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Homework Help: Find torque in the x direction given moment of inertia tensor for plate.

  1. Jun 24, 2012 #1
    This is for prelim study. Just wondering if this solution is correct.


    A thin homogeneous plate lies in the x-y plane. Its moment of inertia tensor in the x,y,z basis is given by

    [itex]\textbf{I}=σl^{4}\begin{pmatrix} 2 & -2 &0 \\ -1 & 2 & 0 \\ 0 & 0 & 4\end{pmatrix}[/itex]

    If the plate is rotated about the [itex]\hat{x}[/itex]-axis with constant angular velocity ω, what torque must be applied to the plate to keep the roation axis pointing in the x direction?

    The attempt at a solution

    We are given [itex]\textbf{I}[/itex] and we know that [itex]\textbf{ω}=(ω,0,0)[/itex].

    [itex]\textbf{L}=\textbf{Iω}=σl^{4}\begin{pmatrix} 2 & -2 &0 \\ -1 & 2 & 0 \\ 0 & 0 & 4\end{pmatrix}\begin{pmatrix}ω \\ 0\\ 0\end{pmatrix}=σl^{4}ω(2,-1,0)[/itex]


    [itex]\Gamma_{x}=\frac{dL_{x}}{dt}=2\sigma l^{4}\dot{\omega}[/itex]

    Thanks for the help.
  2. jcsd
  3. Jun 24, 2012 #2

    Filip Larsen

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    Gold Member

    Not quite.

    As you can see, the angular momentum vector L is not parallel to the rotation vector ω, which means that a torque must be applied to the body in order to keep the rotation axis fixed along the x-axis with the angular momentum vector rotating around it.

    If you imagine a diagram of L as a function of time you can see that the tip of the vector must be going in a circle around the x-axis with the angular speed of ω. In order for it to do so, the torque must be orthogonal to both L (to keep the magnitude of L constant) and ω (to keep the rotational speed around the x-axis constant), meaning that the torque must be parallel to ωxL and its magnitude proportional to ω.
  4. Jun 24, 2012 #3


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    Hmm. Shouldn't the moment of inertia tensor be symmetric?

    Anyway, your calculation gives the angular momentum at the instant when the plate is oriented in the xy plane. At that instant, L has an x component and a y component. So, L lies within the plane of the plate. As the plate rotates, the x-component of L remains constant, but the component that is perpendicular to the x axis will rotate with the plate so that L always remains oriented in the plane of the plate. So, from the point of view of our fixed inertial coordinate system, the angular momentum of the plate is changing with time due to the rotation of the component of L that is perpendicular to the rotation axis.

    Can you find the rate of change of L due to the rotation of the component of L that is perpendicular to the rotation axis?

    [I see now that Filip already gave a similar answer. Sorry]
  5. Jun 25, 2012 #4
    Thank you both for the help.
  6. Jun 29, 2012 #5
    After using the advice given above:


    [itex]\boldsymbol{L}=\sigma l^{4} \omega(2,-1,0)[/itex]

    [itex]\boldsymbol{\tau}=\boldsymbol{\omega}\hspace{1 mm}\times \hspace{1 mm}\boldsymbol{L}[/itex]

    [itex]\boldsymbol{\tau}=\sigma l^{4} \omega^{2} \boldsymbol{\hat{z}}[/itex]

    Just wanted to make sure all looks well. Thanks again for all the help.
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