# Elementary Construction of the Angular Velocity

Physics books seldom contain an accurate definitions of the angular velocity of a rigid body. I believe that the following construction is as simple as possible and also rigorous.

Table of Contents

## Assume we have a rigid body which is moving in the space

Theorem. In each moment there exists a unique (pseudo)vector ##\boldsymbol\omega ## such that for any two points ##A,B## of the rigid body the following formula holds

$$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega\times\boldsymbol {BA},\qquad (***)$$

here ##\boldsymbol v_A,\boldsymbol v_B## are the velocities of the points ##A,B## respectively.

Definition. The vector ##\boldsymbol \omega## is called the angular velocity.

Before proving this theorem we formulate a lemma. Let ##\boldsymbol e_1,\boldsymbol e_2,\boldsymbol e_3## be a right orthonormal basis frozen into the rigid body. Introduce a vector

$$\boldsymbol \omega_*=(\dot{\boldsymbol e}_2,\boldsymbol e_3)\boldsymbol e_1+(\dot{\boldsymbol e}_3,\boldsymbol e_1)\boldsymbol e_2+(\dot{\boldsymbol e}_1,\boldsymbol e_2)\boldsymbol e_3.\qquad (*)$$

Lemma. The following formulas hold $$\dot{\boldsymbol e}_k=\boldsymbol \omega_*\times {\boldsymbol e}_k,\quad k=1,2,3.\qquad (**)$$

Indeed, differentiating the equlities ##({\boldsymbol e}_k,{\boldsymbol e}_j)=\delta_{kj}## we get

##(\dot{\boldsymbol e}_k,\boldsymbol e_j)=-(\dot{\boldsymbol e}_j,\boldsymbol e_k)## and particularly ##(\dot{\boldsymbol e}_k,\boldsymbol e_k)=0##. Due to these formulas one can substitute (*) into (**) to make sure that the lemma is correct.

## Let us prove the theorem

A) Existence. Let ##O## be the origin of some fixed frame. Differentiating the identity ##\boldsymbol{OA}=\boldsymbol{OB}+\boldsymbol{BA}## we have

$$\boldsymbol v_A=\boldsymbol v_B+\frac{d}{dt}\boldsymbol{BA}.$$

On the other hand we can write ##\boldsymbol{BA}=\sum_{i=1}^3 x_i\boldsymbol e_i.## From lemma it follows that

$$\frac{d}{dt}\boldsymbol{BA}=\sum_{i=1}^3 x_i\dot{\boldsymbol e}_i=\sum_{i=1}^3 x_i\boldsymbol\omega_*\times{\boldsymbol e}_i=\boldsymbol\omega_*\times \boldsymbol{BA}.$$ Consequently there exists at least one vector that satisfies the theorem. This vector is ##\boldsymbol \omega_*.## In the sequel we drop the subscript *.

B) Uniqueness. Assume that there is another vector ##\boldsymbol \omega’## such that $$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega’\times\boldsymbol {BA}.$$

Subtracting this formula from the formula (***) we obtain

##(\boldsymbol \omega-\boldsymbol \omega’)\times\boldsymbol {BA}=0.## Since ##\boldsymbol {BA}## is an arbitrary vector we have ##\boldsymbol \omega=\boldsymbol \omega’##.

## The theorem is proved

Example. Let ##OXYZ## be a fixed Cartesian frame. And the rigid body rotates around the axis ##OZ## such that ##\boldsymbol e_Z=\boldsymbol e_3## and

$$\boldsymbol e_1=\cos\phi\boldsymbol e_X+\sin\phi\boldsymbol e_Y,\quad \boldsymbol e_2=\cos\phi\boldsymbol e_Y-\sin\phi\boldsymbol e_X$$ here ##\phi=\phi(t)## is the angle of rotation of the rigid body.

Then differentiate these formulas ##\dot{\boldsymbol e}_1=\dot\phi\cos\phi\boldsymbol e_Y-\dot\phi\sin\phi\boldsymbol e_X=\dot\phi \boldsymbol e_2.## In the same way we get ##\dot{\boldsymbol e}_2=-\dot\phi \boldsymbol e_1.## Substitute these equalities in (*) to get ## \boldsymbol \omega=\dot\phi \boldsymbol e_Z.##

PhD – Interested in differential equations and classical mechanics

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