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What does the principal axes help in rotation probelm?

  1. Dec 17, 2008 #1

    KFC

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    In analytical mechanics, we always need to solve the eigenvalue problem of moment of inertia tensor to find out the principal axes and principal moment of inertia. Well, I know that if we know the principal moment of inertia, the total rotating energy of the system could be written as
    [tex]
    \frac{1}{2}I_x\omega_x^2 + \frac{1}{2}I_y\omega_y^2 + \frac{1}{2}I_z\omega_z^2
    [/tex]

    where [tex]I_x, I_y, I_z[/tex] are principal moment of inertia.

    Except for that, I don't quite understand how to use principal axes and principal moment of inertia in analysis physical problem.

    For understanding more about this topic, I read an example in a textbook. But still feel confusing about the relative concepts between axes. Here it is the example: A dumbbell (with two point mass, each is m, attached to two ends of a weightless bar, length 2L) is rotating around z axis at constant angular momentum [tex]\omega[/tex], the dumbbell make an angle [tex]\theta[/tex] to z axis, at t=0, the dumbbell located on the x-z plane (so no y component at t=0). Find the torque of the system.

    Here is how it solve the problem:

    1) Setting up a coordinate system (call body coordinate system) attaching to the rotating object with z-axis along the rotating axis. Note that the coordinate system rotates exactly the same way as the object is rotating so x and y are relatively unchanged during the rotation.

    [tex]
    x = L\sin\theta, y = 0, z = L\cos\theta
    [/tex]

    2) With x, y and z, according to the definition, we can write the tenor of moment of inertia
    [tex]
    I = \left(
    \begin{matrix}
    2mL^2\cos^2\theta & 0 & -mL^2\sin2\theta \\
    0 & 2mL^2 & 0 \\
    -mA^2\sin 2\theta & 0 & 2mL^2\sin^2\theta
    \end{matrix}
    \right)
    [/tex]

    3) We can find the principal axes and principal moment of inertia by solving the corresponding eigenvalue problem.

    4) From the problem, the body is rotating about z-axis with angular velocity [tex]\vec{\omega}[/tex]. Now, project this angular velocity to the principal axes to get three components [tex]\omega'_1, \omega'_2, \omega'_3[/tex]

    5) Since we already found principal moment of inertia and components of angular velocity along principal axes. We could plug all of these into Euler equation to find the torque.

    The procedure seems quite forward but I don't understand why we could do that. My question is:

    1) when we obtain the principal axes, whether this axes change with respect to time? Does principal axes defined with respect to a space frame or a frame defined and attached to an rotating object?

    2) if we use the following equation to find the torque

    [tex]\vec{L} = I\vec{\omega}, \vec{N} = \frac{d\vec{L}}{dt}[/tex]
    where [tex]\vec{L}[/tex] is the angular momentum, [tex]\vec{N}[/tex] is the torque, both of them are with respect to a origin of a fixed coordinate (or space coordinate). In this way,

    [tex]\vec{N} = mL^2\omega^2\sin2\theta (\sin\omega t \vec{\hat{x'}} - \cos\omega t\vec{\hat{y'}})[/tex]

    and [tex]\vec{\hat{x'}}, \vec{\hat{y'}}[/tex] is the unit vector of fixed (space) coordinate system.

    However, if we apply Euler equation to find the torque, since everything of the equation is defined in principal coordinate system, I wonder what's physical significance of the torque we obtained in this way? Obviously, from Euler equation, we will obtain an time-independent solution of torque, how can I construct a time-dependent torque with that solution?
     
    Last edited: Dec 17, 2008
  2. jcsd
  3. Dec 18, 2008 #2

    tiny-tim

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    Homework Helper

    Hi KFC! :smile:

    The principal axes are fixed in the body

    in this case, the principal axes are the rod itself, with I = 0, and any perpendicular axis, with I = 2ML2

    so the angular momentum is ωsinθ.0 along the rod, and ωcosθ.2ML2 perpendicular to it, and of course rotating with the rod.
    Euler's equations give the same net torque as the fixed-frame equation.

    The only reason why we prefer Euler's equations to the fixed-frame equations is that the moment of inertia tensor is changing in the latter, but only the angular momentum vector is changing in the former …

    and a non-zero derivative of a tensor is a lot nastier to work with than a non-zero derivative of a vector! :smile:
     
  4. Dec 18, 2008 #3

    KFC

    User Avatar

    Thanks a lot tiny-tim :)

    I reorganize all the material and what you said and did the problem again with two methods to find the torque: general method (dL/dt) and Euler equation

    Firstly, in fixed coordinate, the body frame is rotating, it is quite easy to solve the angular momentum

    [tex]
    \vec{L} = \left(
    \begin{matrix}
    -Ml^2\omega\sin 2\theta\cos\omega t \\ -Ml^2\omega\sin 2\theta\sin\omgea t \\
    Ml^2\omega\sin^2\theta
    \end{matrix}
    \right)
    [/tex]

    where [tex]l[/tex] is the half length of the rod. Hence, the net torque with respect to the fixed frame will be
    [tex]
    \tau = \frac{d\vec{L}}{dt} =
    \left(
    \begin{matrix}
    Ml^2\omega^2\sin 2\theta\sin\omega t \\
    -Ml^2\omega^2\sin 2\theta\cos\omega t \\
    0
    \end{matrxi}
    \right)
    [/tex]

    This result is quite make senses. The torque is changing with respect with time.

    Now I apply the Euler to solve the problem. First of all, I solve the eigenvalue problem for the inertia tensor, just same as what you get in last reply, I have three principal inertia as

    [tex]I_1 = I_3 = 2ml^2, I_2=0[/tex]

    and the eigenvectors just give the direction of the principal axes

    [tex]
    \vec{v}_1 = \left(
    \begin{matrix}
    -\cos\theta \\ 0 \\ \sin\theta
    \end{matrix}
    \right),

    \vec{v}_2 = \left(
    \begin{matrix}
    0 \\ 1 \\ 0
    \end{matrix}
    \right)
    ,
    \vec{v}_3 = \left(
    \begin{matrix}
    \sin\theta \\ 0 \\ \cos\theta
    \end{matrix}
    \right)
    [/tex]

    This tells us the z direction of the principal axes is along the rod itself, we could pick y direction of principal axes along with the old y axis of the body frame and x direction of the principal axes just perpendicular to others two principal axes. With this setup, we project the old angular velocity to the principal axes, we have

    [tex]
    \omega_1 = -\omega\sin\theta, \qquad \omega_2 =0, \qquad \omega_3 = \omega\cos\theta
    [/tex]

    One of my friend solve the problem and get [tex]\omgea_1=\omega\sin\theta[/tex] (no minus), but I think it should have negative ...

    Ok, now we can plug everything we get to the Euler equation

    [tex]
    I_1\dot{\omega}_1 - \omega_2\omega_3(I_2 - I_3) = \tau_1, \qquad
    I_2\dot{\omega}_2 - \omega_3\omega_1(I_3 - I_1) = \tau_2, \qquad
    I_3\dot{\omega}_3 - \omega_1\omega_2(I_1 - I_2) = \tau_3.
    [/tex]

    Since [tex]\omega[/tex]'s are constant, all [tex]\dot{\omega}_k[/tex] on the left hand side vanish. and [tex]I_1=I_3[/tex] so the second equation gives [tex]\tau_2=0[/tex]. In addition, [tex]\omega_2=0[/tex], hence, first and third equations also gives [tex]\tau_1=\tau_3=0[/tex], this is totally not consistency with what we get above!!? Please help and tell me what's going one here, two method don't give the same net torque?

     
    Last edited: Dec 18, 2008
  5. Dec 18, 2008 #4

    tiny-tim

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    Hi KFC! :smile:

    I think you've got your directions mixed up …

    if I2 = 0, then ω is a mixture of ω1 and ω2, with ω3 = 0. :wink:
     
  6. Dec 18, 2008 #5

    KFC

    User Avatar

    Yes, you are absolutely right. I am to careless. Now, I rephrase the problem. The principal inertia would be

    Along principal x axis:
    [tex]
    I_1 = 2ml^2, \qquad \omega_1 = -\omega\sin\theta
    [/tex]

    Along principal y axis:
    [tex]
    I_2 = 2ml^2, \qquad \omega_2 = 0
    [/tex]

    Along principal z axis:
    [tex]
    I_3 = 0, \qquad \omega_3 = \omega\cos\theta
    [/tex]

    So applying the Euler equation

    [tex]
    \tau_1 = I_1\dot{\omega}_1 - \omega_2\omega_3(I_2-I_3) = 0
    [/tex]

    [tex]
    \tau_2 = I_2\dot{\omega}_2 - \omega_3\omega_1(I_3-I_1) = -ml^2\omega^2\sin 2\theta
    [/tex]

    [tex]
    \tau_3 = I_3\dot{\omega}_3 - \omega_1\omega_2(I_1-I_2) = 0
    [/tex]

    This time I've got something nonzero. But I am still so confusing this torque is not the torque we get with previous method (directly calculate with dL/dt). But you said "Euler's equations give the same net torque as the fixed-frame equation.", where do I doing wrong?
     
  7. Dec 19, 2008 #6

    tiny-tim

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    Homework Helper

    Hi KFC! :smile:
    But they're the same! :wink:

    Get some sleep! :zzz:​
     
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