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Change of weight with faster Earth rotation.

1. The problem statement, all variables and given/known data

m=55kg
r=6400km
The earths rotation increases so on a bathroom scale, it now reads that you weight 0.
how long is 1 "day" on Earth?


2. Relevant equations

I keep trying to figure this one out, but with different ways Ive tried, I put Fg as 0 and it basically negates my whole equation. The section in the course we are on right now is Centrapetal force and gravitation.



3. The attempt at a solution

This is the one "answer" Ive gotten but it seems a little too easy.

fnet=ma
fc=mac
fc=mv2/r
mg=mv2/r
g=v2/r
√rg=v

plug in the radius in metres, 9.8 for g to get v, and then solve with 2∏r/v which is the same a d/v=t.

any help? :(
 
Quite right.

Initial weight = mg = 55*9.81 Newtons
Centripetal force = mv^2/r.

Since the weight is balanced exactly by this force, we must have
mg =mv^2/r ,which gives v.
 
I keep trying to figure this one out, but with different ways Ive tried, I put Fg as 0 and it basically negates my whole equation. The section in the course we are on right now is Centrapetal force and gravitation.
The reason you can't do [itex] F_{g} = 0 [/itex] is because it isn't 0. What you want is [itex] F_{net} = 0 = F_{g} - F{c} \Rightarrow F_{g} = F_{c} [/itex]

It looks like this is what you ended up doing, and I don't see any problems with your attempt. I don't think it was meant to be a hugely difficult problem.

Edit: To be completely accurate, the normal centripetal force of the earth supplies about 2 Newtons (or so, depending on your mass) of upward force, so to be completely accurate you could show how to find this "usual" centripetal force, and then factor that in. Since a normal person weighs somewhere in the 600-ish Newton range though, it should only make a small difference in your answer.
 
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