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B A question regarding the Centripetal force

  1. Apr 7, 2017 #1
    Hi, if you consider a ball, with mass m, rolling down the outer surface of a hemisphere with radius r, why is it that if you consider the forces acting upon the ball and resolve in a direction, the equation formed is:

    mgcos(x) - R = m(v^2)/r

    If you rearrange that, it would appear that the reaction force is acting in the same direction as the centripetal force and that just doesn't make sense to me. Where am I going wrong here? I'd be grateful if someone could go through this with me :)

    Edit: x is the angle between the ball and the vertical line running through the hemisphere's centre
     
  2. jcsd
  3. Apr 7, 2017 #2

    CWatters

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    Perhaps start with a drawing and show us how you got that equation.

    PS The R term and the 0.5mv^2 term appear to have opposite signs so why do you say they are in same direction?
     
  4. Apr 7, 2017 #3
    They're on opposite sides of the equation... put them on the same side and they have the same sign.
     
    Last edited: Apr 7, 2017
  5. Apr 7, 2017 #4

    A.T.

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    Your equation already assumes certain positive directions / sign conventions (which are opposite for R and centripetal). You get the actual directions by considering the convention and the sign you get form the calculation.

    To avoid this type confusion, use vectors that are all in the same coordinate system.
     
  6. Apr 7, 2017 #5
    Forces.png

    There you go! :)
     
  7. Apr 8, 2017 #6
    Consider when X = 90°. Your equation breaks down because the LHS will be zero but yet there should an 'mg' term still acting. Only the normal force from the hemisphere will have gone to zero.

    The total forces acting on the ball at any time are

    Ftot = Fg + Fn + Ff

    Where Ff is the frictional force on the surface of the ball due to contact with the hemisphere. Fg and Fn are gravity and normal, respectively.

    Because Fn always acts through the centre of the ball, it induces no torque.

    Only Ff will induce a torque to roll the ball, and it follows from Newtons laws in rotational dynamics that for a solid sphere:

    Ff*R = 2/5mR^2*alpha.

    Where R is the radius of the ball NOT the hemisphere. Alpha is the angular acceleration of the balls surface wrt to its axis of rotation.
     
  8. Apr 8, 2017 #7

    PeroK

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    Good diagram. Can you explain your confusion? I might write things the other way round:

    ##R = mg\cos(x) - \frac{mv^2}{r}##

    The normal force is what is left after the circular motion takes a bite out of the normal component of the gravitational force. No motion means the sphere feels the full normal gravitational force; and the faster the ball is rolling, the less force is felt by the sphere; until, at some point, the normal force equals zero and the ball comes off the sphere.
     
  9. Apr 8, 2017 #8

    A.T.

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    The arrows indicate the positive direction of the forces. So if R and Fc are both positive, then they point in opposite directions.
     
  10. Apr 9, 2017 #9
    It helps to establish a coordinate system. While you are on the right track with your formula and your diagram, I feel there are essential bits missing from the diagram, and errors in the formula.

    There is a *possible* coordinate system drawn below and some relations that might help you to see how you may wish to modify your approach.

    20170409_162127.jpeg
     
  11. Apr 9, 2017 #10
    two vector equations in the brekets give three scalar equations for unknowns ##\dot\varphi, T_x,T_y##
     

    Attached Files:

  12. Apr 9, 2017 #11
    Thanks for the reply! Part of the confusion was caused by me thinking that centripetal force (Fc) was meant to be included in the force diagram. I've read elsewhere that Fc is simply just the "net force" and is equal to (reaction force - mgcos(x)) or (mgcos(x) - reaction force). So the only two forces I needed to draw in that diagram is the reaction force and the weight; I should have mentioned that the hemisphere's surface is smooth so there's no friction involved as well. Thanks again to you and the other people who helped me out here, I appreciate it :)
     
  13. Apr 9, 2017 #12
    Hi Ashton, you just said there is no friction. So is the ball rolling, or is it sliding?

    Generally in this kind of problem 'no friction' would imply no rolling friction, meaning no deformation of the surfaces, and no torque imparted by the normal force.
     
  14. Apr 9, 2017 #13

    CWatters

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    No friction has implications for the direction of the reaction force in most problems.
     
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