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About centripetal force

  1. Sep 17, 2014 #1
    Supposed that a man is standing on the edge of a spinning merry-go-round which rotates at an constant angular speed with respect to a stationary fame of reference(the ground), there will be a centripetal force pulling him toward the centre of the marry-go-round and prevents him from sliding off it. Then comes the question, what happens if you take the rotating fame of reference of the marry-go-round as stationary, and it is the earth rotating in the opposite direction at the same angular speed, will there still be any force pulling you toward the centre of the 'stationary' merry-go-round? If we consider the earth as stationary and it is the universe rotating around it, how can we define the centripetal force? As the earth is not spinning, the gravitational pull of the poles will be identical to that of the equator if we consider the earth as an uniform sphere, which contradicts with the reality that in the poles there will be a theoretically greater gravitatinal pull since there is no centripetal force acting. Could anyone explain it?
     
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  3. Sep 17, 2014 #2

    A.T.

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    Yes, the same force that pulls you in the non-rotating frame.

    Gravity, which provides the centripetal force in the non-rotating frame, still acts in the rotating frame of the Earth. But since the Earth doesn't rotate in that frame, the concept of centripetal force is not useful for the Earth. But the universe rotates in that frame, and here the centripetal force is providied by the inertial Coriolis force.
     
  4. Sep 17, 2014 #3

    Doc Al

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    Since the man is moving in a circle, and thus accelerating in the inertial frame of the earth, there must be a net centripetal force on him. That centripetal force is given by the merry-go-round exerting a force on the man.

    In the non-inertial rotating frame of the merry-go-round, Newton's 2nd law must be modified to include non-inertial forces such as an outward centrifugal force acting on the man. Including that force, you can apply Newton's 2nd law. Since the man is not accelerating, the net force must be zero. The inward force from the merry-go-round (which is a 'real' force and thus appears in any frame) is canceled by the outward centrifugal force.
     
  5. Sep 17, 2014 #4
    Thanks for replying!

    But why in the non-rotating frame, no corilis effect could be observed, as the earth rotating around the table and the table rotating with respect to the earth in an opposite direction are literally the same thing? Consider the swing carousel, if we we say it is the earth rotating around the passengers, why people have nausea. There is also a similar question that bothers me, when we are travelling in a accelerating car we would feel a force exerting on us by the seat, as F=ma, but how when we consider it is the road accelerating backward? Will we have the same feeling?
     
  6. Sep 17, 2014 #5

    Nugatory

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    It's not the exact same thing, because the acceleration produces measurable phyical effects.

    Let's say I'm in a windowless room. I cannot tell whether I'm moving at or at rest (no windows, so I can't look out and see if my position relative to other objects in the universe is changing); all I know is that I am at rest relative to the room. However, I can tell whether that frame is inertial or not - if I feel a force pulling towards floor, ceiling, or one of the walls then I know that the frame is not inertial.
     
  7. Sep 17, 2014 #6

    Andy Resnick

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    No, friction is what keeps the person from sliding off- if the merry-go-round were made of ice, the person would indeed slide off. In this case, the friction force balances the cerntripetal force.

    There is still no net force, and no contradiction.
     
  8. Sep 18, 2014 #7

    A.T.

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    No, the friction on the person acts inwards, towards the center, and thus provides the centripetal force necessary to keep the person in circular motion. In the co-rotating frame, where the person is at rest, there is an inertial centrifugal force on the person that balances the friction, and assures zero net force.

    What do you mean by "still no net force"? In the non-rotating there was a net force.
     
  9. Sep 18, 2014 #8
    I'm still confused. What does it mean by acceleration? How can we tell whether an object is accelerating or not? Lets take linear motions as example, as I have learnt from my alevel physics that if a frame is accelerating with respect to another frame, from the stationary frame's point of view there is a net force acting on the objects in the accelerating frame. But if we consider the accelerating frame as stationary, and the objects inside the frame have no relative motion to it, we imagine that there is an inertial force and it is balanced by an opposite force (We are 'pushed' by the seat when we sit in an acelerating car). If the "opposite force' is removed, we will accelerate in the same direction as the inertial force (If we throw a ball in an accelerating train it accelerates backward.) If we consider the accelerating frame stationary, how could it happens? We could say in an accelerating car we are not accelerating with respect to another car which accelerates at the same rate, so why do we experience the inertial force? In a stationary car which other cars accelerate with respect to it, why don't we experience the inertial drag?
     
    Last edited: Sep 18, 2014
  10. Sep 18, 2014 #9

    Andy Resnick

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    The video "frames of reference" does a very good job of explaining why we 'invent' forces to reconcile observations between accelerating frames of reference.
     
    Last edited by a moderator: Sep 25, 2014
  11. Sep 18, 2014 #10

    AlephZero

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    There are two equal and opposite friction forces, one acting on the man, one on the merry-go-round. So there is a net force on the man, which is obviously true. Otherwise he would be moving with constant velocity in an inertial frame, not in a circle.

    There is also a reaction force on the axis of the merrry-go-round, balancing the outward friction force on it. Or, if you consider the combined system of the merry-go-round plus the man, its center of mass is not on the axis of rotation, and you will calculate the same force on the axis doing it that way.
     
  12. Sep 18, 2014 #11

    A.T.

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    The friction acting inwards on the man is a real force, not an 'invented' one. It acts as the centripetal force, not to "balance the centripetal force" as you claimed.
     
    Last edited by a moderator: Sep 25, 2014
  13. Sep 18, 2014 #12
    The friction force *IS* the centripetal force. It keeps the man accelerating towards the center. This is the view from the earth, a rest frame. Centripetal force is real, but it is always, AFAIK, a specific force like friction or gravity. Friction in the merry go round provides the centripetal force/acceleration, whereas gravity does so in an orbit such as moon/earth system. The term "centripetal" simply infers a center oriented force provided by one of several types of force like gravity or friction. Did I help at all?

    Claude
     
  14. Sep 18, 2014 #13

    A.T.

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    Not in general. In rotating frames the inertial Coriolis force can act as the centripetal force in circular motion. See post #2.
     
  15. Sep 19, 2014 #14
    I did not say "in general". Please reread my post. I said in the reference frame of the earth, a stationary frame. Now you invoke rotating frames, which I did not include. Do you get a rush correcting people? You're rebuking a point I never made.

    Claude
     
  16. Sep 19, 2014 #15

    A.T.

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    You said:

    It wasn't clear what you meant by "specific force like friction or gravity", so I merely clarified that it be can a coordinate effect like Coriolis.
     
    Last edited: Sep 19, 2014
  17. Sep 19, 2014 #16
    Coriolis and centripetal are 2 differing entities. Please clarify so that I may understand any situation where centripetal is a mere coordinate effect. I realize that "centrifugal" force is coordinate related, a pseudo-force, a math construct, but I am not aware that centripetal can have that type of status. An example would be appreciated. Thanks.

    Claude
     
  18. Sep 19, 2014 #17

    olivermsun

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    Well suppose you are in a rotating coordinate system, e.g., on the earth.

    Viewed from within your frame:

    You fire an artillery shell far and fast enough that you observe the trajectory curving to one side.

    The acceleration is "centripetal" because it involves a movement on a curved path.

    The cause of the acceleration is the Coriolis force.
     
  19. Sep 19, 2014 #18

    A.T.

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    See post #2. In the rest frame of the Earth distant starts move in circles around the Earth. The centripetal force associated with that circular motion is the Coriolis pseudo-force (reduced by the centrifugal pseudo-force).
     
  20. Sep 19, 2014 #19

    jbriggs444

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    The example I like is from the point of view of a fellow on a carousel who uses the rotating frame in which the carousel is at rest. He observes a child standing on the ground outside the carousel.

    In the rotating coordinate system, the child is revolving around the carousel with a centripetal acceleration of ω2r and is subject to a centrifugal force whose magnitude is also ω2r. The centripetal force required to produce that centripetal acceleration and negate the centrifugal force is provided by the Coriolis force with a magnitude of 2ωv = 2ω2r.
     
  21. Sep 19, 2014 #20
    Bold underlined statement - agreed.

    Bold italicized statement - disagree. In the rotating frame R, the child is revolving with centripetal acceleration rω2, as we agree on that. But this centripetal force is NOT counter-balanced by any centrifugal force. Let me elaborate.

    In R, the rotating frame, the man incurs both a centripetal and centrifugal force, they effectively cancel. In this frame R, Newton's laws hold by creating the math construct known as centrifugal force. Since the man is not moving in R, the net force on him must be 0 in the R frame. Hence his centripetal component of force accelerating him towards the center must be canceled by an equal and opposite force directed away from the center. This force is centrifugal. In R, the man incurs both, which he must since he is at rest. At rest in R, his net forces incurred must sum to 0. Centripetal and centrifugal forces are equal in magnitude and opposite in direction, therefore summing to 0.

    But the child (assume a girl) outside R, in the stationary frame S, when viewed from R, differs from the man in R as far as incurred forces go. In R, the girl is accelerating, not at rest. Since she is accelerating as viewed from R, her net forces incurred are non-zero, wrt R. The force accelerating her towards the center when viewed in R, is indeed centripetal as you stated. But unlike the man, the girl is not at rest wrt R, she is accelerating.

    Therefore the centripetal force on the girl in R frame is not balanced by centrifugal. Unlike the man who is at rest in frame R, the girl incurs a centripetal force unbalanced, resulting in a centripetal acceleration. The man, OTOH, in the frame R, is at rest, not accelerating. Thus any force on him must be counter-balanced to keep him at rest. So the man in R incurs both forces, summing to 0, which agrees with his rest state in R.

    Coriolis is not the same as centripetal nor centrifugal. The 2ωv is not equivalent to rω2. Please recheck your references. Best regards.

    Claude
     
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