# Angular momentum of a rod about hinge

Gold Member

## Homework Statement

A uniform rod (M, L) is rotated about a point L/3 from its left end. Angular momentum about O

## Homework Equations

1) L=I(cm)w for purely rotating body
2) L(orbital)= M*v(cm)*perpendicular distance(r)
3) L(spin)= I*w

## The Attempt at a Solution

I got the correct answer in two ways- in the first one I wrote L= I about O*w. But isn't equation 1 applicable only when axis of rotation passes through COM? or is it applicable here because the body is not translating and thus there are no non-inertial forces on any point?
In the second method I wrote L=L(orbital)+L(spin), L(orbital) is m*v(cm)*distance between O and COM while L(spin) has to be taken as I about com*w...Why is L(spin) taken about an axis through COM even though the body isn't even rotating about that point? please reveal my conceptual flaws

• Delta2

Nathanael
Homework Helper
But isn't equation 1 applicable only when axis of rotation passes through COM?
Nope, equation 1 is general. If the body is purely rotating about the same point that the moment of inertia is taken about, then it will work.

or is it applicable here because the body is not translating and thus there are no non-inertial forces on any point?
Translation has nothing to do with forces if it’s uniform. Anyway no translation is involved in pure rotations, other than circular motions.

Why is L(spin) taken about an axis through COM even though the body isn't even rotating about that point?
You can derive the result “L = L_(orbit) + L_(spin)” from integrating (or summing for discrete distributions) the definition of the angular momentum L of a point particle.

It’s not fun to type latex on a phone, but maybe you can work it out yourself. Just consider the arbitrary position/velocity to be a composition of the CoM position/velocity plus the position/velocity in the CoM frame. Maybe I will type it up tomorrow if you still don’t know.

• Delta2
Gold Member
Nope, equation 1 is general. If the body is purely rotating about the same point that the moment of inertia is taken about, then it will work.

Translation has nothing to do with forces if it’s uniform. Anyway no translation is involved in pure rotations, other than circular motions.

You can derive the result “L = L_(orbit) + L_(spin)” from integrating (or summing for discrete distributions) the definition of the angular momentum L of a point particle.

It’s not fun to type latex on a phone, but maybe you can work it out yourself. Just consider the arbitrary position/velocity to be a composition of the CoM position/velocity plus the position/velocity in the CoM frame. Maybe I will type it up tomorrow if you still don’t know.
I'll try to derive it myself...thanks. I'll take a look at things again and see if I have any more confusions and then let you know

Gold Member
I'll try to derive it myself...thanks. I'll take a look at things again and see if I have any more confusions and then let you know
seems like all my doubts are resolved...thank you very much!