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Centripetal acceleration and outward motion?

  1. Jun 3, 2012 #1
    I'm in my first year of physics and I'm having trouble understanding some aspects of centripetal force.

    When someone is spinning rapidly on, say, an amusement park ride, they tend to be pushed outward the faster they go. Is this the tangential velocity? Or does it involve a different aspect of uniform circular motion? And if so, how do I calculate it?

    EDIT: I know basic equations such as Centripetal acceleration = velocity squared / the radius, and the velocity = the circumference / the time to complete 1 revolution.
    And this is only homework related insofar as I will understand it better; I'm not seeking an easy answer.
    Last edited: Jun 3, 2012
  2. jcsd
  3. Jun 3, 2012 #2
    The centripetal force is a force directed toward the center of rotation. It is a necessary condition for an object to move in a circular path, because objects would prefer to move in straight lines. It is something added to the system, such as a rope, a gravitational attraction or whatever. The act of moving in a circle doesn't produce the centripetal acceleration, the centripetal acceleration causes things to move in a circle.

    What you're describing is a centrifugal force, the reason things want to fly off of an spinning disk for example. This is called a "pseudo force" because there's nothing actually pulling or pushing the object, it's an effect of inertia and has to do with the fact that a rotating object is not in an inertial reference frame. I wouldn't worry too much about this for now, you'll cover it in great detail when you take an upper level course in classical mechanics.
  4. Jun 3, 2012 #3
    Thank you for the clarification. Would you mind explaining how it works for me, though? For example, if someone is in a ride spinning at a certain speed, how fast would they be moving outwards?

    EDIT: Would centrifugal force effectively be an opposition to centripetal force, balancing out the inward force so the path remains circular?
    Last edited: Jun 3, 2012
  5. Jun 3, 2012 #4
    Hi guys, I'm not formaly educated in this very interesting topic. Would I be correct in saying that the force is not in oppostion but rather tangental to, so that the path remains circular. I appoligize if I am futher confusing the OP.
  6. Jun 3, 2012 #5


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    When all forces balance each other, the motion cannot be circular, the velocity vector must be constant.

    In the non-rotating reference frame where the path is circular there is only the centripetal force. It is not balanced, so there is centripetal acceleration: the velocity vector changes direction.

    In the co-rotating non-inertial reference frame where the object is static the centripetal force is balanced by the inertial centrifugal force. So the velocity vector is constant (zero).

  7. Jun 3, 2012 #6


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    No. Centrifugal force is in opposition to centripetal.
  8. Jun 3, 2012 #7
    I'm confused. How does centrifugal force factor in to uniform circular motion? Is it equal but opposite centripetal? If someone is spinning so that there are 20 N of centripetal force, does that mean there are -20 N of centrifugal force?
  9. Jun 3, 2012 #8


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    It is opposite to centripetal. The magnitude depends on the rotation of the reference frame, not the centripetal force.

    Only in the rotating frame where that someone is at rest. Did you read the wiki?

  10. Jun 3, 2012 #9
    So if you knew the centripetal force and acceleration, along with the radius and tangential velocity, how would you calculate the effects of centrifugal force? If the someone were spinning around something at a certain speed, how fast would they be moving outwards?
  11. Jun 3, 2012 #10


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    Again: The magnitude of centrifugal force is independent of the centripetal force, acceleration or velocity of the object. It depends only on the angular velocity of the reference frame and the distance from the rotation axis.
    I don't understand your question. Is he just spinning, or moving away? In which reference frame?
  12. Jun 3, 2012 #11
    Oh, it's angular velocity.

    If someone is riding on a merry-go-round, or teacups, or something similar, they move outwards, due to "centrifugal force". Is there a way to calculate how much of that force is being applied to them as they are pushed against a wall or something similar? And how fast are they moving away?
  13. Jun 3, 2012 #12


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    The force being applied to them by the merry-go-round is the centripetal force:
    Forces are not related to velocity, but acceleration. When you add all the forces and divide by the mass, you get the acceleration:
  14. Jun 3, 2012 #13
    But isn't the centripetal force going inward? I want to know how you figure out how much force is pushing you outwards. Not that it actually is, since centrifugal force isn't an actual force. But there must be a way to determine how much outward motion would be experienced by someone moving at a certain speed on a merry-go-round.

    EDIT: Is it likely that, while the merry-go-round is attached to the axis of rotation and thus maintaining centripetal acceleration and force, the human inside of it isn't? And thus, wouldn't he be propelled by the tangential velocity?
    Last edited: Jun 3, 2012
  15. Jun 3, 2012 #14
    You may be getting confused because there are two "centrifugal" forces, and they are distinct. One is simply a reaction force directed outwards, for example if you tie a ball to a string and twirl it around, you are pulling inward on the ball, and there is a reaction force of the ball pulling "outward" on your hand.

    The other is the effect of non-inertial acceleration. This force can be calculated by:

    [tex] \vec{F} = m\vec{\omega}×(\vec{\omega}×\vec{r})[/tex]

    Notice this force has to point radially outward.

    See http://en.wikipedia.org/wiki/Reactive_centrifugal_force
    and http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame)
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