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Suppose a motorcyclist is driving at constant speed along a circular track and, suppose further, that a local coordinate system is associated with the motorcycle, where

  • $x$ is tangent to the circle and oriented in driving direction
  • $z$ is orthogonal to the circle in the point of contact and tilted by an angle $\phi$ against the vertical
  • $y$ is the cross procuct of $z$ and $x$, i.e. $y=z\wedge x$

Question:

how can I calculate the angular velocities around the individual axes?

Maybe the question is to simple, but I'm no physicist and perplexed by the peculiarities of mechanics.

1 Answers1

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You construct a 3×3 rotation matrix from the components of the local axes as columns.

$$R = \left( \begin{array}{c|c|c} \hat{x} & \hat{y} & \hat{z} \end{array} \right) $$

I think for your case you have

$$R = \left( \begin{array}{c|c|c} \cos \theta & -\sin\theta \cos\phi & \sin\theta\sin\phi \\ \sin\theta & \cos\theta \cos\phi & -\cos\theta \\ 0 & \sin\phi & \cos\phi \end{array} \right) $$

where $\phi$ is the inclination angle, and $\theta$ is the angle along the arc of track.

Usually you construct this from a sequence of rotations

$$ R = \mathrm{RotZ}(\theta)\,\mathrm{RotX}(\phi) $$

This is then used to transform vectors from local coordinates to world coordinates.

Since you are given that in world coordinates $\boldsymbol{\omega} = \pmatrix{0 \\ 0 \\ \dot{\theta}} $ (ground is along the xy plane), you can transform this to local coordinates with

$$ \boldsymbol{\omega}_{\rm local} = R^\top \boldsymbol{\omega} $$

But in the case where the tilt of the track changes, you can construct the rotational velocity vector, as a sequence of rotations

$$ \boldsymbol{\omega} = \hat{k} \dot{\theta} + \mathrm{RotZ}(\theta) \hat{\imath} \dot{\phi} = \pmatrix{ \dot{\phi} \cos\theta \\ \dot{\phi} \sin\theta \\ \dot{\theta} } $$

See also related answer post.

John Alexiou
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