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).

  • $\begingroup$ so it's correct what I wrote? It's written so through the whole book without any explanation about mass. The same for potential and kinetic energy. $\endgroup$
    – gov
    Jul 21 '13 at 15:20
  • $\begingroup$ You are correct, maybe your book calls it specific angular momentum, since that is equal to the equation from your book. $\endgroup$
    – fibonatic
    Jul 21 '13 at 16:46
  • $\begingroup$ @fibonatic yes, exactly. It's written specific, I didn't know that this term exists (I am not physics student). Thanks. $\endgroup$
    – gov
    Jul 21 '13 at 18:39

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