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Homework Help: Rotation of a cone rolling on its side without slipping on a plane

  1. Apr 23, 2014 #1
    1. The problem statement, all variables and given/known data

    A uniform right circular cone of height [itex]h[/itex], half angle [itex]α[/itex], and density [itex]ρ[/itex] rolls on its side without
    slipping on a uniform horizontal plane in such a manner that it returns to its original position in
    a time [itex]\tau[/itex]. Find expressions for the kinetic energy and the components of the angular momentum
    of the cone. Hint: If [itex] \vec{v} = \vec{ω}\times\vec{r}[/itex] in the inertial frame, points on the cone instantaneously at rest
    in this frame will lie in the direction of [itex]\vec{ω}[/itex].

    2. Relevant equations


    [itex] T_{rot}=\frac{1}{2}I_{ij}\omega_{i}\omega_{j} = \frac{1}{2}L_{j}\omega_{j}[/itex]

    3. The attempt at a solution

    Setup: I'll start off by defining an inertial coordinate system [itex]\hat{x}'[/itex] Also successive rotations of this coordinate system will be given by [itex] \hat{x}'',\hat{x}''',...[/itex] etc. Eventually I want to build to a body frame [itex] \hat{x} [/itex].

    Rotation 1: about the [itex] \hat{x_3}' [/itex] axis by an angle [itex] \theta [/itex] given by the rotation matrix [itex]\mathbf{A}[/itex]

    [itex]\begin{pmatrix}x''_1\\x''_2\\x''_3\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}\begin{pmatrix}x'_1\\x'_2\\x'_3\end{pmatrix}[/itex]

    Rotation 2: about the [itex] \hat{x_2}'' [/itex] axis by an angle [itex] \alpha [/itex] given by the rotation matrix [itex]\mathbf{B}[/itex]

    [itex]\begin{pmatrix}x'''_1\\x'''_2\\x'''_3\end{pmatrix} = \begin{pmatrix}\cos\alpha&0&-\sin\alpha\\0&1&0\\\sin\alpha&0&\cos\alpha\end{pmatrix}\begin{pmatrix}x''_1\\x''_2\\x''_3\end{pmatrix}[/itex]

    Rotation 3: about the [itex] \hat{x_1}'''[/itex] axis by an angle [itex] ψ[/itex] given by the rotation matrix [itex]\mathbf{C}[/itex]

    [itex] \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1&0&0\\0&\cos ψ&\sin ψ\\0&-\sin ψ&\cos ψ\end{pmatrix}\begin{pmatrix}x'''_1\\x'''_2\\x'''_3\end{pmatrix}[/itex]

    So now I have [itex] \vec{x}=\mathbf{CBA}\vec{x}'[/itex] also [itex][\mathbf{CBA}]^{T}\vec{x}=\vec{x}' [/itex] which is a relationship between the inertial and body frames.

    I can also get [itex] \vec{\omega} = \dot{\theta}\hat{x_3}'+\dot{ψ}\hat{x_1}''' [/itex]

    Putting [itex] \vec{\omega} [/itex] into the body frame: [itex] \vec{\omega} = \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}[/itex]

    Okay now before I go any further I want to make sure what I have is correct. Also please if you find a mistake please explain in detail why it is wrong. I really want to get an understanding of rigid body rotations. I'm also attaching my drawing of the various transformations too, thanks in advance for your help.

    Attached Files:

  2. jcsd
  3. Apr 25, 2014 #2
    The Inertia tensor for a rotatons about the apex of the cone is:

    [itex]\mathbf{J} = \begin{pmatrix}\frac{3MR^2}{10}&0&0\\0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}&0\\0&0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}\end{pmatrix}[/itex]

    The the expression for energy is: [itex] T_{rot} = \frac{1}{2}\begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}&\dot{\theta}\cos\alpha\sin ψ &\dot{\theta}\cos\alpha\cos ψ \end{pmatrix} \begin{pmatrix}\frac{3MR^2}{10}&0&0\\0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}&0\\0&0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}\end{pmatrix} \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}[/itex]

    [itex] = \frac{1}{2}[\frac{3MR^2}{10}(-\dot{\theta}\sin\alpha+\dot{\psi})^2+(\frac{3Mh^2}{5}+\frac{3MR^2}{20})(\dot{\theta})^2\cos^2\alpha] [/itex]

    [itex] = \frac{1}{2}[\frac{3MR^2}{10}[(\dot{\theta})^2\sin^2\alpha-2\dot{\theta}\dot{\psi}\sin\alpha+\dot{\psi}^2]+(\frac{3Mh^2}{5}+\frac{3MR^2}{20})(\dot{\theta})^2\cos^2\alpha] [/itex]

    Also [itex] \vec{L} = \begin{pmatrix}\frac{3MR^2}{10}&0&0\\0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}&0\\0&0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}\end{pmatrix} \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}[/itex]

    [itex] = \begin{pmatrix}\frac{3MR^2}{10}(-\dot{\theta}\sin\alpha+\dot{\psi})\\\dot{\theta}(\frac{3Mh^2}{5}+\frac{3MR^2}{20})\cos\alpha \sin\psi \\\dot{\theta}(\frac{3Mh^2}{5}+\frac{3MR^2}{20})\cos\alpha \cos\psi \end{pmatrix}[/itex]
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