# Earths Centripetal Acceleration Effects

1. Aug 3, 2006

### NotMrX

It's my belief that if some one was at the equator or the south pole there weight might be differernt

Sum of F = m a = mg- N
a = v^2/r

at the pole the distance from person to rotational axis is 0 so a= 0 and then mg = N

On the equator the velocity can be calculated from 24 hours to go 2pi R and the normal is a little less there.

However a problem that does confuse is if we had a pendelum on a string what angles it would make depending on what degree from the horizontal it is at. What angle does the pendulem make with line going to the center of the earth assuming the earth is spherical and it is not moving relative to the stuff around it?

2. Aug 3, 2006

### Saketh

Calculate it theoretically and see for yourself. You'll find that the centripetal factor is negligible compared to the gravitational factor.
I'm not exactly sure what you're asking. How does the degree from the horizontal affect the angle the pendulum makes with the earth's surface normal? If you draw a diagram, the angle from the horizontal is complementary to the angle to the earth's surface normal, so their sum is 90 degrees. Are you asking if this is related to the centripetal acceleration?

If you are interested in this sort of physics, take a look at a Foucault pendulum.

3. Aug 3, 2006

### Andrew Mason

Why not try make a pendulum that is free to move in any horizontal direction and set it going. Make it heavy with a very long string. See what happens to the direction of swing over time.

AM

4. Aug 3, 2006

### Gelsamel Epsilon

Actually at the equator G is less because of the centrifugal forces acting against gravitational accelration (So hence less weight). While at the poles it acts less, so you're heavier there.

9.789 m/s² at the Equator and 9.823 m/s² at the poles.

5. Aug 3, 2006

### Andrew Mason

The centripetal acceleration does reduce the normal force but it is very small compared to the reduced gravity at the equator due to the earth's bulge. Since gravity is a $1/r^2$ force, mass at the equator is farther from the earth's centre so gravity is less (about .5% less) at the equator than at the poles. Incidentally, the August Discover magazine article on gravity has this wrong (it states that gravity is greater at the equator because of the bulge).

AM

6. Aug 3, 2006

### Gelsamel Epsilon

I'm suprised, I learnt that it was weaker at the equator (not the reason at that point in time, just the fact) when I was in year 10. Yet a magazine made a blunder on that.

7. Aug 4, 2006

### NotMrX

For simplicity sake i was just going to leave the bulge out of it and assume the earth was a perfect sphere. Thanks for the info, I didn't realize the bluge has such an effect. I think maybe my problem is with reference frames and fictatious forces. For instance a pendulem hanging from a train could be described Fx= ma = TcosZ to someone watching outside train but if the person on train didnt feel it accellerate then they might describe by Fx = 0 = TcosZ - ma where ma is a fictatious force.

At 30 degrees from the equator we have centripetal acceleration or a fictatious force pushing on the pendulem in a direction different from gravity so it must not be pointing exactly to the center of earth. I realize it is just a fraction of a degree but I was wondering I could get any help in figuring out how to calculate it. It has been weeks and I keep getting a different answer then the back of the book and I think I am making some systematic error.

8. Aug 4, 2006

### Staff: Mentor

Just set up a coordinate system and calculate the net force on the pendulum bob, including the inertial force (which acts outward from the earth's axis of rotation). (I would use a coordinate system that has its y-axis parallel to the radial line that connects the earth's center to a point on the surface.) Without rotation, the net force points straight down along the radial line (y-axis). With rotation, the force is at an angle. Hint: Find the components of the inertial force parallel and perpendicular to the radial line (y-axis). Then find the net force and the angle it makes with the radial line.

9. Aug 10, 2006

### NotMrX

Thanks for the help. I got it to work out. Originally I solved it for fictatious force antiparrelel the way it should be. I just need to point the vector arrow the other way. Reference frames can be confusing. I guess the inertia makes it lag behind at angle.