I do not understand this because angular momentum is $L=I\omega$ ($I$ is moment of inertia;$\omega$ is angular velocity) but it I have also seen equations where $L= rmv\sin(x)$. I do not understand how these are related, could someone please explain the connection?
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The angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$, so the value of angular momentum is $L = r \cdot mv \sin(x)$, where $x$ is the angle between direction of $\vec{r}$ and momentum $\vec{p}$. |
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$\vec{L}=I\vec{\omega}$ if for a rigid body, whereas $\vec{L}=\vec{r}\times\vec{p}\implies|\vec{L}|=|\vec{r}||\vec{p}|\sin\theta$ is for a point particle. Note that, for a point particle, $I=mr^2,\vec{\omega}=\frac{\vec{v}\times\hat{r}}{\vec{r}}\implies |\vec{\omega}|=\frac{|\vec{v}|}{|\vec{r}|}\sin\theta$. Substituting in $\vec{L}=I\vec{\omega}$, we get $\vec{L}=mr^2\frac{|\vec{v}|}{|\vec{r}|}\sin\theta=mvrsin\theta=\vec{t}\times\vec{p}$ |
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In general, the angular momentum for a rigid body is $\vec{L}=I\vec{\omega}$. For the special case of a point particle $\vec{r}$ from the axis of rotation, we have $I=mr^2$ and $\vec{\omega}=\frac{\hat{r}\times\vec{v}}{r}$, or $\omega=\frac{v}{r}\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{r}$. In this case, the angular momentum becomes $L=mrv\sin\theta$. |
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