Calculating Mass from a Ferris Wheel Scale Reading

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SUMMARY

The problem involves calculating the mass of a woman riding a Ferris wheel with a radius of 16 m, using readings from a digital bathroom scale. The scale shows a minimum force of 510 N and a maximum force of 666 N during the ride. By applying the equations of motion and centripetal force, the woman's mass is determined to be 60 kg. The key equations used include F_net = ma and the relationship between gravitational force and centripetal acceleration.

PREREQUISITES
  • Understanding of centripetal acceleration (Ac = v^2/R)
  • Knowledge of gravitational force (w = mg)
  • Familiarity with the concept of net force (F_net = ma)
  • Ability to interpret scale readings in a physics context
NEXT STEPS
  • Study the derivation of centripetal force equations
  • Learn how to analyze forces in circular motion
  • Explore the implications of varying forces in different positions on a Ferris wheel
  • Investigate the use of digital scales in physics experiments
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for practical examples of centripetal motion and force analysis.

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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