# Principal of conservation of angular momentum in uniformly moving body

Consider a point mass object moving along a straight line with constant velocity. As the velocity is constant so is its linear momentum. If we measure the angular momentum of the body with respect to an origin O, then we realize that the angular momentum of the body will change as the body moves because angular momentum $\textbf{L}=\textbf{r}\times\textbf{P}$ and $\textbf{r}$ varies from one point to other. But according to equation $$\frac{d\textbf{L}}{dt}=\textbf{r}\times \frac{d\textbf{P}}{dt}=\textbf{r}\times\textbf{F}$$ This would be equal to zero since $\textbf{P}$ is a constant or because there is no force acting on a uniformly moving body. How is this possible if $\textbf{L}$ is different at different points with respect to the origin

• Why is there X and x? – QuIcKmAtHs Jan 1 '18 at 6:45
• @XcoderX X or x indicates cross multiplication between two vectors – Siddharth Prakash Jan 1 '18 at 6:47
• Hmm then since r and p are constant throughout what... – QuIcKmAtHs Jan 1 '18 at 6:51
• P is a constant with respect to any point in a fixed frame of reference since the object is in uniform motion. But r which is the position vector of the object with respect to origin O changes as the object is moving. So the angular momentum of the body about O should also change with time – Siddharth Prakash Jan 1 '18 at 6:59
• You might want to properly typeset your question as per physics.meta.stackexchange.com/questions/804/… – ZeroTheHero Jan 1 '18 at 7:11

This is the same Rod's answer but let me take a more graphical approach: The angular momentum when the particle is at position (1) is just:

$$L = rmv$$

because at this point the direction of motion is at right angles to the line joining the object to the reference point. So far so good.

Now consider the point (2). The angular momentum at this point is:

$$\mathbf L = \mathbf d \times m\mathbf v$$

Considering just the magnitude and using the definition of the cross product we get:

$$L = d\,mv\,\sin\theta = \frac{r}{\sin\theta} mv\,\sin\theta = rmv$$

So we find the magnitude of the angular momentum is constant. And the direction is constant because it's normal to the diagram in both cases. So even though it looks as if the angular momentum should be changing in fact it is constant.

The formula $\vec{L}=\vec{r}\times\vec{p}$ superficially looks as though angular momentum of a coasting body does change with time. But the change of $\vec{r}$ is parallel to the momentum, so $\vec{L}$ is in fact constant. In symbols, applying the Leibniz rule:

$$\mathrm{d}_t \vec{L} = (\mathrm{d}_t \vec{r}) \times \vec{p} + \vec{r}\times \mathrm{d}_t\,\vec{p}$$

but $\vec{p}$ is constant, by assumption, so the second term vanishes. Also $\mathrm{d}_t \vec{r}=\vec{v}$, the velocity, which is parallel to the momentum, so the first term vanishes as well.