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Rotation Problem: Earth lengths/Apparent Weights

  • Thread starter a Vortex
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  • #1
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Homework Statement


A friendly Brazilian has a mass of 150 kg. Being in Brazil, he rotates in a circle around the center of the earth once per day, The radius of this circle (which is essentially the radius of the Earth) is 6.40 x 10^6 m.

I have found that the normal force is 1.4649 x 10^3.

Homework Equations


ac = v2/r , centripetal acceleration
[itex]\Sigma[/itex]F=ma
v = (2[itex]\pi[/itex]r)/T , T is period of one revolution


The Attempt at a Solution


I only got to mg - Fn = mac
I know what the apparent weight is but I'm not sure about the normal force. If I can find this, then I can then find ac, v, and then T to find about the length of day.
 

Answers and Replies

  • #2
ehild
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What is the question?

ehild
 
  • #3
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Sorry forgive me. The question is: How long would a day have to be for the Brazilian's apparent weight to be 1.46 x 10^3 N?

Btw, the answer should be about 6.16 x 10^4 sec or 17.1 hours.
 
  • #4
ehild
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1,854
The weight is equal to the force the man presses a horizontal support - scales, ground, chair ... So its is the same as the normal force.

ehild
 

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