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Let a particle be moving in space with constant velocity. We are required to show that for that particle , angular momentum is constant throughout the motion irrespective of origin we choose.

MY PROOF:-

Let's choose an origin $O$ and $3$-dimensional Cartesian coordinate system.

Let the position vector of required particle from origin be $\vec{r}$ and constant velocity vector at that point be $\vec{v}$. And let it's mass be $m$ then angular momentum is $$\vec{L}=\vec{r}×m\vec{v}$$

Differentiating on both sides, $$\frac{d\vec{L}}{dt}=\frac{d\vec{r}}{dt}×m\vec{v}+m\vec{r}\frac{d\vec{v}}{dt}$$

Since the velocity is constant $\frac{d\vec{v}}{dt}$ term goes to $0$. The first term obviously goes to $0$ as $\vec{v}=\frac{d\vec{r}}{dt}$. So, $$\frac{d\vec{L}}{dt}=0$$ So $\vec{L}$ is constant. Since the origin we chose was arbitrary, it works at any position.

Hence proved.

I was just wondering if my solution is true.

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1 Answer 1

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Yes.

You can also show that $\vec{r} = \vec{r}_0 + \vec{v}\, t$ and that

$$ \require{cancel} \vec{L} = \vec{r} \times m \vec{v} = \vec{r}_0 \times m \vec{v} + ( \cancel{ \vec{v} \times m \vec{v} }) t = \text{(const)}$$

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