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Circular Motion / Newton's Laws

  1. Oct 1, 2007 #1
    A very small cube of mass m is places on the inside of a funnel rotating about a vertical axis at a constant rate of W revolutions per second. The wall of the funnel makes an angle A with the horizontal. The coefficient of static friction between the cube and funnel is us and the center of the cube is at a distance r from the axis of rotation. Find the largest and smallest values of W for which the cube will not move with respect to the funnel.

    I'm not really sure how to even begin this one
    The Fnet in the y direction must = 0 for it to not move
    and the fnet in the x direction must move with the funnel
    we have the force of gravity acting on the mass
    Fg = mg
    we need force of friction
    normal force will be perpindicular to the funnel
    so Fn = cosA*Fg = cosA*mg
    Ff = usFn = us*cosA*mg

    is this correct so far?

    Thanks
     
  2. jcsd
  3. Oct 1, 2007 #2

    Astronuc

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    correct.

    Also don't forget the horizontal force due to rotation.
     
  4. Oct 2, 2007 #3
    Fn = cosA*Fg = cosA*mg
    Ff = usFn = us*cosA*mg

    Horizontal force due to rotation would be a centripedal force towards the centre of the circle?
    Fc = mv^2/r

    So I have all the forces acting on the mass (I believe)

    I need to find a way to incorporate W revolutions per second into this.
    Could I use W revolutions per second instead of velocity in the Fc equation?
    Fc = m(W/s)^2/r
    Or would that cause the units to go bonkers?
     
  5. Oct 2, 2007 #4

    learningphysics

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    In this particular case I don't think Fn = mgcos(theta).

    You do know that in the vertical direction net force = 0. And you know that in the horizontal direction, net force = mv^2/r.

    What are the components of the normal force and the frictional force in the vertical diretion?

    What are their components in the horizontal direction?
     
  6. Oct 2, 2007 #5
    for normal force,
    would Fn in the vertical direction be Fn = sinA/mg
    and the the horizontal Fn = cosA/Fc = cosA/(mv^2/r)

    and frictional force being usFn
    vert Ff = us(sinA/mg)
    hori Ff = us[cosA/(mv^2/r)]
    ?
     
  7. Oct 2, 2007 #6

    learningphysics

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    never mind about mg for now... You have Fn. What is its horizontal component and vertical component (in terms of Fn and theta)?
     
  8. Oct 2, 2007 #7
    I'm not sure I'm following you
    How can I find Fn without mg?

    would I want the verticle component to be
    y = Fn*sin(theta)

    and horizontal:
    x = Fn*cos(theta)
     
  9. Oct 2, 2007 #8

    learningphysics

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    yes, you will need that afterwards.

    yes, sorry but can you write y and x in terms of the angle A? If you were using theta as A, then the components aren't right.
     
  10. Oct 2, 2007 #9
    y = Fn*cosA
    x = Fn*sinA
     
  11. Oct 2, 2007 #10

    learningphysics

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    exactly. And how about the x and y components of friction using A?
     
  12. Oct 2, 2007 #11
    for friction wouldnt just be the us of Fn?
    Ff = usFn
    Fn = Ff/us

    y = (Ff/us)*cosA
    x = (Ff/us)*sinA
     
  13. Oct 2, 2007 #12

    learningphysics

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    Yes, this is true at the limit, when it is just about to slip.

    ok. but these aren't the vertical and horizontal components of friction... these are still the vertical and horizontal components of the normal force.

    what are the vertical and horizontal components of friction... do it the same way you did it with the normal force.
     
  14. Oct 2, 2007 #13
    for Fn
    y = Fn*cosA
    x = Fn*sinA

    for Ff
    y = Ff*sinA
    x = Ff*cosA
     
  15. Oct 2, 2007 #14

    learningphysics

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    exactly. now here's the main trick to the problem... the friction can act either way... either up the plane or down the plane... depending on the rate of W. meaning the components for friction could be:

    y = Ff*sinA (upward)
    x = Ff*cosA (rightward)

    or

    y = -Ff*sinA (downward)
    x = -Ff*cosA (leftward)

    in the moment when the cube is about to slip Ff = us*Fn.

    you need to solve the problem for the two situations... when the friction is up the plane with a magnitude of us*Fn... that gives one W. And then when the friction is down the plane with a magnitude of us*Fn. that gives the other W.

    you know that the y-components of friction, normal force cancel with gravity...

    And you know that the x-components of friction and normal force give rise to centripetal motion... hence [tex]\Sigma{F_x} = mv^2/r[/tex]
     
  16. Oct 2, 2007 #15
    for Fn
    y = Fn*cosA
    x = Fn*sinA

    for Ff
    y = Ff*sinA (upward)
    x = Ff*cosA (rightward)
    or
    y = -Ff*sinA (downward)
    x = -Ff*cosA (leftward)

    in the moment when the cube is about to slip Ff = us*Fn.

    So first situation with Friction up the plane
    y = Ff*sinA (upward)
    x = Ff*cosA (rightward)

    Fnet,x = mv^2/r = Ff*cosA + Fn*sinA
    mv^2/r = us*Fn*cosA + Fn*sinA
    mv^2/r = Fn(us*cosA + sinA)

    I'm not sure how to incorporate the W into the expression, can I use W/s to replace velocity?
     
  17. Oct 2, 2007 #16

    learningphysics

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    careful here. we're looking for the centripetal force... the x-component of the normal force is acting inward... so it should be positive. But the x-component of the frictional force acts outward so it should be negative. (unless I'm misunderstanding the problem)

    yes. get the velocity in terms of W.
     
  18. Oct 2, 2007 #17
    ah yes you're right

    So first situation with Friction up the plane
    y = Ff*sinA (upward)
    x = -Ff*cosA (rightward)

    Fnet,x = mv^2/r = -Ff*cosA + Fn*sinA
    mv^2/r = -us*Fn*cosA + Fn*sinA
    mv^2/r = Fn(-us*cosA + sinA)
    v = rFn(-us*cosA + sinA) / m
    v = W/s = rFn(-us*cosA + sinA) / m
     
  19. Oct 2, 2007 #18

    learningphysics

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    The W/s isn't right. W is the number of revolutions per second.

    so [tex]2\pi{r}W[/tex] is distance travelled per second....

    ie [tex]v = 2\pi{r}W[/tex]
     
  20. Oct 2, 2007 #19

    learningphysics

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    don't forget the square.
     
  21. Oct 2, 2007 #20
    ah, alright, let me try again

    y = Ff*sinA (upward)
    x = -Ff*cosA (rightward)

    Fnet,x = mv^2/r = -Ff*cosA + Fn*sinA
    mv^2/r = -us*Fn*cosA + Fn*sinA
    mv^2/r = Fn(-us*cosA + sinA)
    v^2 = [rFn(-us*cosA + sinA) / m]
    v = (2pi*rW)^2= [rFn(-us*cosA + sinA) / m]
    W^2 = Fn(-us*cosA + sinA) / mr4pi^2
    W = sqrt[Fn(-us*cosA + sinA) / mr4pi^2]
     
    Last edited: Oct 2, 2007
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