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Nonuniform Disk Frequency

  1. Dec 22, 2009 #1
    Well, I had a couple problems on my final I was hoping to go over- hope nobody minds. Here's the second.

    1. The problem statement, all variables and given/known data
    A nonuniform disk of radius R and mass m has the center of mass at a distance A from the geometrical center. Its moment of inertia about the axis passing through the center of mass and perpendicular to it's plane is I. Assuming that the cylinder rolls without sliding, find the angular frequency of small oscillations about a position of stable equilibruim on a horizontal plane.


    2. Relevant equations
    Parallel Axis Theorem
    I = i + mr^2, where i is the moment of inertia through the geometrical center.

    Kinetic Energy of A Rotating Object
    T = T(translational) + T(rotational)
    = (m/2)*(x'^2 + y'^2) + (1/2)*I*omega^2

    Lagrangian
    L = T - U
    And
    d( dL/dq' ) / dt - dL/dq = 0, where ' is short hand for the time derivative.

    Small Angle Approximation
    cos(phi) is about equal to 1 - (1/2)*(phi)^2

    3. The attempt at a solution

    First, we put the pieces of the Lagrangian together.

    U = Potential Energy, in this case we only worry about Gravitational. Basing it off of the center of mass of the cylinder I got this.

    U = m*g*(R - A*cos(phi)) where phi is the angle of the center of mass of the disk to the center of the disk to the vertical axis.

    T(rotational) = (1/2)*I*phi'^2.

    By parallel-axis theorem I = (mR^2)/2 + mA^2

    so T(rotational) = (m/2)*((R^2)/2 + A^2)*phi'^2.

    And T(translational) = (m/2)*(x'^2 + y'^2)

    x = D + A*sin(phi) where D is displacement from the origin in x.
    y = R - A*cos(phi)

    So x' = A*phi'*cos(phi) and y' = A*phi'*sin(phi)

    Which gives T(translational) = (m/2)*((A*phi'*cos(phi))^2 + (A*phi'*sin(phi))^2)
    Which simplifies to T(translational) = (m/2)*A^2*phi'^2


    Thus L = (m/2)*A^2*phi'^2 + (m/2)*((R^2)/2 + A^2)*phi'^2 - m*g*(R - A*cos(phi))

    We limit ourselves to small oscillations so. . .

    L = (m/2)*((R^2)/2 + 2A^2)*phi'^2 - m*g*(R - A*(1 - (1/2)*phi^2))

    Thus dL/d(phi') = m*(2A^2 + (R^2)/2)*phi' so d( dL/dq' ) / dt = m*(2A^2 + (R^2)/2)*phi''

    And dL/d(phi) = -m*g*A*phi

    So, phi'' + (g*a)/(2A^2 + (1/2)*R^2) * phi = 0

    and omega = sqrt((g*a)/(2A^2 + (1/2)*R^2)) from the characteristic equation of such a lagrangian.
     
  2. jcsd
  3. Dec 28, 2009 #2
    Still no bites?

    Well, if it helps I just want to see how the professor got this.

    A.) L = (1/2)*[m*(R^2 + A^2 - 2AR*cos(phi)) + I]*phi'^2 + m*g*A*cos(phi)

    B.) omega = sqrt( (m*g*A) / ( m*(R-A)^2 + I) )
     
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