# Angular momentum problem

rtsswmdktbmhw

## Homework Statement

At time t, particle with mass m has displacement ##\vec r (t)## relative to origin O. Write a formula for its angular momentum about O and discuss whether this depends on choice of origin.

The second part is what I'm more unsure of.

## The Attempt at a Solution

##\vec l=\vec r\times\vec p## so at time t, ##\vec l=\vec r(t)\times m\vec v##

For the second part, I wasn't sure what to do. I tried and got as far as this:
##\vec l_2=\vec r_2\times m\vec v_2##
Then ##\vec l-\vec l_2=(\vec r\times m\vec v)-(\vec r_2\times m\vec v_2)##
What now?

## Answers and Replies

Gold Member
The point you're missing is that when you change the origin, the velocity doesn't change(of course if the translation is independent of time):
$\vec v '=\frac{d}{dt}\vec r'=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v$

tonyxon22
What happens if you try to describe the displacement of the particle r(t) from other point as a combination of motions (using the relative movement).

Something like this:
From origin (point O) you know the displacemente is r(t)
From other origin (point A) you know the displacement is rA(t)
Point A does not moves relative to Oringin O

So r(t) can be described also as OA + rA(t)
What does this say to you?

Gold Member
I don't understand what you're asking!

rtsswmdktbmhw
The point you're missing is that when you change the origin, the velocity doesn't change(of course if the translation is independent of time):
$\vec v '=\frac{d}{dt}\vec r'=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v$
Hm, well the magnitude doesn't change but wouldn't the direction with respect to the second origin be different? But, I gave it a shot (below).

What happens if you try to describe the displacement of the particle r(t) from other point as a combination of motions (using the relative movement).

Something like this:
From origin (point O) you know the displacemente is r(t)
From other origin (point A) you know the displacement is rA(t)
Point A does not moves relative to Oringin O

So r(t) can be described also as OA + rA(t)
What does this say to you?
Okay, so let ##\vec r_2=\vec r+\vec a## and ##\vec v_2=\vec v## then ##\vec l_2=(\vec r+\vec a)\times m\vec v## and so ##\vec l_2-\vec l=(\vec r+\vec a)\times m\vec v- \vec r\times m\vec v=\vec a\times m\vec v## so depends on where the origin is with respect to the first origin?

tonyxon22
Gold Member
Hm, well the magnitude doesn't change but wouldn't the direction with respect to the second origin be different? But, I gave it a shot (below).
No, both the direction and the magnitude of the velocity vector remain unchanged. My proof was vectorial!
Okay, so let r⃗ 2=r⃗ +a⃗ \vec r_2=\vec r+\vec a and v⃗ 2=v⃗ \vec v_2=\vec v then l⃗ 2=(r⃗ +a⃗ )×mv⃗ \vec l_2=(\vec r+\vec a)\times m\vec v and so l⃗ 2−l⃗ =(r⃗ +a⃗ )×mv⃗ −r⃗ ×mv⃗ =a⃗ ×mv⃗ \vec l_2-\vec l=(\vec r+\vec a)\times m\vec v- \vec r\times m\vec v=\vec a\times m\vec v so depends on where the origin is with respect to the first origin?
That's true but because in general, $\vec v$ changes from point to point in both direction and magnitude, you can say what happens so in general Angular momentum does change in a time independent translation of origin.

rtsswmdktbmhw
Ok, thank you very much for the help!