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Ball rolling off a hemisphere

  1. Feb 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A ball is rolling along a frictionless hemisphere with radius R. The question asks about when will the ball rolls off the hemisphere. I understand that this happens when the normal force vanishes. But I am also wondering what if the normal force provided by the hemisphere balance off the component of weight of the ball along it s.t. the centripetal force is zero. Will the ball rolls off the hemisphere too?



    2. Relevant equations




    3. The attempt at a solution
    I think at that moment, the net force acting on the ball will be solely the tangential component of weight acting on the ball so the ball would roll off the hemisphere too? This really confuses me. Or this situation actually never happens? Thanks you!
     
  2. jcsd
  3. Feb 14, 2016 #2
  4. Feb 14, 2016 #3

    cnh1995

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    Here's what I think about this problem..
    Here, mv2/R will be the net force in the radial direction as long as the ball is on the hemisphere. If θ is the angular displacement of the ball from the top of the hemisphere and v is its tangential velocity,
    mv2/R=mgcosθ-Normal reaction
    At the break-off point, Normal reaction=0.
     
  5. Feb 14, 2016 #4
    It will fly off at a point angular displacement (theta) where cos(theta) =( R-h) /R where R is the radius and h is the height through which it has fallen from top- but the body can not roll -it will rather slip along the surface and the normal reaction will be equal to component of m.g along radius - the centripetal force required
     
  6. Feb 14, 2016 #5

    haruspex

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    The way you ask the question suggests to me you are confused about what centripetal force is. It is not an actual force, acting on the object along with other actual forces such as gravity and normal force.

    Centripetal acceleration is that component of an object's acceleration that accounts for its curvilinear motion, that is, it is the component normal to its velocity. Thus centripetal force is the component of the net force that acts normal to the velocity. Students get a bit confused because mostly the calculation runs backwards. Instead of calculating a net force, finding the normal component and deducing the radius of curvature, we usually start with the known speed and radius of curvature and deduce the centripetal force required to achieve it.

    This can lead to the perception that circular motion somehow causes the force. Indeed, in the case of an object spun in a centrifuge that is more or less the case. The normal force grows as necessary to prevent the object penetrating the wall, so it is reasonable to think of the rotational motion as causing the centripetal force. But in the present case that would be backwards. The rolling/sliding ball is not constrained to stay on the surface. The normal force cannot contribute to the centripetal force (beyond cancelling the excess contribution from gravity) because it acts the wrong way. When the speed and angle get to the point where gravity is no longer sufficient to supply the necessary centripetal force, the ball flies off.
     
    Last edited: Feb 14, 2016
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