Centripetal Force of an amusement park ride

In summary, the conversation discusses the calculation of the force exerted by an amusement park ride on a rider standing inside a rotating cylinder. The correct answer is the net force exerted by the ring and gravity on the rider, rather than just the centripetal force provided by the ring. This is due to the use of imprecise language in the question, which could have been clarified if asked for a clarification.
  • #1
kevin17ym
7
0

Homework Statement


In an amusement park ride, passengers stand inside an 8m radius cylinder. Initially, the cylinder rotates with its axis oriented along the vertical. After the cylinder has acquired sufficient speed, it tilts into a vertical plane, that is, the axis tilts into the horizontal. Suppose that, once the axis has tilted into the horizontal, the ring rotates once every 4.5 seconds. If a rider's mass is 40kg, with how much force does the ring push on her at the top of the ride?


Homework Equations


F = ma, W = mg, Centripetal Force = mvv/r


The Attempt at a Solution


Since it's asking for the force that the wall of the ride is pushing on the person, shouldn't it just be mvv/r? So the answer is about 620N? Yet the correct answer is mvv/r - mg, which is 230N.
Why is that? Why does gravity matter if the question is asking "how much force does the ring push on her at the top of the ride?
 
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  • #2
It's a question of semantics - how hard the ring pushes on the person is the force felt by the person is what they are after. In general, "how hard you get pushed" refers to the net force - what you get pushed by is whatever is in the direction the net force appears to come from.
 
  • #3
So, the correct answer should be mvv/r? Not the net force, since the question is asking how hard the ring pushes on her, not how hard does the person get pushed? Do you get 620N if you do mvv/r?

also, v = wr
 
  • #4
That's what I mean - it's semantics: you have to be able to tell when the question is using sloppy everyday language and when it is using careful physics jargon. In this case it was sloppy everyday ... they wanted the net force.

It's very borderline though.
If the question were asked here, I'd have asked for a clarification before answering.
If I were you I'd complain: it is not clear from the wording if the question wants the effective force from the ring or if it wants the contribution to the effective force from the ring. i.e. is the question to be answered from an inertial or a non-inertial POV?

If you look at it when the person is on the other side of the ring, then they feel the ring exert a stronger force - part of the stronger force comes from gravity.

I think the writer was relying on the shift to talking about the force that the person feels would clue the reader to realizing that the persons POV was needed.
 
  • #5
kevin17ym said:
So, the correct answer should be mvv/r?
No. How hard the ring pushes on the person is the same as how hard the person pushes on the ring - action and reaction are equal and opposite. Pace Simon, but I see no sloppiness here.
If the ring pushes on the person with force F, what is the sum of the vertical forces on the person? What is the person's vertical acceleration? What equation does that give you?
 
  • #6
haruspex said:
No. How hard the ring pushes on the person is the same as how hard the person pushes on the ring - action and reaction are equal and opposite. Pace Simon, but I see no sloppiness here.
OH riight - I misread it.
The question asks for the force of the ring on the person - the centripital force on the person is provided by the ring and by gravity.
 
  • #7
Okay thx.
 

What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle and is necessary to maintain the object's constant speed and direction.

How does centripetal force work in an amusement park ride?

In an amusement park ride, centripetal force is generated by the ride's motion. As the ride moves in a circular path, it produces a force that acts towards the center of the circle, keeping the riders moving along the path. This force is what allows the riders to experience the sensation of being pressed against the ride's walls.

What factors affect the centripetal force of an amusement park ride?

The speed of the ride, the radius of the circular path, and the mass of the riders all affect the centripetal force of an amusement park ride. The faster the ride moves, the greater the centripetal force will be. A smaller radius or a heavier mass will also increase the centripetal force.

Is centripetal force dangerous in amusement park rides?

While centripetal force can feel intense and may even cause discomfort, it is not inherently dangerous in amusement park rides. The rides are designed to ensure that the force does not exceed safe levels and that riders are not at risk of being thrown from the ride.

How is centripetal force calculated in an amusement park ride?

The formula for calculating centripetal force is F = mv²/r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path. In an amusement park ride, the force is typically measured in G-forces, with 1 G equal to the force of gravity. Most amusement park rides will not exceed 3-4 Gs, as higher forces can be uncomfortable or even dangerous for riders.

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