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Rigid body motion

  1. Nov 8, 2006 #1
    I am having problems figuring out this problem. A rigid body having an axis of symmetry rotates freely about a fixed point under no torques. If alpha is the angle nbetween the axis of symmetry and hte instantous axis of rotaiton, show that he angle between the axis of rotaiton(omega) and the invariable line(L) is tan^-1((Is-I)*tan(alpha)/(Is+ I*tan(alpha)^2)).

    Using Euler equations:

    omega y'=omega*sin(alpha)
    omega z'=omega*cos(alpha)
    Ly'=I*omega*sin(alpha)
    Lz'=Is*omega*cos(alpha)

    and Ly'/Lz'=tan(theta)=I/Is=tan(alpha)

    Would the cross product of omega and L give me the angle between the axis of rotation and the invariable line? I am not sure how to begin this problem or if this is the right start.
     
  2. jcsd
  3. Nov 9, 2006 #2

    OlderDan

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    What exactly are I and Is? I see the possibility of another angle in the problem. If alpha is the angle between the symmetry axis and omega (it looks like you have used z as the symmetry axis), then there is some angle beta between the symmetry axis and the angular momentum. That would make

    Ly'=L*sin(beta)
    Lz'=L*cos(beta)

    The angle theta would be (alpha - beta). The components of L are related to the components of ω by the moment of inertial tensor. If you know the components of L in terms of ω then you can find the magnitude of L and cross product L x ω as you suggested to find sin(theta) and theta.
     
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