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Normal Force While Sliding Off Ice Dome

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-3-12_16-11-51.png

    2. Relevant equations


    3. The attempt at a solution

    I know that the solution involves setting the normal force to zero and equating the radial component of gravity to centripetal force. But why is it not possible for the normal force to decrease at a sufficiently low rate such that even before the normal force reaches zero, the net force (normal+radial component of gravity) is insufficient for supplying the centripetal force, thus resulting in the boy falling off?

    Thanks for your time.
     
  2. jcsd
  3. Mar 12, 2015 #2

    ehild

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    The normal force N acts outward, opposite to the radial component of gravity, Gr. Gr-N=Fcp, N≥0, the ice dome only pushes, it can not pull. The boy flies off if negative normal force would be needed to keep him on track.
     
  4. Mar 12, 2015 #3
    Thanks for the reply.

    But what I don't quite understand is why the normal force takes whatever value it does at a particular angle. Why is it impossible for the normal force at a given angle to be so large that it counteracts much of the radial component of gravity and leaves a small net force which is insufficient to provide the centripetal force?
     
  5. Mar 12, 2015 #4

    ehild

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    The normal force just prevents the sliding body to break through the surface. You can not choose it. If the body is on the surface and moves along the circle determined by the surface, the resultant force must be equal to mv2/R, that is, the normal force must be N=Gr-mv2/R.
     
  6. Mar 12, 2015 #5
    I can accept that. But besides appealing to common sense to see that the body follows a circular path, is there any other way to account for why the normal force conveniently adopts the correct magnitude to maintain circular motion? I know that this magnitude is related to the compression of the interacting surfaces but why does it compress in just the right way for circular motion?

    Thanks for your time.
     
  7. Mar 12, 2015 #6

    ehild

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    Forget about circular motion. You stand on the horizontal ground. What is the normal force exerted by the ground on you?
    You know, it is equal to your weight, mg and it acts upward. So the net force is zero, you stay in rest.

    Why? If it was less you would sink into the Earth. If you tried to stand on water surface, you would sink, as the water surface is not able to exert so big force.
    The soil is somewhat elastic, so it exerts force about proportional the amount it is compressed, and at the end it exerts just the right magnitude of force to prevent further sinking.
    If the upward force from the ground would be greater than mg, you would fly upward. (That could happen if there was a bomb under the surface just exploding, or an earthquake or you stood on the top of a volcano, but you stay in rest during normal circumstances, so the resultant force on you is zero, and the resultant force is your weight and the normal force in opposite direction: N-mg=0)
    The same when a body moves along the circle. We know that it can only move along that circle of radius R and with speed V if the resultant force is the centripetal force inward: which is the inward radial component of the external force - the outward normal force. mV2/R=Gr-N. The normal force set its magnitude by itself, through the deformation the body causes.
     
  8. Mar 12, 2015 #7
    I agree. But suppose that we're dealing with more complex problems where the path of motion is less obvious, is there no way of finding the path since the magnitude of the normal force is unknown?

    And also, isn't the normal force greater than weight when a falling object rebounds from the ground and when an object rises in an elevator?

    Thanks.
     
    Last edited: Mar 12, 2015
  9. Mar 12, 2015 #8

    ehild

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    It is on the other way round. You can find the normal force from the path and motion of the body. In an accelerating elevator, the body accelerates with the elevator, and the force accelerating it is the resultant of mg and the normal force from the floor of the elevator.
    When a falling object rebounds from the ground, it exerts force on the ground during its deceleration during the impact. The ground is compressed, and this compression can be somewhat elastic. During rebound, the compression releases and the elastic energy is transferred to the object.
     
    Last edited: Mar 12, 2015
  10. Mar 12, 2015 #9

    PeroK

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    Deleted.
     
    Last edited: Mar 12, 2015
  11. Mar 12, 2015 #10

    PeroK

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    Deleted.
     
    Last edited: Mar 12, 2015
  12. Mar 12, 2015 #11

    PeroK

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    Okay, I've deleted all my posts on this thread. The floor is yours.
     
  13. Mar 12, 2015 #12

    haruspex

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    It isn't the normal force that provides for circular motion, it's the net force resulting from gravity and the normal force. The normal force is whatever is sufficient to prevent penetration of the ice, and no more. It would be more accurate to think of gravity as providing the centripetal force, with the normal force opposing it as necessary.
     
  14. Apr 11, 2015 #13
    Sorry for the bump. But I gave this problem some further consideration and I'm just wondering how one would proceed with the problem if the ice dome were elliptical in shape with precisely defined semi-major and minor axes.

    Thanks
     
  15. Apr 11, 2015 #14
    You would need an equation to describe the shape of the ice mound. You would also need an equation to describe the centripetal acceleration of a point mass moving through the new ellipse shape. Here is a thread discussing just that. https://www.physicsforums.com/threads/centripetal-acceleration-of-an-ellipse.709317/ It would certainly make the problem more difficult. If you could find those two equations, however, I think the approach to solving the problem would be the same. Did you find the solution under the circumstance that the ice mound is perfectly circular?
     
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