# Angular momentum problem

## 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

ShayanJ
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):
$\vec v '=\frac{d}{dt}\vec r'=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v$

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?

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

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
ShayanJ
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.

Ok, thank you very much for the help!