Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
A particle of mass $m$ is at a very large distance $p$ from origin $O$ and is moving with velocity $\vec{V}$ which is perpendicular to $\vec{OP}$. I have to calculate angular momentum $L$ of the particle.
I know that $\vec{L}=\vec{r}\times m\vec{\dot r}$.
Since $|\vec{r}|=p$ and $|\dot r|=V$ and $\alpha=90^{\circ}$ is angle between $\vec{r}$ and $\vec{\dot r}$, I got that $L=|\vec{r}||m\vec{\dot r}|sin\alpha=pmV\cdot 1=pmV$.
In the book, it's written that $L=pV$. What happened with mass?
Your book is incorrect. Since $p$ is a length, $pV$ and $L$ cannot be equal by dimensional analysis alone.
The specific angular momentum, however, does equal $pV$, though it is very misleading to use the letter $L$ for it. A body's specific angular momentum is its angular momentum divided by its mass, i.e. its angular momentum per unit mass. It captures the interesting kinematics of angular momentum (i.e. how bodies move in space through time) but it is quite useless when it comes to dynamics (i.e. how bodies interact).
