How to derive the formula for angular velocity?

Question

In my mechanics course, we entered rotational mechanics. We were previously introduced to the idea of angular velocity in 2-dimensions as follows, $$v=\omega r$$ Where $v$ stands for angular velocity.

$\omega$ stands for $\frac{d\theta}{dt}$. And $r$ stands for radius.

But suddenly were told that $\omega$ is a vector and it points along the direction of axis of rotation.

Using the diagrams i was able to show that, $$v=\frac{d\Phi}{dt}rsin(\theta)$$

Where $ \omega=\frac{d\Phi}{dt}$. And $\theta$ is the angle between $r$ and $\omega$. But I am unable to proceed further because I want to show that, $$v=\omega × r$$ Where $\omega$ and $r$ are vectors and their vector product is considered.

Answer 1

Any point of a rigid body can be taken as the center of rotation. The reason is that the distance between the points is constant. So, if there is any velocity regarding to the point of reference it must be orthogonal to the radial distance, otherwise the distance would be changing. So: $$\mathbf{v_k.r_k} = 0$$ Because they are orthogonal, it is possible to define a vector $\boldsymbol{\omega_k}$ such that $$\mathbf{v_k} = \boldsymbol{\omega_k\times r_k}$$

This definition is useful because it can be proved that $\boldsymbol{\omega_k}$ is the same vector for all $k$, (that means all points of the rigid body have the same vector $\boldsymbol \omega$) at any given time $t$.

See https://physics.stackexchange.com/a/680029/195949

Answer 2

The motion we are considering is as described in the figure.

The position vector at time $t$ is $\vec r(t)$, while that at time $t+dt$ is $\vec r(t+dt)$. The linear displacement vector is $\vec {dr}(t) \equiv \vec r(t+dt) - \vec r(t)$. The angular displacement is shown in the figure as $d\phi$. The linear velocity is defined simply as the rate of change of the linear displacement:

$$ \vec v \equiv \frac{\vec {dr}(t)}{dt} $$

Now, the angular velocity $\vec \omega$ is defined in two parts:

• its magnitude is given by the rate of change of angular displacement

• its direction is perpendicular to that of $\vec r$ and $\vec{dr}$, as determined by the thumb-rule.

To relate the angular displacement $d\phi$ to the position and displacement vectors, we make another construction.

We have drawn a line perpendicular to $\vec r(t)$ and it makes an angle $\alpha$ with $\vec {dr}$. Note that the angle between $\vec r(t)$ and $\vec{dr}$ is $\frac{\pi}{2} - \alpha$. Let the length of this perpendicular line segment be $h$. The length $h$ is given by $h/|\vec {dr}| = \cos(\alpha) \implies h = |\vec {dr}|\cos\alpha$. The angular increment $d\phi$ should be equal to the ratio of the arc length $h$ and the radius $|\vec r(t)|$. This allows us to write $d\phi = h/|\vec r(t)| = \cos\alpha \frac{|\vec {dr}|}{|\vec r(t)|}$. The magnitude of $\vec \omega$ is therefore

$$ |\vec \omega| = \frac{d\phi}{dt} = \frac{\cos\alpha}{|\vec r(t)|} \frac{|\vec {dr}|}{dt} $$

The direction of $\vec \omega$ is given by the thumb rule, and it is expressed by the cross-product:

$$\hat \omega = \frac{\vec r(t)\times \vec{dr}}{|\vec r(t)\times \vec{dr}|} = \frac{\vec r(t)\times \vec{dr}}{|\vec r(t)||\vec{dr}|\cos\alpha}$$

Note that the denominator has $\cos\alpha$ instead of $\sin\alpha$ because the angle between $\vec r(t)$ and $\vec{dr}$ is $\pi/2-\alpha$.

Combining the direction and the magnitude, we get

$$\vec \omega = \frac{\vec r(t)\times \vec{dr}}{|\vec r(t)|^2 dt}$$

Using the vector triple product formula, we get

$$\vec \omega \times \vec r = \frac{|\vec r(t)|^2 \vec{dr} - (\vec r(t)\cdot \vec{dr})\vec r(t)}{|\vec r(t)|^2 dt} = \vec v - (\hat r(t)\cdot \vec v)\hat r(t)$$

For circular motion, the linear velocity is always perpendicular to the position vector, so the second term vanishes. In that special case, we can conclude that

$$\vec \omega \times \vec r = \vec v$$

Answer 3

Consider a body rotating about the origin, and an arbitrary point A on the body.

At some instant, the location of the point is $\vec{r}_A$, the velocity of the point $\vec{v}_A$ and because it is a rigid body the distance to the origin is constant, $\| \vec{r}_A \| = \ell$.

Use the constant distance condition in the form of $ \vec{r}_A \cdot \vec{r}_A = \ell^2$ and take the time derivative. Here $\cdot$ is the dot product.

$$ \tfrac{\rm d}{{\rm d}t} ( \vec{r}_A \cdot \vec{r}_A ) = \vec{v}_A \cdot \vec{r}_A + \vec{r}_A \cdot \vec{v}_A = 0 $$

or

$$ 2 ( \vec{v}_A \cdot \vec{r}_A ) = 0 $$

So what kind of motion obeys the above constraint. Whatever produces a velocity that is perpendicular to the position vector. Since the velocity vector is a function of the arbitrary position of A only, this means that a vector field would be sufficient to describe the motion of the entire body.

Any vector $\vec{\omega}$ that generates the velocity field with

$$ \vec{v}_A = \vec{\omega} \times \vec{r}_A $$

is consistent with rigid body motion, since $\vec{r}_A \cdot ( \vec{\omega} \times \vec{r}_A) = 0 $ by definition of the cross product.

Answer 4

Start with Rodrigues Rotation Formula

The derivation in the page is quite comprehensible.

$$ r_{rot} = r cos\phi + (n\times r)sin\phi + (r\cdot n)n(1-cos\phi) $$

Where $r_{rot}$ is the vector after the rotation and $r$ is the initial vector, $n$ is the normal vector at the origin of the rotation plane. We can calculated the displacement of the vector by subtracting $r$ on both sides, leading to:

$$ (r_{rot} -r) = r (cos\phi -1) + (n\times r)sin\phi + (r\cdot n)n(1-cos\phi) $$

Now, by using the Small angle approximation we can say that a circumference arc can be approximated by a straight line if $\phi$ is small enough. The approximation is exact if $\phi -> 0$. Therefore, if $\phi$ goes to zero, $(r_{rot} -r)$ represents an infinitesimal circular movement step and $(r_{rot} -r) = dr$ . Taking the differential element on both sides of the equation, the cosine terms will cancel with the ones and the sine term will remain according to the small angle approx:$

$$ dr = (n\times r)d\phi $$

We can bring the angle inside the brackets leading to:

$$ dr = (nd\phi\times r) $$

Now, we want to find the velocity, to do so we divide both sides by $dt$

$$\frac{dr}{dt} = (n\frac{d\phi}{dt}\times r)$$

$\frac{d\phi}{dt}$ is the familiar angular speed, but notice it's multiplied by the normal vector, thus, we define the product $(n\frac{d\phi}{dt})$ as the angular velocity vector, $\omega$

$$\frac{dr}{dt} = (\omega \times r)$$

Answer 5

I want to show that $~\vec v=\vec \omega\times \vec r$

The components of the a vector $~\vec r~$ that given in body fixed coordinate system (index B)

can be transferred to inertial system (index I). with this equation:

$$\vec{r}_I=\mathbf R\,\vec r_B$$

where $~\mathbf R~$ is the transformation matrix between body system and inertial system

take the time derivative ($~\vec r_B=~$ constant) $$ \frac{d}{dt}\left(\vec{r}_I\right)=\mathbf{\dot{R}}\,\vec r_B$$

with $$\mathbf{\dot{R}}= \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \,\mathbf R$$

you obtain $$ \vec v_I=\frac{d}{dt}\left(\vec{r}_I\right)=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \,\mathbf R\,\vec r_B=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \,\vec r_I=\vec \omega_I\times \vec r_I$$

so $$\boxed{\vec v=\vec \omega\times \vec r}$$

---

with

$$\vec \omega=\frac{d\theta}{dt}\vec n\quad ,\parallel\vec n\parallel=1\\ \vec r=r\,\vec e_r\quad \parallel\vec e_r\parallel=1$$

thus $$\vec v=\vec\omega\times \vec r=\vec\omega\times \vec r_\perp= \frac{d\theta}{dt}\vec n\times (\vec e_r\,r \,\sin(\theta))=r\sin(\theta)\frac{d\theta}{dt}\vec e_v$$

where $~\vec e_v=\vec n\times \vec e_r~$

$$\Rightarrow\\v=r\,\sin(\theta)\,\dot\theta$$