Canonical definition of Angular Momentum,

[guv]

Let's start with an arbitrary solid body rotating around a fixed axis of rotation with angular velocity $\vec \omega$ in the $\hat z$ direction. For simplicity, let's say the origin O is on the axis of rotation. Take a look at the picture I sketched in the next post. Tried my best to be clear about the notations I used in my derivation.

By definition, $$\vec L = \int_{\Omega} \vec r \times (dm \; \vec v) = \int_{\Omega} \vec r \times (\vec \omega \times \vec r) dm$$

Where $\vec r$ is is position position vector of dm from origin O and $\Omega$ is the domain of integration (the entire solid body).

If the angle between axis of rotation and position vector is $\theta$ (imagine $\hat z$ points upward from O, $\vec r$ points to the upper right from O), then
$$\vec L = \int r^2 \omega sin(\theta) dm \hat L$$

where $\hat L$ is a unit vector from dm perpendicular from $\vec r$ (imagine this unit vector from dm going upper left from dm)

We can decompose this $\hat L$ into its z direction component and a radial component.
$$\vec L = \int r^2 \omega sin^2\theta \hat z dm + \int r^2 \omega sin\theta cos\theta (-\hat p) dm$$

Here is $\hat p$ is a unit vector pointing outward from dm perpendicular to axis of rotation.

If $\omega$ is a constant, then
$$\int r^2 \omega sin^2\theta \hat z dm = \omega \int r^2 sin^2\theta dm \hat z = I_{zz} \omega \hat z$$

This is the z component of the angular moment since the $\vec \omega = (0, 0, \omega)$.

My problem is I couldn't really simply $\int r^2 \omega sin\theta cos\theta (-\hat p) dm$ into 0 when center of mass is on the axis of rotation. Can someone help? Thanks,

guv

 

[guv]

 

[guv]

The Center of Mass $\vec R_{cm}$ is defined by $\int_{\Omega} (\vec r - \vec R_{cm}) \; dm \equiv 0$.

My question is inspired by the derivation in Louis Brand's "Vector Analysis" book, p.176-177 when the author jumps from eq. (2) to eq. (3). I couldn't complete this derivation.

 

[Tahmeed]

Is the body rotating?? In that case doesn't sin$$\theta$$ and cos$$\theta$$ becomes sin pi/2 and cos pi/2? so one of the trigonometric ratio becomes 0. Not sure though, got to think. Better wait for someone else's reply.