Mainly centripetal forces, please check (esp #4)

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    Centripetal Forces
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Homework Help Overview

The problem involves a Ferris wheel with a radius of 7.0m and explores concepts related to centripetal forces, linear speed, and apparent weight. The original poster presents a series of questions regarding the linear speed of the ride, the centripetal force acting on a person, and how these relate to the sensation of weightlessness and apparent weight at different points of the ride.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculations for linear speed and centripetal force, with some verifying the original poster's math for the first three parts. There is a focus on the interpretation of apparent weight at the bottom of the Ferris wheel and the forces acting on Sasha.

Discussion Status

Some participants have provided feedback on the calculations, particularly questioning the interpretation of apparent weight in relation to the forces acting on Sasha. There is ongoing exploration of how to correctly account for the forces involved, with no explicit consensus reached yet.

Contextual Notes

Participants are navigating the definitions of forces, particularly distinguishing between weight and centripetal force, and how these contribute to the apparent weight experienced by Sasha at different points on the ride.

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Homework Statement



Sasha's favorite ride at the fair is the Ferris wheel that has a radius of 7.0m.

1. If the ride takes 20.0s to make one full revolution, what is the linear speed of the wheel?

2. What centripetal force will the ride exert on Sasha's 50.0kg body?

3. In order for Sasha to feel weightless at the top of the ride at what linear speed must the Ferris wheel turn?

4. At this speed, how much will she appear to weigh at the bottom of the Ferris wheel?

use 10 m/s^2 for gravity



Homework Equations



T = 1/f
f = 1/T

speed = (2)(pi)(r)/period

centripetal acceleration = (linear speed)^2 / radius

centripetal force = mass * centripetal acceleration ------- mv^2 / r

angular momentum = mvr



The Attempt at a Solution



1. v= 2pir/t
v = (2)(pi)(7)/(20) = 2.199 m/s

2. centripetal force = mv^2 / r
Fc = (50)(2.199)^2 / 7 = 34.544 N

3. centripetal force i think must equal the same as Sasha's weight

(50)(v^2) / 7 = (50)(10)
50v^2 / 7 = 500
50v^2 = 3500
v^2 = 70
v = 8.366 m/s

4. at 8.366 m/s, she will weigh 418.33 N because
(8.366)(50) = 418.33
 
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i didnt check the math but 1,2,3 look good.

number 4 however is incorrect, what you've calculated is Sasha's linear momentum.

Sasha's apparent weight will be equal to the normal force exerted on her at the bottom fo the ride.
 
For part 4. Consider the at the bottom of the ferris wheel and then consider the direction of the weight and the centripetal force.
 
so does that mean i have to find her normal weight then times it by the centripetal force?
 
No. Her weight acts downwards.The centripetal force acts upwards. The resultant of the two forces is her apparent weight
 
Her weight downward is 500 N and her centripetal force is 34.544N so 500 - 34.544 = 465.465 N. is that the answer?
 
Fnet = mv^2/r, where mv^2/r is the centripetal force.

There are two forces acting on Sasha, gravity and a normal force in opposing directions, so taking the radial direction as positive.

Fn - Fg = mv^2/r

Sasha's apparent weight will equal the normal force, so

Fn = mv^2/r + Fg
 

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