My teacher gave this as a fact $\vec{v}=\vec{\omega} \times \vec{r}$(cross product) where $\vec{\omega}$ and $\vec{r}$ are angular velocity and position vectors respectively. He also said that this relation held true for any reference point on the body.
But I was confused as I knew the definition of angular velocity only when $\vec{v}$ was perpendicular to $\vec{r}$.
Also I guess that the definition of magnitude of $\vec{\omega}$ is:
$$=\frac{\text{component of velocity perpendicular to r }}{\text{magnitude of r}}$$
But how do we define its direction then?
Also my teacher made another claim: He told that the "angular velocity" observed in a frame on any particle on the rigid body is the same and that itself is defined as the angular velocity of the rigid body.
To understand the above claim and the equation $\vec{v}=\vec{\omega} \times \vec{r}$, I needed the definition of angular velocity.
Note: All I need is the definition of angular velocity about any point on the rigid body with magnitude and direction. I looked up related links to questions on Phys.SE itself but was unable to find one with complete relevancy.
Best Answer
In terms of magnitude is $$\omega=\frac{d\theta}{dt}=\lim_{\Delta t\rightarrow 0}\frac{\theta(t+\Delta t)-\theta(t)}{\Delta t}=\frac{v}{R}$$
In vectorial terms $$\vec{v}=\vec{\omega} \times \vec{r}$$ $$|\vec{v}|=|\vec{\omega} \times \vec{r}|=\omega r \sin{\alpha}=\omega R$$ the result doesn't depend on the origin of $\vec{r}$ as long as is chosen along the rotation axis.
Then, because the relative velocity between any 2 points in the rigid body is null all the points are rotating around the same axis with the same angular velocity.
Deriving the velocity vector using the Leibniz rule $$\frac{\partial \vec{v}}{\partial t}=\frac{\partial }{\partial t}(|\vec{v}|\hat{v})=\frac{\partial |\vec{v}|}{\partial t}\hat{v}+|\vec{v}|\frac{\partial \hat{v}}{\partial t}$$ You get two terms: the first one is the tangential velocity $\frac{\partial |\vec{v}|}{\partial t}$ directed in the velocity (tangential) direction while the second term is the modulus of the velocity times the derivative of the velocity versor. Then, deriving the versor you get $$\frac{\partial \hat{v}}{\partial t}=\frac{\partial \theta}{\partial t}\hat{n}$$, where $\hat{n}$ is the normal vector (with respect to the tangential one). Thus, collecting the results for the second term you get $|\vec{v}|\frac{\partial \hat{v}}{\partial t}=|\vec{v}|\frac{\partial \theta}{\partial t}\hat{n}=v\omega\hat{n}=\omega^2R\hat{n}$.
Finally $$\vec{a}=\vec{a_t}+\vec{a_n}=\frac{\partial |\vec{v}|}{\partial t}\hat{v}+\omega^2R\hat{n}$$