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I have an object moving with linear velocity V. It is also rotating about the axis perpendicular to V, at speed z/t where z is an angle in radians. How can I calculate the radius of the circle its motion will create?

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Instant Centre of Rotation

A rotating body with angular velocity vector ${\boldsymbol \omega} = \pmatrix{\omega_x & \omega_y & \omega_z}$ has a point A attached to it. This point at some instant has linear velocity ${\boldsymbol v} = \pmatrix{v_x & v_y & v_z}$.

The body is rotating about an axis located relative to A with a position vector found by $$ {\boldsymbol r} = \frac{ {\boldsymbol \omega} \times {\boldsymbol v}}{ \| {\boldsymbol \omega} \|^2} $$

Here $\times$ is the vector cross product, and $\| {\boldsymbol \omega} \|^2 = \omega_x^2+\omega_y^2+\omega_z^2$.

Proof

Use the velocity transformation rule to find the linear velocity at the location ${\boldsymbol r}=\frac{ {\boldsymbol \omega} \times {\boldsymbol v}}{ \| {\boldsymbol \omega} \|^2}$ and show that it is zero, or parallel to ${\boldsymbol \omega}$.

$$ {\boldsymbol v} + {\boldsymbol \omega} \times {\boldsymbol r} = {\boldsymbol v} + \frac{{\boldsymbol \omega} \times( {\boldsymbol \omega} \times {\boldsymbol v})}{ \| {\boldsymbol \omega}\|^2} = {\boldsymbol v}+ \frac{{\boldsymbol \omega}({\boldsymbol \omega}\cdot {\boldsymbol v}) -{\boldsymbol v}({\boldsymbol \omega} \cdot {\boldsymbol \omega})}{\| {\boldsymbol \omega}\|^2} = {\boldsymbol v} + 0 - {\boldsymbol v} = 0$$

Use the vector triple product identity $a \times (b\times c) = b(a \cdot c) - c (a \cdot b)$, and note that ${\boldsymbol \omega}\cdot {\boldsymbol \omega} = \| {\boldsymbol \omega}\|^2$.

The above assumes that ${\boldsymbol \omega}\cdot {\boldsymbol v} =0$ which is obvious since ${\boldsymbol v} = -{\boldsymbol \omega} \times {\boldsymbol r}$.

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  • $\begingroup$ FYI - I added related posts. $\endgroup$ Feb 6, 2018 at 18:22

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