# Angular momentum about an Axis and a Point

1. Apr 8, 2012

### the-ever-kid

I was wondering if the angular momentum of any body with constant $\omega$ (angular velocity) about a point on the centre of the circular path and about any point on the axis of rotation is the same....

Last edited: Apr 8, 2012
2. Apr 8, 2012

### tiny-tim

hi the-ever-kid!

remember the parallel axis theorem

moment of inertia about any axis = moment of inertia about a parallel axis through the centre of mass + md2

where d is the distance between the two parallel axes …

so how does that apply to the angular momentum in the x y and z directions through your two points?

3. Apr 8, 2012

### the-ever-kid

i realise what what you just said but what i really wanted to know is like if the direction of the angular momentum is dependent on the point of reference

like suppose an object is rotating in a circle then i know that the angular momentum is $I\omega = L$ and if that point was in the plane itself but still paralell to the axis of rotation then we apply the paralell axis theorem and add the extra md2
term

yes?

but what if that point was on the axis but not in th plane of rotation but either above or below it .....the will the angular momentum be in the same direction?

4. Apr 8, 2012

### tiny-tim

d will be different for different directions (through the same point)

(and i'm not sure whether you're envisaging a general case, or eg a rotationally symmetric body rotating about it axis of symmetry)

5. Apr 8, 2012

### the-ever-kid

http://puu.sh/oyqf [Broken] see O and P

Last edited by a moderator: May 5, 2017
6. Apr 8, 2012

### BruceW

If it is a point particle moving in a circle at constant speed (which is what the picture looks like), then there is a simple answer to your question. Use the equation for angular momentum, calculate it around either point and see if there is a difference.

EDIT: which equations do you know for the angular momentum?

7. Apr 8, 2012

### the-ever-kid

thank you every one got my answer ......... i just remembered that $$\mathcal{L}= r \times~m\mathcal{v}$$ the $\times$ means cross product
$\therefore$ , it will be $\bot$ to the direction of the distance from reference thus as its direction changes it will not be the same as the O

LIKE THIS: http://puu.sh/oyDM [Broken]

Last edited by a moderator: May 5, 2017
8. Apr 9, 2012

### BruceW

very good. This result surprised me at first, but it is right. Note that the vertical component of angular momentum stays constant, since the magnitudes of horizontal position and momentum are constant and are always perpendicular to each other. Or speaking more generally, the direction of the shift of the origin is vertical, so the vertical component of angular momentum is unchanged.