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Thread: cylindrical coordinates

  1. #1
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    cylindrical coordinates

    Angular momentum is defined as L = r x mv , where v is contained in the xy plane. show that in cylindrical coordinates angular momentum takes the form $\displaystyle L = mr^{2}\dot{\phi} $

    i know v = $\displaystyle \dot{r} \hat{r} + r \hat{\phi} \dot{\phi} $
    $\displaystyle L =r \times m (\dot{r} \hat{r} + r \hat{\phi} \dot{\phi} ) $

    am not sure what i get when you take the corss product of

    $\displaystyle r \times m\dot{r} \hat{r} $ ?
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  2. #2
    MHF Contributor
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    Re: cylindrical coordinates

    $\displaystyle \vec{L} = \vec{r} \times m\vec{v}$

    Setting $\displaystyle \vec{v} = \frac{dr}{dt} \hat{r} + r \frac{d\phi}{dt} \hat{\phi}$, and $\displaystyle \vec{r} = r \hat{r}$ gives:

    $\displaystyle \vec{L} = r \hat{r} \times m(\frac{dr}{dt} \hat{r} + r \frac{d\phi}{dt} \hat{\phi})$

    $\displaystyle \vec{L} = r \hat{r} \times m\frac{dr}{dt} \hat{r} + r \hat{r} \times mr \frac{d\phi}{dt} \hat{\phi}$

    and we have $\displaystyle \hat{r} \times \hat{r} = 0$ and $\displaystyle \hat{r} \times \hat{\phi} = \hat{z}$, so

    $\displaystyle \vec{L} = mr^2 \frac{d\phi}{dt} \hat{z}$

    The radial component of velocity doesn't contribute to the angular momentum.

    - Hollywood
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