# Centripetal Force Calculations

1. Dec 15, 2013

### Xuvaze

Hello! Finals are coming up in less than a week and I received some sample problems as review for the exam. I have my work layed out and a draft for it, but I would like some reassurance, if that's okay. :-)

Anyway, here is the sample problem:
You have been contacted by StM.org (Stop the Myth) to aid them in an upcoming campaign. They are asking you to create a presentation to compare the weight of a person standing on the equator to the centripetal force experienced by that person due to the Earth's rotation. The results of your work will be used to support their effort to dispel the myth that, "People on the equator feel a lot lighter due to the rotation of the Earth."

- You may use any mass for a person in your calculations.
- You must clearly show how you calculated the magnitudes of both forces and how they compare to each other.
- You need to explain whether your answers depend on the mass of the person or would the anyone feel the same way.
- You should clearly state whether the centripetal force makes a significant impact on the weight.

Here is what I got:
The centripetal acceleration at the equator = 4*pi^2*r/T^2
Where T = time period. For Earth, this is 24hours = 86400 seconds.
R = radius of Earth. This is 6400 km = 6400000 meters
Therefore, 4*pi^2*6400000/86400^2 = 0.03 m/s^2
Since the acceleration due to gravity is approx. 9.81m/s^2, then you would weigh only about 0.3% less at the equator than at either the North or South poles.

Is that right? I feel like I am missing something. Any help would be much appreciated! Thanks so much!

2. Dec 15, 2013

### haruspex

Your working is fine, but the OP specifically asks you to answer in terms of forces, so you should do that.
I would also go through a bit of a discussion about what it is that makes you feel the weight you do, and how the centripetal force makes this less.

3. Dec 15, 2013

### Xuvaze

That's kind of where I'm stuck at, I'm not quite sure what they mean by forces. Do you mean the magnitude and direction? Or do they mean forces such as gravity?

4. Dec 15, 2013

### Xuvaze

I'm sorry I'm not a master at Physics. /:

5. Dec 15, 2013

### haruspex

No, I mean that you worked entirely in terms of accelerations. That would satisfy me, but what they asked you to do was assume a mass, 100kg say, and discuss the change in the forces. The answer will be the same of course, just 0.3%.
For the question of how heavy people feel, you have to deal in terms of the person's frame of reference, which is of course non-inertial. Your sense of your weight comes from the compression in your body, with gravity pushing down and the ground pushing up. If you feel less heavy at the equator it's because that compression is less. We know (for the purposes of the question - it's not discussing the oblateness of the Earth) that the gravitational force is actually the same, so you need to explain why the compression is less.

6. Dec 17, 2013

### Xuvaze

Oh!

Ah! I see! So in this case I should just put in any value of mass into my equation? Lets see. .
The centripetal acceleration at the equator = 4*pi^2*r/T^2
Where T = time period. For Earth, this is 24hours = 86400 seconds.
R = radius of Earth. This is 6400 km = 6400000 meters
Therefore, 4*pi^2*6400000/86400^2 = 0.03 m/s^2
Since the acceleration due to gravity is approx. 9.81m/s^2, then you would weigh only about 0.3% less at the equator than at either the North or South poles.
So then 0.3 x 100kg = 0.003 x 100kg = 3kg less at the equator than at either of the poles.
Or would I use the formula to calculate weight? Which is Fg = mg?
Fg = (100kg) x (9.81m/s/s) = 981 N . . Scratch that I don't think that's right. I'll just stick with the first equation: 0.3% x 100kg = 0.003 x 100kg = 3kg less at the equator than at either of the poles.

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.0337 m/s2, whereas a point at the poles experiences no centripetal acceleration. If you stand on the equator then you must be accelerating at of 0.0337 m/s2 down, so have a net force of 100*0.0337 = 3.37N.
So that would mean -2.3N.
The two forces that actually are acting must add up to that answer:
N + mg = mac
So, N = mg - mac.
When you consider that g = 9.81, and ac = 0.0337, you would see that the effects of rotation make a 0.3% difference from equator to pole.

I believe I figured it out, thanks to the help of this forum. I hope it is correct! I am now awaiting reassurance to submit it, if that is okay. :-) Thanks a lot for your help!

Last edited: Dec 17, 2013
7. Dec 17, 2013

### haruspex

You were doing so well until there. 0.003 x 100kg = ?
I wouldn't say 'experiences' there. It requires a centripetal acceleration to stay on the surface of the earth. What it experiences are gravity and the normal force from the surface. It's the difference between these that provides the necessary centripetal acceleration.
Yes.