disappears too, leaving the rider to continue in a straight line along their current momentum, which is tangential. This means they will not fly directly outward but will move away from the center in a path that appears radial as they gain distance. Both linear and angular momentum are conserved, and the trajectory depends on the frame of reference, whether from the ride or the ground. Understanding these dynamics clarifies misconceptions about being "flung" outward when the ride's walls are removed.
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Hi just have a basic question. Just went to the local fair yesterday and road the Gravitron ride. The ride that sucks you to the walls. I became curious on the physics that cause you to get stuck to the wall so I have been researching centripetal force. Now it seems that the seats are exerting a force on me directly towards the center of a circle. I also understand that if hypothetically all of a sudden the walls disappeared on the ride I would move perpendicular to the force (tangental to the point on the circle). To me this makes little sense as I always thought I would move in a straight line directly in the opposite direction of the center of the circle. IF you have a force pushing you towards the center wouldn't that mean you would fly away from the center in a straight line assuming the walls suddenly disappeared. Please clarify this? Thank you.
 
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If the wall behind you suddenly disappears, then the force it is exerting on you goes to zero as well. That leaves you with the linear momentum (and angular momentum) that you had just before the wall disappeared. That linear momentum was tangent to the circular path you were traveling, so that is the direction you will fly off the (disintegrated) ride. There will likely be a rough landing!
 
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A basic law of motion is that you will move in a straight line if there are no forces on you. While you are in the ride, you feel the force on your back pushing you in radially, but your momentum is sideways (tangential). Momentum is conserved--force is not. When the walls disappear, it doesn't matter where the force used to be. It matters what your current momentum is. You will continue that trajectory until you hit something.
 
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People love to state that you are not 'flung outwards' in such circumstances. But it is not totally untrue because your path takes you ever further from the centre of rotation and, as the distance increases, the path becomes more an more approximately radial `(at least parallel to a radius).
 
There is a sense in which you do, in fact, fly directly away from the center of the ride. It's all about reference frames.

Suppose that instead of the wall completely disappearing, a hatch opens up that is big enough for you to fall through. From the point of view of the rotating ride, you will begin to drop radially outward. If you switch to the ground point of view, both you and the hatch will be moving tangentially. The hatch will accelerate radially inward away from you. Both descriptions are correct. You could continue to look straight through the hatch and see the hub of the ride. At least for a moment or two.

If you continue to maintain the viewpoint of the rotating ride, your trajectory would begin to be curved by Coriolis force, receding backwards compared to the rotating ride. If you continue to maintain the viewpoint of a ground observer, the hatch, of course, does not accelerate in a straight line but continues its circular motion while you move in a straight line. Both descriptions are correct. Either way you would lose sight of the hub of the ride. Your position would no longer fall on a line drawn radially outward from hub through hatch.
 
sophiecentaur said:
People love to state that you are not 'flung outwards' in such circumstances. But it is not totally untrue because your path takes you ever further from the centre of rotation and, as the distance increases, the path becomes more an more approximately radial `(at least parallel to a radius).

I think that this misses the point in that both linear and angular momentum are conserved, and the relevant moment arm for the "moment of momentum" calculation remains the radius of the rid, no matter how far the rider is thrown.
 
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Dr.D said:
I think that this misses the point in that both linear and angular momentum are conserved, and the relevant moment arm for the "moment of momentum" calculation remains the radius of the rid, no matter how far the rider is thrown.
I agree. But the difference becomes less and less - if you are considering the angle between the radial line to the projectile and the tangential line. The emphasis always seems to be in the effect at the point of release, where that anglular difference is 90 degrees and that really does conflict with what we often experience.
 
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