Is Angular Momentum Dependent on Choice of Origin?

[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.

Homework Equations

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?

 

[ShayanJ]

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):
<br /> \vec v &#039;=\frac{d}{dt}\vec r&#039;=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v<br />

 

[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?

 

[ShayanJ]

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):
<br /> \vec v &#039;=\frac{d}{dt}\vec r&#039;=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v<br />

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?

 

[ShayanJ]

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!