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Centrifugal force units.

  1. Jun 24, 2010 #1
    I found this on Wikipedia for the equation relating to centrifugal force.

    F = d/dt (m*v)

    So I have my mass and velocity but I just multiplied the two, that can't be correct can it. You have to derive something I imagine. Also would the units just be in lbs or Newtons. I'm using feet/s and lbs.

    Help I'm pretty rusty on my calculus.
     
  2. jcsd
  3. Jun 24, 2010 #2

    berkeman

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    If your velocity is constant, there is no net force (no acceleration).

    A better way to write it (assuming constant mass) is F = m dv/dt

    So the units in mks are Newtons = kg m / s^2
     
  4. Jun 24, 2010 #3
    The odd thing I found is was this equation for centripetal force. where v^2 = F * R/M

    Radius and mass.

    Couldn't I just use this equation for centrifugal force since all centrifugal and centripetal are is the name for the direction which the force is going on the wheel?
     
  5. Jun 24, 2010 #4
    what is mks. how do you perform that equation?
     
  6. Jun 24, 2010 #5

    berkeman

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    mks = meters, kilograms, seconds. It's one of the two metric systems of units. The other is cgs = centimeters, grams, seconds.

    If you mean the F = m dv/dt equation, the dv/dt is a timie derivative of velocity. That is, how much the velocity chages over time, which is called the acceleration.

    v = dx/dt (velocity is change in position with time, in mks units of m/s)
    a = dv/dt (acceleration is change of velocity with time, in mks units of m/s^2)
     
  7. Jun 24, 2010 #6

    K^2

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    SI is the more common name for MKS.

    Centrifugal force is given by F = m*v²/R. Thats kg*(m/s)²/m = kg*m/s² = N.

    So the units are exactly the same as these of F = ma.
     
  8. Jun 24, 2010 #7
    Sorry I'm a little lost on figuring out the derivative. How do you set that up?
     
  9. Jun 24, 2010 #8

    Doc Al

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    What exactly are you trying to calculate?
     
  10. Jun 25, 2010 #9
    The centrifugal force of an object based upon the radius, weight or mass of the object.

    In the long run I would like to calculate the force exerted by the object at any point in the radius of the circle while rotating around at a constant velocity.

    In this case would we now be talking about acceleration?
     
  11. Jun 25, 2010 #10

    Doc Al

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    Realize that 'centrifugal force', at least in standard physics usage, is a fictitious force that only exists when analyzing motion from a rotating frame. You don't need centrifugal force to understand rotation.

    OK. You can easily calculate the centripetal force acting on the object. And for each force acting on the object there will be a corresponding force exerted by the object.

    Anything moving in a circle will have a centripetal acceleration.
     
  12. Jun 25, 2010 #11
    I understand the rotation, I would just like to know what the force of that object would be at different points from the center as it rotates. Like when you twirl a rock on a string around your finger, as you bring it in I'm sure the force changes and so does the speed. Am I making any sense?

    I figured that the centripetal force would be ficticious if any, how do you even get an object to go to the inside of a rotation. Don't want to get off topic, but maybe this is a goodtime for a description of the two.
     
  13. Jun 25, 2010 #12

    K^2

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    It's just a choice of the coordinate system. If your coordinate system rotates, you have centrifugal force. If it doesn't, there isn't one.

    If you are standing on a rotating disk, it's convenient for you to define your position relative to disk. That's your choice of coordinate system, and you then end up observing outwards pull you call Centrifugal Force. That being canceled by the friction with the ground, you remain stationary in that system.

    If, however, you decide to take your bearings relative to an object that's outside the disk, you note that you aren't just standing still. You are moving. In fact, you are constantly accelerating towards the center of the disk. The force that provides that acceleration is still the same friction with the ground that was countering Centrifugal Force in the rotating frame.

    So as long as you choose coordinate system without rotation, you really don't need centrifugal force.
     
  14. Jun 28, 2010 #13
    Simply put I just would like to find the force of a single object rotating in a circle relating to weight, RPM or any omega, and distance from center.

    Would the equation below work for what I am trying to do.


    Or do I need to derive something, If this is the case, I could use some help.
     
  15. Jun 28, 2010 #14

    uart

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    Yes the equation [itex]F= m v^2/r[/itex] will work just fine if you use SI units of kg, meters and seconds (which of course means the force is in Newtons).

    Feel free to ask if you need to know how that equation is derived.
     
  16. Jun 28, 2010 #15

    Doc Al

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    For a mass m moving in a circle of radius r at constant speed v (or angular speed ω), the net force on it (also known as the centripetal force) is given by:

    Fc = mv²/r = mω²r
     
  17. Jun 28, 2010 #16
    What units would that be, I would initially think lbs. but it how does the ft/s cancel.

    lbs * (ft/s)^2/ft = lbs*ft/s^2 that can't be right. what am I forgetting?
     
  18. Jun 28, 2010 #17

    Doc Al

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    You are forgetting that a pound is a unit of force (for example, weight) not mass. In the English system the unit of mass is the slug.

    slug * ft/s^2 ≡ lbs

    At the earth's surface, an object's weight in pounds is related to its mass in slugs via: Weight = mass*g = mass * (32 ft/s^2)

    In the SI system:

    kg * m/s^2 ≡ Newtons
     
  19. Jun 28, 2010 #18
    So for example:

    you have an item which ways 5lbs which would equal 5/32.2 = .15528 slugs.

    your velocity is 50ft/s and your radius is 2ft.

    So according to this equation. Fc = mv²/r = mω²r

    Fc = (.15528 slugs*50ft/s ^2)/2 ft = 194 (lb*f*s^2/ft)(ft^2/s^2)/(ft) = 194 lb right.

    And this force would be towards the center of the cirlce?
     
  20. Jun 28, 2010 #19

    Doc Al

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    Right.

    Yes. This is the net force towards the center required to make it go in a circle at that speed and radius.
     
  21. Jun 28, 2010 #20
    To determine the centrifugal force, would it be the same as the centripetal? equal opposing reacting force?
     
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