Calculating Mass from a Ferris Wheel Scale Reading

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Homework Help Overview

The problem involves calculating the mass of a woman based on readings from a digital scale while she rides a Ferris wheel. The scenario includes centripetal forces acting on the woman as she experiences varying scale readings during the ride.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal forces and the readings on the scale at different points of the Ferris wheel's motion. There are attempts to derive the mass using equations related to forces and acceleration.

Discussion Status

Some participants have suggested methods for calculating the mass by analyzing the forces at the top and bottom of the Ferris wheel. There is a mention of different approaches to arrive at the same result, indicating a productive exploration of the problem.

Contextual Notes

The discussion includes references to specific values from the scale readings and the radius of the Ferris wheel, which may influence the calculations but are not strictly necessary for determining the mass.

bertoline
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Homework Statement


A woman rides on a Ferris wheel of radius 16 m that maintains the same speed throughout its motion. To better understand physics, she takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of scale reading versus time. Note that the graph has a minimum value of 510 N and a maximum value of 666 N .

What is the woman's mass?

Homework Equations


Ac=v^2/R
V=(2piR)/T
w=mg
Period = 22s

The Attempt at a Solution


Woman velocity/acceleration:
Velocity= 2pi(16)/22 = 4.57 m/s
Acceleration=(4.57)^2/16 = 1.31 m/s^2

i am stuck here, can't figure what equation to use next.
i think this might be it
Fnet=ma = m (v^2/R)
 
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Talking in terms of centripetal 'forces', the force on the scale at the top is mg-mv^2/r=510N and the force at the bottom is mg+mv^2/r=666N. If you subtract those you can solve for mv^2/r and then use your value for v^2/r to solve for m. Alternatively, you could add them and solve for mg. You should get about the same thing. So you never actually needed to know the radius and rotation rate of the wheel. I'd suggest you do the former though, since that's probably what they want to do.
 
Dick said:
Talking in terms of centripetal 'forces', the force on the scale at the top is mg-mv^2/r=510N and the force at the bottom is mg+mv^2/r=666N. If you subtract those you can solve for mv^2/r and then use your value for v^2/r to solve for m. Alternatively, you could add them and solve for mg. You should get about the same thing. So you never actually needed to know the radius and rotation rate of the wheel. I'd suggest you do the former though, since that's probably what they want to do.

i solved it.
answer: 60kg.
 
Right.
 

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