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- Thread starter physicsss
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ShawnD

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That is one hell of a question!

If you draw what it's asking, and line the equator up as your x axis, you will see that gravity is pulling at a 33 degree angle, but centrifugal force is pushing along the x axis.

For g, use that gravity formula with the G constant, the mass of earth, and the radius of earth. For centrifugal force, it's just v^2 / r, or [itex]m *\omega^2 * r[/itex] (i think).

Hopefully you can see this picture

http://myfiles.dyndns.org:8080/math/circle_gravity1.jpg

If you draw what it's asking, and line the equator up as your x axis, you will see that gravity is pulling at a 33 degree angle, but centrifugal force is pushing along the x axis.

For g, use that gravity formula with the G constant, the mass of earth, and the radius of earth. For centrifugal force, it's just v^2 / r, or [itex]m *\omega^2 * r[/itex] (i think).

Hopefully you can see this picture

http://myfiles.dyndns.org:8080/math/circle_gravity1.jpg

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I can't...

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Chi Meson

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ShawnD said:That is one hell of a question!

If you draw what it's asking, and line the equator up as your x axis, you will see that gravity is pulling at a 33 degree angle, but centrifugal force is pushing along the x axis.

For g, use that gravity

I'm sure ShawnD meant "centripetal force."

As anything goes in a circle, there must be some force that pulls inward. THis force is supplied by the gravitaional force. At a latitude of 33 deg, gravity is pulling toward the center of the earth. BUt the componant along the x-axis (as ShawnD has described it) is "being used as" the centripetal force. "Effective weight" ("w," this is the weight you feel, rather than the actual gravitational force on you) is reduced by this "centripetal force requirement" becuase this componant is causing an acceleration.

in general: w = mg - ma where a (here) is centripetal acceleration, (v^2)/r

Remember that the circle being traveled in is NOT the equator.

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ShawnD

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No. Centripetal means towards the center. Centrifugal means away from the center. The spinning motion of earth tries to throw people off; away from the center.Chi Meson said:I'm sure ShawnD meant "centripetal force."

I changed the link. See if it works now.

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- #6

vsage

Since centripetal force acts toward the center of rotation (the axis of the earth NOT the center), the centrifugal force acts in the exact opposite direction. So according to newton's law, Fnormal + Fgravity + Fcentrifugal = 0. From this we get Fgravity + Fcentrifugal = -Fnormal

Fgravity - Fcentripetal = -Fnormal.

Your Fnormal is your effective force of gravity (the part that isn't cancelled out by lack of centripetal force).

I'd argue this isn't a highschool-caliber question :) Good luck to you if you solve this entirely because it takes a great deal of cleverness.

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ShawnD

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physicsss said:

First of all, the angles. You're trying to put things in terms of gravity which is towards the center of earth, so say gravity is acting at an angle of 0. The force due to the earth's rotation is 33 degrees from that point; the tricky part is figuring out if it's sine or cosine. Since the rotation lowers the effective gravity more when the angle is smaller, it's cosine (I think). The effective gravity should be something like this:

effective gravity = (gravity formula) - (centripetal/centrifugal force formula)*cos(33)

For the velocity.... Substitute [tex]\frac{V^2}{r}[/tex] with [tex]\omega^2r[/tex]. For omega (w), remember that the earth spins once per day; that's [itex]2\pi[/tex] radians in however many seconds a day is. Get a value for omega (w) in radians/second.

To sum it up, here is the equation I

[tex]gravity = \frac{Gm}{r^2} - \omega^2r \cos(33)[/tex]

G is the gravity constant. m is the mass of the earth in kilograms. r is the radius of the earth in meters. omega (w) is the rotation speed of earth in radians per second.

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Do you know what the angle is that they are looking for?

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Chi Meson

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ShawnD said:No. Centripetal means towards the center. Centrifugal means away from the center. The spinning motion of earth tries to throw people off; away from the center.

you say to-ma-to, I say to-mah-to.

you say centrifugal force throws you away from the center of a circle,

I say what's the object that applies centrifugal force on you?

Seriously, this topic has gotten out of hand on other threads before and I am writing this with a sense of humor;

So is that detail about forces (that some object must apply a force on another object) satisfied with this use of "centrifugal force," or is that detail found to be no longer necessary (especially in light of GR considerations)?

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Please do not confuse these concepts.

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Chi Meson

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