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Rotation of the Earth and Apparent Weight?

by wmrunner24
Tags: centripetal, force, rotation, weight
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wmrunner24
#1
Oct22-10, 11:43 AM
P: 57
1. The problem statement, all variables and given/known data

Because of Earth’s rotation about its axis, a point on the Equator experiences a centripetal acceleration of 0.034 m/s2, while a point at the poles experiences no centripetal acceleration.
What is the apparent weight at the equator of a person having a mass of 118.1 kg? The
acceleration of gravity is 9.8 m/s^2 .

Answer in units of N.

2. Relevant equations
F[tex]_{centripetal}[/tex]=ma[tex]_{centripetal}[/tex]

F[tex]_{gravity}[/tex]=mg

3. The attempt at a solution

So, at first this problem greatly confused me. Now I think I have an idea of how to approach it.

F[tex]_{centripetal}[/tex]=F[tex]_{gravity}[/tex]-F[tex]_{normal}[/tex]

This equation describes the net force toward the center of the Earth. Gravity acts toward the center of the Earth, while the normal force will resist it. If a person were to be standing on a scale, it would be the normal force that would produce the "apparent" reading. So then:

F[tex]_{normal}[/tex]=F[tex]_{gravity}[/tex]-F[tex]_{centripetal}[/tex]

F[tex]_{normal}[/tex]=mg-ma[tex]_{centripetal}[/tex]

F[tex]_{normal}[/tex]=1153.3646 N

Is this line of thinking correct? Thanks in advance for any help provided.
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Xerxes1986
#2
Oct22-10, 11:55 AM
P: 50
Yes that is correct.


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