Circular motion problem (A2 level)

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Homework Help Overview

The discussion revolves around a circular motion problem involving a girl standing at the equator of the Earth. The problem includes calculating angular speed, resultant force, and the reading on weighing scales due to gravitational and centripetal forces.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between gravitational force, centripetal force, and the reading on the scales. Questions arise regarding why the resultant force should be subtracted rather than added when calculating the scale reading.

Discussion Status

The discussion is active, with participants clarifying the forces acting on the girl and how they relate to the net centripetal force. Some guidance has been provided regarding the nature of the forces and their contributions to the scale reading.

Contextual Notes

Participants are navigating the implications of Earth's rotation on the forces experienced by the girl, and there is an emphasis on understanding the definitions and roles of different forces in the context of circular motion.

breen155
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Hey, I am having a problem understanding something to do with circular motion, its the last part of a question.

Homework Statement


part 1 - Calculate the angular speed of a girl standing at the equator, i used angular speed = 2pi/(60x60x24) = 7.3x10^-5 rad/s

part 2 - The radius of the Earth is 6400km and the girl has a mass of 60kg, calculate the resultant force on the girl necessary for this circular motion. I used
F=mass x radius x angular speed ^2 and got 2.03N

Part 3 - (The problem part) If the girl was standing on weighing scales calibrated in Newtons, what reading would she get?
I used reading = mg + resultant Force which got me an answer of 590.6N (taking g = 9.81) However the mark scheme stated that i had to take away the resultant force rather than add it, I was just wondering why this was the case? I thought that the force of the girl would be towards the centre of the Earth and therefore must be added rather than taken away :confused:

Thanks in advance
 
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breen155 said:
I thought that the force of the girl would be towards the centre of the Earth and therefore must be added rather than taken away :confused:
The net force on the girl is towards the center of the earth. What individual forces act on the girl?

Hint: If the Earth didn't rotate, the forces on the girl would add to zero: The upward normal force (which is what the scale reads) and the downward gravitational force (mg) would be equal. How does that change when you consider the Earth's rotation?
 
well the forces acting downwards are her weight and the centripetal force? and upwards is the contact force between her and the scales?
 
breen155 said:
well the forces acting downwards are her weight
Right.
and the centripetal force?
The "centripetal force" is just the name we give to the net force, not a separate force in itself. The net force is downward, though.
and upwards is the contact force between her and the scales?
Right.

So the two actual forces on her are her weight and the contact force. They must add up to equal the net centripetal force. (Set up that equation.)
 
so is it Force = weight (mg) - contact force?
 
breen155 said:
so is it Force = weight (mg) - contact force?
Right! (By "Force" I assume you mean "centripetal force".)

So solve for the contact force. Is it more or less than her weight?
 
so its (9.81 x 60)-2.03 = 586.57N
 
breen155 said:
so its (9.81 x 60)-2.03 = 586.57N
Good!
 
:) oh i get it now, thanks for all the help :)
 

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