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Normal force on slipping hemisphere

  1. Apr 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A hemispherical bowl of mass m and radius R is placed on a rough horizontal surface. Initially the centre C of the bowl is vertically above the point of contact with the ground(see figure).
    Now the bowl is released from rest. Find the normal force acting on the hemisphere initially if the coefficient of static friction is 1/2 of that required to prevent initial slipping of the bowl.
    2017-04-09 14.48.04.jpg

    2. Relevant equations
    a =R(w), w= angular acceleration
    Centre of mass of hollow hemisphere = R/2 distance from centre

    Max = nN (n = coeff. Of friction)
    Mg - N = May

    MgR/2 = 5/3MR^2 (w) ...Torque about lowermost pt
    w = 3g/10R
    3. The attempt at a solution
    I already figured out the minimum coefficient of friction for no slipping to be 6/17. This matches with the solution. But I am unable to find the normal force corresponding to half of this value(3/17).
    I have 2 equations relating horizontal and vertical accelerations of the CM to N and Mg (above) but this is not enough as I need another to eliminate variables.
     
  2. jcsd
  3. Apr 9, 2017 #2

    haruspex

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    You have not used the fact that the slipping bowl maintains contact with the surface, and does not penetrate it.
     
  4. Apr 9, 2017 #3
    The vertical acceleration of the lowermost point should be 0, then at this instant?
     
  5. Apr 9, 2017 #4

    haruspex

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    Yes.
     
  6. Apr 9, 2017 #5
    But that would mean the normal is the same as before(no slipping) as

    Vertical acceleration of lowest pt=
    Accn.(CM) - component of angular acceleration
    = (Mg-N)/M - (Rw)(2/√5) =0

    This gives N = 17mg/20 as nothing has changed from when there was no slipping. So what am I doing wrong?
     
  7. Apr 9, 2017 #6
    Let's see your free body diagram.
     
  8. Apr 9, 2017 #7
    this is absolutely not evident and it should be obtained from the equations of motion (if it is true indeed). Why do not you write the equations in vector form first?
     
  9. Apr 9, 2017 #8
    FBD attached
     

    Attached Files:

  10. Apr 9, 2017 #9
    I have doubts regarding this equation. These all should be carefully written from the theorems of dynamics
     
  11. Apr 9, 2017 #10
    Oh, I think i found a possible error - am I not allowed to apply torque = I*alpha about that point as it is a noninertial frame which is not the instantaneous axis of rotation or the centre of mass
     
  12. Apr 9, 2017 #11
    Ok yes that was the error. I solved it taking torque about the center of mass and got the correct value of alpha. Then I got the right answer(N = 85mg/118).
    Thanks for the help
     
  13. Apr 9, 2017 #12
    actually you are not allowed to use this formula about instantaneous axis of rotation also
     
  14. Apr 9, 2017 #13

    haruspex

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    The normal force between two bodies is the minimum force normal to the plane of contact that prevents interpenetration. The point of contact is initially at rest. Therefore it neither accelerates downwards into the floor nor up away from the floor.
     
  15. Apr 9, 2017 #14

    haruspex

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    There are two standard points on a moving object that can be used as axes for this equation. One is the mass centre and the other is the instantaneous centre of rotation. (In fact, it turns out that the set of axis points for which the equation holds forms a circle passing through those two points, but that's not very useful since it is quite hard to calculate.)
     
  16. Apr 9, 2017 #15

    haruspex

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    You can take moments about the lowest point provided you use the point of contact on the ground, i.e. a fixed point in space. The angular acceleration about that point includes a component from the linear acceleration of the mass centre.
     
  17. Apr 9, 2017 #16
    Another informal explanation. All assertions must follow from the equations of motion. I do not claim that it is wrong I just say that formulas are needed, not words.

    The last part is wrong. General formulation of corresponding theorem for planar case is as follows. Let ##A## be a point of the rigid body and ##S## be its center of mass. Then
    $$J_A\dot{\boldsymbol\omega}+m \boldsymbol{AS}\times \boldsymbol a_A=\boldsymbol M_A,\qquad (*)$$
    here ##J_A## is the moment of inertia about an axis that is perpendicular to the plane and passes through the point ##A##,
    ##m## is a mass of the body
    ## \boldsymbol a_A## is an acceleration of the point ##A##,
    ##\boldsymbol M_A## is a torque that the rigid body experiences about the point ##A## .
    So from (*) it is clear that your assertion holds if only vectors ##\boldsymbol a_A## and ##\boldsymbol{AS}## are collinear.
     
  18. Apr 9, 2017 #17

    haruspex

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    Yet the equations themselves follow from observations about the physical circumstances. My argument in post #13 seems sufficiently rigorous to me.
    We may be at cross purposes here. I agree that the contribution of the linear acceleration must be included. See my post #15. As long as you do that it is clearly valid to use the point in the body which is the instantaneous centre of rotation since that is effectively an inertial frame. What is perhaps surprising is that is not valid to use an arbitrary point fixed in the body, even if the linear contribution is included.
     
  19. Apr 9, 2017 #18
    Now I see post #15:
    Sorry, this phrase does not make sense: angular acceleration of a rigid body does not depend on points. I also do not understand how angular acceleration can include components of a linear acceleration. What does exactly this mean?

    Anyway the second term in the left side of (*) has not disappeared yet.
     
  20. Apr 9, 2017 #19

    haruspex

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