How to get rotation components in each axis?

Question

To explain what I mean let's say I have an object that is moving and I know its speed and direction (angle). I can get the X and Y components of speed using speed*cos(angle) and speed*sin(angle).

Is this possible with rotation?

I have a sphere that is rolling. Each 2*PI*R that it moves forwards, it rotates 2*PI, obviously. The general formula is angle = distance/R.

If it's rolling along the X axis (for example, from point (0,0) to (x,0)) I know it rotates angle along the Y axis.

But what if it's rolling along the (1,1) vector? That's a 45º degree angle but doing rot_x = cos(45) * angle and rot_y = sin(45) * angle doesn't work, does it? Is it possible to get each rotations component to know how the sphere should be rotating?

Answer 1

The transformation of the $\{x,y,z\}$ components of the vector is

$$\{\hat x, \hat y,\hat z\} = \{x,y \cos (\alpha)-z \sin (\alpha),y \sin (\alpha)+z \cos (\alpha)\}$$

for the rotation around the $x$-axis by an angle $\alpha$,

$$\{\bar x, \bar y,\bar z\} = \{x \cos (\alpha )+z \sin (\beta),y,z \cos (\beta)-x \sin (\beta)\}$$

for the rotation around the $y$-axis by an angle $\beta$ and

$$\{ \acute{x}, \acute{y},\acute{z}\} = \{x \cos (\delta)-y \sin (\delta),x \sin (\delta)+y \cos (\delta),z\}$$

for the rotation around the z-axis by an angle $\delta$.

Answer 2

Rotational quantities, θ, ω, and α, are usually represented by vectors along the axis of rotation. For a rolling object, they are perpendicular to the translational velocity and parallel to the surface on which rolling occurs. The arc length, s = r θ, does not change if the surface is tilted.

Answer 3

If the location of where you want to measure speed is $\pmatrix{x & y}$ from the instant center of rotation, then

$$ \pmatrix{v_x \\ v_y} = \pmatrix{ -y\,\omega \\ x \, \omega} $$ where $\omega$ is the angular speed.

In the case of a rolling ball, the center of rotation is at the contact point if there is no slipping, and thus the center of the ball is at $\pmatrix{0 & R}$ relative to the contact point. So the velocity vector of the ball (center) is

$$ \pmatrix{v_x \\ v_y} = \pmatrix{-R\, \omega \\ 0} $$

The above are just the 2D projection of the transformation laws for velocity

$$ \vec{v} = \vec{\omega} \times \vec{r} $$ where $\times$ is the vector cross product.