How Does Angular Speed Affect Scale Readings on a Ferris Wheel?

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The discussion focuses on calculating the difference in scale readings on a Ferris wheel as a function of angular speed (w), mass (m), radius (R), and gravitational acceleration (g). At zero angular speed (w=0 rad/s), the scale should read the same at both the top and bottom, resulting in a delta F of zero. As angular speed increases, the difference in scale readings (delta F) will also increase due to the additional centripetal force acting on the woman at the bottom of the wheel. The equations provided indicate that the scale readings at the top and bottom can be expressed as n1 = mg - mv^2/R and n2 = mv^2/R + mg, respectively. The key takeaway is that the change in scale readings is determined by the difference between these two values.
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Homework Statement



A woman of mass m rides in a Ferris wheel of radius R. In order to better understand physics, she takes along a bathroom scale and sits on it. Determine the difference in scale readings between the bottom and top of the Ferris wheel (“delta” F of scale) as a function of the constant angular speed of the ferris wheel (w),m,R,and g.

If w=0 rad/s, what should “delta” F of scale equal? Does your dunctoin agree with this observation?
If w was twice as large, what would happen to “delta” F of scale?

Homework Equations


mv^2/R
mg

The Attempt at a Solution


Top (n1): mg - mv^2/R
Bottom (n2): mv^2/R + mg
m = (n2 + n1)/2g

Not very sure...
 
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will21 said:

Homework Statement



A woman of mass m rides in a Ferris wheel of radius R. In order to better understand physics, she takes along a bathroom scale and sits on it. Determine the difference in scale readings between the bottom and top of the Ferris wheel (“delta” F of scale) as a function of the constant angular speed of the ferris wheel (w),m,R,and g.

If w=0 rad/s, what should “delta” F of scale equal? Does your dunctoin agree with this observation?
If w was twice as large, what would happen to “delta” F of scale?



Homework Equations


mv^2/R
mg



The Attempt at a Solution


Top (n1): mg - mv^2/R
Bottom (n2): mv^2/R + mg
m = (n2 + n1)/2g

Not very sure...
Your equations are correct, but solving for m does not give you the desired solution. The difference in the scale readings is (n2) - (n1), where (n2) and (n1) represent the scale readings at the bottom and top of the ferris wheel, respectively..
 
Yes, recall that what you are looking for is the change, and thus the difference, between the two quantities.
 

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