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Rotation and force

  1. Jun 11, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]R=0.2m [/tex]


    [tex]\mu _s=0.5 [/tex] between the iron cube and the surface


    2. Relevant equations

    Fx=Fsin(α) or cos(α)

    3. The attempt at a solution
    in pink I drew the gravity force and the normal force.
    Which other force acts on the body? the friction force is directed along the surface, but in which direction going towards the center of the cone or outward? Why?

    Attached Files:

    Last edited: Jun 11, 2007
  2. jcsd
  3. Jun 12, 2007 #2


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    I can't make out the setup from your diagram. Centripetal force is directed towards the center of rotation.
  4. Jun 12, 2007 #3
    the exercise is a little difficult to explain in words, it would be much easier with a picture.
    An iron cube is on the inside rough surface ( [tex]\mu_s=0.5 [/tex]) of a cone which is upside down. this cone rotates around its centre axis with a [tex] \omega=constant[/tex]. between the axis and the surface there's an angle of [tex] \theta=30[/tex]°. the cube is at a distance R=20cm from the centre axis. find the [tex]\omega_{max}=?[/tex], maximum angular velocity, so that the cube remains still on the cone's surface.

    I drew 2 imaginary axis x and y with origin in the cube. I am sure there are at least 2 forces:gravity,pointing down, and normal pointing away form the surface. which is the direction of the friction force? is it toward the centre of the cone or outward? why?
    Last edited: Jun 12, 2007
  5. Jun 12, 2007 #4


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    OK, I think I understand the setup now. You need the sliding forces to cancel out. The downward sliding is due to gravity, and the counterforce is the component of the centripetal force that is parallel to the surface. Remember that the normal force will be made up of two components - gravity and centripetal force.
  6. Jun 12, 2007 #5
    [tex]m (\omega^2R) = N+G[/tex] where N and G stand for Normal force and Gravity force.

    omega² ∙R is the centripetal acceleration.
    How do I fit in the friction force?

    Is it right? I'm pretty much confused so please give me clear hints.
    Last edited: Jun 12, 2007
  7. Jun 12, 2007 #6


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    Ignoring friction for the moment, do you agree that

    downward sliding force = mgcos(theta)

    upward sliding force = mrw^2sin(theta)

  8. Jun 13, 2007 #7
    I agree with the 1st statement, but I'm still having doubts on the direction of the centrifugal force. Is it parallel to the ground (not the cone's surface) point out?
    Could you just show it with a picture post? This will clear out all my doubts. Thank you.
  9. Jun 13, 2007 #8


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    There is a component of the centripetal force that is parallel to the surface.
    I suggest you draw a diagram.
  10. Jun 13, 2007 #9
    Could you just tell me if the axis x and y are right as I drew them?
    Did I draw all the forces acting on the object correctly ?

    Attached Files:

  11. Jun 13, 2007 #10


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    Yes. That's what I drew. Now, I could be wrong, but in equilibrium those sliding forces are equal. Also friction is irrelevant because nothing moves.

    If you equate the forces, mass cancels out as it should, and you can solve for omega. Mass cancels out because gravity and centripetal 'force' are actually accelerations not forces.
  12. Jun 14, 2007 #11
    ok, thanks a lot for your help
  13. Jun 14, 2007 #12

    D H

    Staff: Mentor

    If it weren't for static friction, there would be no range of rotation rates over which the iron cube is stationary. There would instead be one equilibrium rate. If the cone rotates slow than this rate the cube falls down the cone. For a fixed rotation rate w, the w cross r term decreases as the cube falls down the cone and increases as the cube works its way up the cone. The equilibrium rate thus represents an unstable equilibrium.

    When you account for static friction (nothing is moving, so this is a static friction problem), the unstable equilibrium point becomes a range of rates.
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