# Homework Help: UM Motion

1. Jan 20, 2009

1. The problem statement, all variables and given/known data
Suppose the earth is a perfect sphere with R=6370 km. if a person weighs exactly 600.0 N at the north pole how much will the person weigh at the equator? (hint: the upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case) ans: 597.9 N

2. Relevant equations

Fg=Gm1m2/r^2
Fg=ma
Fc=Mv^2/r?

3. The attempt at a solution

I have no real clue on how to solve such a question, i plugged in for Fg but the answer i seem to get is higher than that of 597.9 N. Need help for a solution please!!!!!

2. Jan 20, 2009

I will give you a hint.......A person standing on the equator experiences a centripetal force towards the centre of the earth .Work out what this is and draw a free body force diagram for the person.

3. Jan 20, 2009

### Staff: Mentor

Apply Newton's 2nd law. What's different about being on the equator compared to being on the north pole?

4. Jan 20, 2009

### chrisk

Think about the angular velocity at the north pole and the equator. Then calculate the centripetal acceleration at the equator and use this value to find the effective acceleration of gravity at the equator.

5. Jan 20, 2009

how could i determine the angular velocity though? i am not given a time

6. Jan 20, 2009

### chrisk

The time is 24 hours for one revolution! Check out this web site for an explanation

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/AccOfGravity.html [Broken]

Last edited by a moderator: May 3, 2017
7. Jan 20, 2009

so i should use Fc=4pie^2Mr/ T^2 to solve for Force centripetal

8. Jan 20, 2009

### chrisk

Yes, use this equation but it's a little easier to find the effective acceleration of gravity at the equator, g - acentripetal, then multiply by the mass.

9. Jan 20, 2009

okay so it would be easier to just get the centripetal acceleration, then take gravity- acceleration centripetal then multiply by mass for F=ma right? but how would i get the acceleration centripetal when Ac=V^2/r and i don't have V?

10. Jan 20, 2009

### chrisk

V= earth's angular velocity times the earth's radius.

11. Jan 20, 2009

i am not familiar with earths angular velocity?

12. Jan 20, 2009

### gabbagabbahey

Really? You don't know how long a day is? Or you don't know how to calculate angular velocity from knowledge of the period of rotation?

13. Jan 20, 2009

don't know how to calculate angular velocity from knowledge of the period of rotation? I've never heard of the term for one so maybe that's why im lost with what your saying

14. Jan 20, 2009

### gabbagabbahey

Angular frequency is what you really need for this problem; and it is defined by:

$$\omega\equiv2\pi f=\frac{2\pi}{T}=\frac{|v|}{r}$$

Angular velocity is a vector quantity with magnitude $\omega$ and direct determined by a cross product.

15. Jan 20, 2009