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Block sliding on the inside of a cone

  1. Oct 29, 2006 #1
    Hello there, this is my first posting on this board. I am a third year physics major and I've taken a lot of courses but I've always just scraped by with a large amount of help from small groups. I'm thinking about reconsidering my major because I feel like I no longer can understand the concepts.. but anyways, I came across this problem and I am stumped.

    Using {r, phi, z} coordinates write the equation of motion for a small block that is free to move on the inside surface of an upper cone under the influence of gravity. Repeat this problem with the addition of simple friction.

    I have done the free body diagram but I don't know how to post that on here. Anways this is what i got:

    r^=-Ncos(phi)=-mv^2/r
    z^=Nsin(phi)-mg=0

    Then divide z^ by r^ and get this:
    tan(phi)=rg/(v^2)

    ???

    I really have no idea what I am doing. Any help or sense of direction would be greatly appreciated.
     
    Last edited: Oct 29, 2006
  2. jcsd
  3. Oct 30, 2006 #2

    Office_Shredder

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    What's an upper cone?
     
  4. Oct 30, 2006 #3
    It's an inverted cone, it looks like this:

    \/
     
  5. Oct 30, 2006 #4

    OlderDan

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    It looks like you are using phi as the angle of the cone. Is that what is intended? I would think there has to be an azimuthal angle in your coordinate system, and some people use phi for that angle. Does your text or prof use it that way?. The angle of the cone establishes a fixed ratio for r and z (taking the origin to be the vertex of the cone). I take it the by r^ you mean the force in the r direction and z^ is the force in the z direction. There has to be something about the azimuthal direction in the equations of motion. Unless the mass moves only in a vertical plane (two dimensional problem of an inclined plane) the azimuthal motion is an important component.

    Let's make sure the variables are correctly identified, and then take it from there.
     
  6. Oct 30, 2006 #5
    Okay i'll try to explain the setup I did as well as I can

    \|/_
    the | between the \/ represents the Z axis.
    The underscore represents the R axis
    and the angle between |/ is Phi.

    Does that help explain my setup at all? I'm trying to make it as clear as possible..
     
  7. Oct 30, 2006 #6

    OlderDan

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    Then if the vertex of the cone is at the origin, phi is just a parameter of the cone such that tan(phi) = r/z = constant. How are you going to describe the position of the block in three dimensions when you only have two position coordinates (r and z) that have a fixed ratio? Your problem has been reduced to two dimensions with a constraint that reduces it to a single independent position variable. The real problem is a three dimensional problem with a constraint that relates two of the variables, reducing the number of independent variable to two. You need another position angle in the problem.
     
  8. Oct 30, 2006 #7
    \|/_
    the | between the \/ represents the Z axis.
    The underscore represents the R axis
    and the angle between |/ is Alpha, Like you said tan(alpha)=r/z.
    Phi is the angle of where the block is located inside of the cone, and it can be integrated from 0 to 2Pi to find the location of the block.

    Does that sound better?
     
  9. Oct 30, 2006 #8

    OlderDan

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    I would not use the word "integrated" but I know what you mean, and yes this sounds better. Phi is the azimuthal angle coordinate. Now try writing your equations of motion including the motion in the azimuthal direction.
     
  10. Oct 30, 2006 #9
    \|/_
    the | between the \/ represents the Z axis.
    The underscore represents the R axis
    and the angle between |/ is Alpha, Like you said tan(alpha)=r/z.
    Phi is the angle of where the block is located inside of the cone, and it can be integrated from 0 to 2Pi to find the location of the block.

    phi^=[m(v^2)]/r
    r^=-Ncos(alpha)
    z^=Nsin(alpha)-mg=0
    F=phi^ + r^ + z^

    Thanks for sticking with me. I know this is elementary to most everyone, I've taken E&M I and II, Modern, Thermo, and now I'm in classical and I feel like I've lost a grip on the fundamentals.
     
  11. Oct 31, 2006 #10

    OlderDan

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    There is nothing elementary about this.

    In the absence of friction, there is no force in the phi direction. The normal force is the only force acting other than gravity. The radial force is the centripetal force, as you originally had it, but you want to express it in terms of angular velocity rω = rdφ/dt, v^2 = r^2ω^2. Any velocity components in the r or z direction do not contribute to the centripetal acceleration.
     
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