Ferris wheel problem - vertical circles and centripetal acceleration

[therest]

Homework Statement

Given a Ferris wheel that rotates 5 times each minute and has a diameter of 19 m, with the acceleration of gravity as 9.8 m/s^2, what is the centripetal acceleration of a rider? Answer in units of m/s^2.

(There's a diagram that shows a ferris wheel with radius 9.5 m spinning at ω radians per second.)

Homework Equations

centripetal acceleration = v² / r = ω²r

The Attempt at a Solution

one rotation was equal to 2pi * r, so
5 rotations = 5(2pi * r) radians in one minute
or 10pi * r / 60 radians/second = ω

I used centripetal acceleration = ω²r = (10pi * r / 60)² * 9.5. I got 4.97 for centripetal acceleration. But this was wrong! I don't know where I made a mistake.

Oh, also, how does the force due to gravity figure into this problem? Shouldn't it be ignored since the ferris wheel is being powered by a motor?

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While I'm at it, I might as well ask about the other three parts of the problem.
Part b

Homework Statement

What force does the seat exert on a 72 kg rider at the lowest point of the ride? Answer in units of N.

Homework Equations

F_net = F_centripetal, so since net force must be upwards to maintain a centripetal force, normal force will be stronger than net force.
So: F_normal - F_g = F_net = F_c = m * v²/r
F_normal = F_g + F_c = m(g + a_c)

The Attempt at a Solution

Isn't the force exerted by the seat the normal force?
I think it would just be the mass times the sum of acceleration due to gravity and the centripetal acceleration, or m(g + a_c). I can't do this though as I have not found centripetal acceleration yet. That was part one.

Part c

Homework Statement

What force does the seat exert on a 72 kg rider at the highest point of the ride? Answer in units of N.

Homework Equations

At the highest point, F_net = F_c = F_g - F_normal
F_normal = mg - ma_c

The Attempt at a Solution

Operation seems wasy with the information from part a.

Part d

Homework Statement

What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom? Answer in units of N.

Homework Equations

F_g = mg

The Attempt at a Solution

Wouldn't this just be the weight of the rider since there would be no centripetal force pulling him up or down? Or would it be the equilibrant to the force due to gravity nd the centripetal force?

Thanks for help in advance!
Bonus information: I have only two hours to submit the answer to this problem electronically. Because I am a procrastinator. :(

 

[Unsigned]

The way I solve it, I consider that the Ferris wheel seat is a part of the wheel, and since the seat is at the rim of the wheel, we can say that the centripetal acceleration of the seat is (w^2)r, where w=(2pi x no. of rotations per min)/ 60.

we get a = 2.6018 m/s^2.
This gives us part (a).

Now for (b) and (c) :
We consider, the person to be in a non inertial frame that is accelerating at 2.6018 m/s^2 towards the centre. As a result, i consider, he is facing a pseudo force of the same, in a dir'n opposite to the dir'n of the centripetal force.

At the highest point, we use the equation of equilibrium.
We know Normal force, N is in the upward direction, and the centrifugal force F is in the upper direction too.
The weight of the person, W, is in the downward direction now.

This gives us an equation : N=W-F.

Now substitute values, and the answer is N=518.2676 N.

(d) halfway between top and bottom: the seat is again vertical, but the centripetal force is in the direction of the centre. Hence the centrifugal force (non inertial frame again) is in the horizontal outward direction ! As a result the net force experienced by the man is N=W. This solves it.

Now, we are considering the man sits still, and does not cause the seat to oscillate. :)

part (b) again : see, we consider the centripetal force as the net force.
hence : the net force is inward.
this means that the weight is greater than the normal reaction by the seat.
as a result : centripetal force=W-N.

therefore : N=W-ma ( considering that we subsequently note centripetal acc- centrifugal acc, only dir'ns are opp. as the frame of references are not the same.)
Also, centripetal force is real, while centrifugal force is pseudo, a concept :)

Somak.

 

[kerimek]

I would like to first commend you for taking the time to understand and solve this problem. It shows a great level of dedication and curiosity towards science. Now, let's break down the problem and address each part individually.

Part a:
Your approach to finding the centripetal acceleration is correct. However, the mistake you made was in the units. In your calculation, you used meters for the radius, but in the given answer, the units should be m/s^2. This is because the final answer should be in terms of acceleration, which is measured in m/s^2. So, the correct answer for the centripetal acceleration would be 4.97 m/s^2.

Part b:
You are correct in thinking that the force exerted by the seat is the normal force. And you are also correct in using the equation Fcentripetal = m * v^2/r. However, in this case, the rider is at the bottom of the ride, where the net force is equal to the centripetal force, so the equation becomes Fnet = Fcentripetal = Fnormal + Fg. Plugging in the values, we get Fnormal = 72 kg * (9.8 m/s^2 + 4.97 m/s^2) = 1,044.24 N.

Part c:
At the highest point of the ride, the net force is equal to the centripetal force, which is equal to the weight of the rider. So the equation becomes Fnet = Fcentripetal = Fg - Fnormal. Plugging in the values, we get Fnormal = 72 kg * (9.8 m/s^2 - 4.97 m/s^2) = 356.16 N.

Part d:
At the halfway point, the rider is experiencing both the force due to gravity and the centripetal force. So the equation becomes Fnet = Fg + Fcentripetal = Fnormal. Plugging in the values, we get Fnormal = 72 kg * (9.8 m/s^2 + 2.485 m/s^2) = 846.72 N.

Bonus information:
I would also like to remind you that procrastination is not a good habit and it can lead to mistakes and stress. It's always better to start early and give yourself enough time to understand and solve problems. Good