How Long for One Revolution on a Ferris Wheel?

In summary, the normal force acting on the person at the bottom of the ferris wheel is equal to their apparent weight, which is the sum of their actual weight and the centripetal force. By setting the net force in the centripetal direction equal to the centripetal force, the time it takes for the ferris wheel to go one revolution can be calculated.
  • #1
phoenix133231
9
0

Homework Statement


Suppose that you're on a circular ferris wheel with a radius of 22 m. You know that your mass is 64 kg on the very bottom part of the ferris wheel, even though your actual mass is 58 kg. How long does it take for the ferris wheel to go one revolution?

Homework Equations


I believe centripetal force? F = mv^2/r
Gravitational force: F = mg
Circumference: C = 2*pi*r


The Attempt at a Solution


So, what I did was set the centripetal force equal to the gravitational force:
mv^2/r = mg

which implies:

v = Sqrt[g*r] which gave me the value 14.6 m/s. I then found the circumference of the ferris wheel to be 2*pi*22 = 138.23 m, and used the relation t = displacement/velocity to obtain 9.41 s.

This, however, came out to be wrong... is my problem setup incorrectly?
 
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  • #2
phoenix133231 said:

Homework Statement


Suppose that you're on a circular ferris wheel with a radius of 22 m. You know that your mass is 64 kg on the very bottom part of the ferris wheel, even though your actual mass is 58 kg. How long does it take for the ferris wheel to go one revolution?

Homework Equations


I believe centripetal force? F = mv^2/r
Gravitational force: F = mg
Circumference: C = 2*pi*r


The Attempt at a Solution


So, what I did was set the centripetal force equal to the gravitational force:
mv^2/r = mg

which implies:

v = Sqrt[g*r] which gave me the value 14.6 m/s. I then found the circumference of the ferris wheel to be 2*pi*22 = 138.23 m, and used the relation t = displacement/velocity to obtain 9.41 s.

This, however, came out to be wrong... is my problem setup incorrectly?
The centripetal force is not in general equal to the gravitational force. It is equal to the net force in the centripetal direction (inward toward the center of the circle). You have to identify both forces acting on you. One is the weight. What's the other? And what therefore is the net force? Incidentally, the problem incorrectly states that there is a change in mass. I hope you catch this error.
 
  • #3
PhanthomJay said:
The centripetal force is not in general equal to the gravitational force. It is equal to the net force in the centripetal direction (inward toward the center of the circle). You have to identify both forces acting on you. One is the weight. What's the other? And what therefore is the net force? Incidentally, the problem incorrectly states that there is a change in mass. I hope you catch this error.

I made a mistake, the problem stated that you measured yourself on a scale on the very bottom of the circle to be 64 kg. I don't see the other force...
 
  • #4
phoenix133231 said:
I made a mistake, the problem stated that you measured yourself on a scale on the very bottom of the circle to be 64 kg. I don't see the other force...
If the scale reads 64 kg, you must first convert it to force units in Newtons. That force, sometimes called your 'apparent' weight, is the normal contact force of the scale acting upward on you. The other force acting on you is the gravity force (your actual weight (in Newtons!) acting always downward on you). The vector sum of these 2 forces is the net force acting on you.
 
  • #5
Oh... I think I see it. So, the normal force would be (64kg)(g) and the gravitational force would be (59kg)(g), thus the net force is 5*g upwards, which implies that this is what the centripetal force is equal to?

Is my reasoning correct?
 
  • #6
That sounds right to me. (Actually, it is the centripetal acceleration, if you check the units).
 
  • #7
phoenix133231 said:
Oh... I think I see it. So, the normal force would be (64kg)(g) and the gravitational force would be (59kg)(g), thus the net force is 5*g upwards, which implies that this is what the centripetal force is equal to?

Is my reasoning correct?
Yes, that is the net force upward (toward trhe center of the wheel) in Newtons, where g = 9.8 m/sec^2, although note you mentioned m = 58 kg in your original post, not 59 kg.. Now you must solve for the time it takes the ferris wheel to go one revolution at the constant speed you will first calculate.
 

1. What is the force that keeps the riders on a Ferris wheel from falling off?

The force that keeps the riders on a Ferris wheel from falling off is the centripetal force. This force acts towards the center of the circular motion and is caused by the tension in the wheel's spokes.

2. How does the speed of the Ferris wheel affect the force on the riders?

The speed of the Ferris wheel affects the force on the riders by increasing or decreasing the centripetal force. As the speed increases, the centripetal force also increases, making the ride feel more intense. When the speed decreases, the force decreases, making the ride feel more relaxed.

3. What is the difference between the forces experienced at the bottom and top of the Ferris wheel?

The forces experienced at the bottom and top of the Ferris wheel are different due to the change in direction of the motion. At the bottom, the riders experience a greater force because they are accelerating towards the center. At the top, the riders experience a lower force because they are moving in a straight path tangent to the circle.

4. How does the weight of the riders affect the overall force on the Ferris wheel?

The weight of the riders affects the overall force on the Ferris wheel by increasing the centripetal force needed to keep the riders in circular motion. The heavier the riders, the greater the centripetal force required, making the ride feel more intense.

5. Can the force on the riders be greater than the weight of the riders?

Yes, the force on the riders can be greater than the weight of the riders. This happens when the speed of the Ferris wheel is high enough to create a centripetal force greater than the riders' weight. In this case, the riders will experience a feeling of weightlessness at the top of the wheel, similar to the experience of astronauts in space.

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