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Mass on a Rotating Parabolic Dish

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    A 3-kg block is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Knowing that the coefficient of static friction is 0.66 and that r = 2 m, determine the maximum allowable velocity v of the block.

    Picture attached below

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    I began by drawing a FBD on the mass:
    W: Weight, pointing vertically downwards
    N: Normal, pointing perpendicular to the surface
    f: Friction, pointing 90 degrees clockwise from N.

    From there I calculated the angle from the origin:
    y = (2^2)/4 = 1
    Θ = tan^-1(1/2), with the adjacent arm being 2, vertical arm being 1, and the hypotenuse being sqrt5

    Then I determined that the net force should be horizontally pointed inwards, with the formula: F = mv^2/r

    So Fy is 0, meaning:
    (3)(9.81) = 0.66N[1/sqrt(5)] + N[2/sqrt(5)]
    --> N = (29.43sqrt[5])/(2.66)

    Now for Fx:
    [3v^2]/2 = N[1/sqrt(5)] - 0.66N[2/sqrt(5)]
    The problem is this resolves to be:
    [3v^2]/2 = -3.54...
    which has no solution.

    So, what am I supposed to do from here?
     

    Attached Files:

    Last edited: Feb 5, 2017
  2. jcsd
  3. Feb 5, 2017 #2
    Its rotating so I think there should be centripital force maybe ?
     
  4. Feb 5, 2017 #3
    Wouldn't it just be the net force, mv^2/r? Or would it be a fourth force?
     
  5. Feb 5, 2017 #4
    İt would be the net force...Its really hard to undersatnd from your work.Did you add it ? oh ok I saw it
     
  6. Feb 5, 2017 #5
    Yeah, that's how I ended up with an indeterminate equation (I would have had to square root a negative).

    I pointed the centripetal force horizontally to the left, and attempted to solve it this way (N was calculated by setting the net force in y equal to 0):
    [3v^2]/2 = N[1/sqrt(5)] - 0.66N[2/sqrt(5)]
    The problem is this resolves to be:
    [3v^2]/2 = -3.54...
     
  7. Feb 5, 2017 #6

    gneill

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    Where did you get this equation for y? Check the source.
     
  8. Feb 5, 2017 #7
    Sorry, I meant to write 2^2/4, but that would actually equal 1.
     
  9. Feb 5, 2017 #8
    Do you know the answer ?
     
  10. Feb 5, 2017 #9

    gneill

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    Staff: Mentor

    Of what use is the angle from the origin?

    More interesting than the angle of the line from the origin to the block would be the local slope of the dish where the block is resting. The problem will then look more like a block resting on an inclined plane with various forces acting. How might you determine the slope and hence the angle of incline?
     
  11. Feb 5, 2017 #10
    No.

    Differentiate y, with respect to r:
    dy/dr = r/2
    Then with r being two, that would imply a 45 degree angle.

    So, knowing that, would my solution otherwise be the same, just using 45 degrees instead? Or am I still missing something?
     
  12. Feb 5, 2017 #11
    I agree I think its true
     
  13. Feb 5, 2017 #12

    gneill

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    I kind of phased out looking at your solution after I noticed the angle issue. But essentially you need to resolve the centripital and gravitational forces into normal and slope-wise components to continue.

    Hint: If you work in the non-inertial frame of reference of the rotating dish you're allowed to speak of centrifugal force as though it were an actual force and use it in your FBD.

    upload_2017-2-5_16-4-49.png
     
    Last edited: Feb 5, 2017
  14. Feb 5, 2017 #13
    Okay, I'm still having difficulty with this. So, what I did was rotate the coordinate system 45 degrees (CCW).

    Now, on your diagram I'm mostly confused as to how the Normal and Frictional forces were resolved into one centrifugal force, unless I misunderstood the diagram.

    Because in the direction of the tangent, there's no net force, which implies that Fc = W. From there, both W and Fc point in direction of the net force (along the normal axis) which implies:
    [3v^2]/[2] = [1/sqrt(2)][2*3*9.81]
    From there, v can be solved to be 5.27 m/s, but that was marked as wrong.

    So what am I supposed to do with the contact forces?
     
  15. Feb 5, 2017 #14

    gneill

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    My diagram does not include the friction force, only the two "external" forces due to gravity and rotation. It shows how they resolve into a net external force. I suppose I should have made that clear. Sorry about that.

    You need to resolve that net force into normal force and force along the surface (incline), then include friction (which requires that normal force component) to write the net force acting along the slope.
     
  16. Feb 5, 2017 #15
    I still don't know what to do, from here. I'm still confused about the centripetal force (Fc). Is it a result of the other 3 forces (normal, friction, gravity) or is it a 4th independent force resulting from the rotation?
     
  17. Feb 5, 2017 #16

    gneill

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    When you move to the rotating frame of reference you can use the centrifugal force in the same way that you use the gravitational force. The only difference is that the centrifugal force is position dependent (radial distance from the axis of rotation). In my figure I've drawn in the gravitational and centrifugal forces that act on the block at its current location. It's up to you to break them down into normal and slope-wise components.
     
  18. Feb 5, 2017 #17

    haruspex

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    Centripetal force is just the radial component of the resultant force. It is not an additional force.

    Edit: on the other hand, using the block's non-inertial frame, you have centrifugal force acting like an applied force (and a zero resultant for equilibrium).
     
    Last edited: Feb 5, 2017
  19. Feb 5, 2017 #18
    I'm really confused right now. To project Fc and Fg onto tangental/normal components, you'd just multiply it by 1/sqrt(2) (sin or cos 45) which will give both the tangental and normal components for both forces (and then pointing in their respective directions).

    But then, I still don't see how the frictional force comes back into play.
     
  20. Feb 5, 2017 #19

    gneill

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    What direction will friction act? Normal to the slope or parallel to it? What has to occur for the block to move?
     
  21. Feb 5, 2017 #20
    It'd be parallel to the slope.
     
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