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    I know that angular momentum is constant in circular motion, however is there any better justification than what I post below as to why it is constant?

     \dfrac{d\vec{L}}{dt}=\vec{\tau}
     \vec{\tau}=\vec{r} \times \vec{F}=\vec{r} \times m\vec{a}
    In the case of circular motion,  \vec{\tau}=\vec{r} \times -m\omega^2\vec{r}=0

    Thus, the rate of change of angular momentum is 0 and hence angular momentum is constant.
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    (Original post by KeyFingot)
    I know that angular momentum is constant in circular motion, however is there any better justification than what I post below as to why it is constant?

     \dfrac{d\vec{L}}{dt}=\vec{\tau}
     \vec{\tau}=\vec{r} \times \vec{F}=\vec{r} \times m\vec{a}
    In the case of circular motion,  \vec{\tau}=\vec{r} \times -m\omega^2\vec{r}=0

    Thus, the rate of change of angular momentum is 0 and hence angular momentum is constant.
     \dfrac{d\vec{L}}{dt}= \dfrac{d[\vec{p}\times \vec{r}]}{dt}=\dfrac{d[ rpsin(\pi/2)]}{dt}=0
    dunno if it´s better
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    Thanks
 
 
 
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