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(This is a high school–level problem, so no air resistance, etc.) A person is sitting on a Ferris wheel of radius $r$ moving at a constant speed. What is the force from the seat acting on the person when the person is at the bottom of the ride? When the person is at the top?

My attempt at a solution:

When the person is at the top, the forces acting on the person are his weight and an equally large normal force from the seat pushing him upwards. Since the problem involves uniform circular motion, at the top of the ride, there must be some force pulling the person towards the center of the circle with magnitude $\frac{mv^2}{r}$.

The cause of this centripetal force must be the seat belt on the person, pulling him downwards?

When the ride is at the bottom, the normal force from the seat both counteracts the weight of the person and applies a centripetal force of $\frac{mv^2}{r}$ upwards.

Centripetal force kind of confuses me since my professor says a proof of it is beyond the scope of the course.

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  • $\begingroup$ You can think of centripetal force as the sum of a bunch of radial forces rather than its own standalone force. In this case, at the top of the wheel, the sum of the normal force, the force provided by the seat belt, and the gravitational force must be a net force with magnitude $\frac{mv^2}{r}$ pointing towards the center of the wheel. Note that centripetal force is dependent on speed, meaning the seat belt may not necessarily need to exert any downwards force if the wheel is spinning slowly. $\endgroup$
    – Rations
    Sep 8, 2015 at 19:54
  • $\begingroup$ @Rations Ok. So the net force acting on the person when he is at the top of the wheel Fs = v^2/2 * m...and this force consists of the gravity minus the normal force from the seat...right? $\endgroup$
    – Carefullcars
    Sep 8, 2015 at 20:01
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    $\begingroup$ Gravity minus the magnitude of the normal force is only true when (1) the person is at the top of the ride, (2) the direction pointing towards the center has been defined as positive, and (3) when you know the Ferris wheel is moving slowly enough that the direction of the normal force must be opposite the direction of gravity. $\endgroup$
    – Rations
    Sep 8, 2015 at 20:12

1 Answer 1

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Assuming that you mean a "ferris" wheel:

In a ferris wheel, $\frac{m*v^2}{r}$ is very small, because ferris wheels move slowly.

Also, on the wheel, all of the cars with people remain upright. This means that the force of gravity is always pulling downwards on people as they ride.

So, there are three cases that you can look at to explain this:

  1. You are at the top.

In this case, the centripetal force (which is required to keep you moving within the circle is provided by gravity. Gravity pulls you down towards the center of the wheel.

  1. You are at the bottom.

In this case, the force provided is an upward force provided by the metal structure of the wheel. The metal beams that support the car as it travels along at this point.

  1. You are on the side.

In this case, the force towards the center of the wheel is provided by a combination of the structure of the wheel (if you are on the bottom/side, and gravity if you are more on the top)

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  • $\begingroup$ Yeah i think i understand now...Given that v = k = 1m/s and r = 70m...then when the wheel is at the top Fc(centripetalforce) = 1/70... so 1/70 = G-N(normal force of seat). So N = G-1/70 $\endgroup$
    – Carefullcars
    Sep 8, 2015 at 20:04
  • $\begingroup$ Right, for a person weighing 100kg, going 1 m/s on top of a 70m wheel, They feel 1.43N of centripetal force, and 981N of gravitational force. I also edited the answer to explain where this centripetal force actually comes from, even though it is relatively insignificant. $\endgroup$
    – Stack Tracer
    Sep 8, 2015 at 20:18
  • $\begingroup$ OH yeah i forgot the mass when i calculated the centripetal force, but I feel like i understand now, thanks. $\endgroup$
    – Carefullcars
    Sep 8, 2015 at 20:25

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