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

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