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Homework Help: Dynamics of Circular Motion

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A small block with mass m is placed inside an inverted cone that is rotating about a vertical axis such that the time for one revolution of the cone is T. The walls of the cone make an angle β with the vertical. The coefficient of static friction between the block and the cone is [itex]\mu_{s}[/itex]. If the block is to remain at a constant height H above the apex of the cone, what are the maximum and minimum values of T?

    2. Relevant equations

    [itex]\Sigma[/itex]F=ma

    a[itex]_{rad}[/itex]=[itex]\frac{V^{2}}{R}[/itex]

    V=[itex]\frac{2\pi R}{T}[/itex]

    f[itex]_{s}[/itex]=[itex]\mu_{s}[/itex]n

    R=H tan[itex]\beta[/itex] (From diagram)

    3. The attempt at a solution

    As far as I can tell, the only problem I'm having is with my diagram. I first placed the x-axis along the side of the cone with the friction force parallel to it, and then moved it clockwise until the weight was parallel to the y-axis. The angles that form can be seen in the attached file, along with my work.

    What I come up with is T[itex]_{max}[/itex]=2[itex]\pi[/itex][itex]\sqrt{\frac{h tanβ(cosβ-\mu_{s}sinβ)}{g(sinβ+\mu_{s}cosβ)}}[/itex] and T[itex]_{min}[/itex]=2[itex]\pi[/itex][itex]\sqrt{\frac{h tan(cosβ+\mu_{s}sinβ)}{g(sinβ-\mu_{s}cosβ)}}[/itex]

    The answer that the book gives is T[itex]_{max}[/itex]=2[itex]\pi[/itex][itex]\sqrt{\frac{h tanβ(sinβ+\mu_{s}cosβ)}{g(cosβ-\mu_{s}cosβ)}}[/itex] and T[itex]_{min}[/itex]=2[itex]\pi[/itex][itex]\sqrt{\frac{h tanβ(sinβ-\mu_{s}cosβ)}{g(cosβ+\mu_{s}sinβ)}}[/itex]

    I can only come up with this solution if I switch the angles around that the normal and friction forces make.

    Also, the question comes from Young and Freedman 11th edition. Chapter 5, problem 5.119
     

    Attached Files:

    Last edited: Jan 25, 2012
  2. jcsd
  3. Jan 25, 2012 #2
    In the first force diagram β is in the wrong place?
     
  4. Jan 26, 2012 #3
    Oh man, you're right!!! Thank you.
     
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