# How to get rotation components in each axis?

Physics Asked on January 6, 2021

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?

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 $$delta$$.

Answered by Gendergaga on January 6, 2021

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.

Answered by R.W. Bird on January 6, 2021

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.

Answered by John Alexiou on January 6, 2021