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Determine the magnitude and direction

  1. Sep 30, 2004 #1
    Determine the magnitude and direction of the effective value of g at a latitude of 33.0° on the Earth. Assume the Earth is a rotating sphere.
     
  2. jcsd
  3. Sep 30, 2004 #2

    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
    [​IMG]
     
    Last edited: Oct 1, 2004
  4. Sep 30, 2004 #3
    I can't...
     
  5. Sep 30, 2004 #4

    Chi Meson

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    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.
     
  6. Sep 30, 2004 #5

    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.

    I changed the link. See if it works now.
     
    Last edited: Sep 30, 2004
  7. Sep 30, 2004 #6
    I think ShawnD said it nicely but I just have to add in my interpretation. Since you are stationary and you know the earth is rotating and is massive, three forces are acting on you: gravity, the normal force and centrifugal force but centrifugal force is not really a force but rather lack of centripetal force (equal to -Fcentripetal).

    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.
     
  8. Sep 30, 2004 #7
    What is the velocity? And how do I find the angle? the hint says "from a line directly toward the center of the earth" I need some more help :cry:
     
  9. Oct 1, 2004 #8

    ShawnD

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    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 think you should use.

    [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.
     
    Last edited: Oct 1, 2004
  10. Oct 1, 2004 #9
    Do you know what the angle is that they are looking for?
     
  11. Oct 1, 2004 #10

    Chi Meson

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    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)?
     
  12. Oct 1, 2004 #11
    Using the concept of "centrifugal force" in any physics problem is in my opinion a very bad approach. Centrifugal force does not exist. When you draw a free-body diagram of a potato in circular motion, do you draw a force vector pointing out of the circle? No (I hope not, anyway!). No force is trying to throw people away from the centre of the earth! Please do not say these sorts of things to emerging physicists! The reason things seem to be pushed away from the centre of a circular orbit is that their instantaneous velocity vector is tangental to the circle at any point (perpendicular to the centripetal force vector). If you were to let go, the object would continue to follow this vector because it would no longer be held in circular motion by the centripetal force. That's why things seem to fly outwards from the circle when you let them go after spinning them around.

    Please do not confuse these concepts.
     
  13. Oct 1, 2004 #12

    Chi Meson

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    Well I agree with Sirus. Yet I've agreed with just about everything else ShawnD has ever said. Is there another take on this?
     
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