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If you have a disk which is rotating on a surface without slipping then we can view the motion in two different ways. Firstly as a rotation about the centre of mass of the disk along with a translational velocity of the centre of mass. Alternatively, the motion can be viewed as just a rotation about the instantaneous centre, in this example it is the point of contact between the surface and the disc.

Is it possible to select an arbitary axis (perpendicular to the plane of the disc) such that we can view the motion as a rotation about that axis plus a translational component? If so, is the angular velocity always the same regardless of the chosen axis?

Oliver Smith
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1 Answers1

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Yes and Yes.

The rotational velocity ${\boldsymbol \omega}$ is shared among all parts of a rotating body. Linear velocity is obviously location dependent, so it usual we designate it with a letter or a symbol indicating the location where it is measured. For example, considering an arbitrary point A we state the transformation from the center C as

$$\mathbf{v}_A = \mathbf{v}_C + {\boldsymbol \omega} \times (\mathbf{r}_A - \mathbf{r}_C)$$

  • Lemma 1 - Knowing the rotational velocity and the linear velocity at any point, allows us the find the velocity at any other point with the equation above, and thus we know the entire state of motion.

  • Lemma 2 - Given the rotational motion and the location of the instant centre, allows us to find the velocity at any other point with the equation above, and thus we know the entire state of motion.

The is identical to Lemma 1 except the liner velocity is zero by definition at the instant centre.

  • Lemma 3 - Knowing the rotational velocity and the linear velocity at any point A, allows us to find the location B of the instant centre** with the equation $$\mathbf{r}_{B} = \mathbf{r}_A + \frac{{\boldsymbol \omega} \times \mathbf{v}_A}{\|{\boldsymbol \omega}\|^2} $$

  • Proof - Consider the velocity of the arbitrary point A described as a rotation about the instant centre B with $\mathbf{v}_A = {\boldsymbol \omega} \times \mathbf{r}$ where $\mathbf{r}=\mathbf{r}_A-\mathbf{r}_B$ is the relative location of the point to the centre. Now consider the following $$ {\boldsymbol \omega} \times \mathbf{v}_A = {\boldsymbol \omega} \times ({\boldsymbol \omega} \times \mathbf{r}) = {\boldsymbol \omega} \left( {\boldsymbol \omega} \cdot \mathbf{r}\right) - \mathbf{r} \left( {\boldsymbol \omega} \cdot {\boldsymbol \omega}\right)$$

There is such an $\mathbf{r}$ to solve the above equation which has ${\boldsymbol \omega} \cdot \mathbf{r}=0$, or in layman's terms the location of the centre of rotation is perpendicular to the rotation axis. This value is $$\mathbf{r} = \mathbf{r}_A-\mathbf{r}_B=-\frac{{\boldsymbol \omega} \times \mathbf{v}_A}{{\boldsymbol \omega} \cdot {\boldsymbol \omega}}$$ Note that ${\boldsymbol \omega}\cdot {\boldsymbol \omega} = \| {\boldsymbol \omega} \|^2$. The triple vector product identity $\mathbf{a} \times ( \mathbf{b} \times \mathbf{c}) = \mathbf{b} ( \mathbf{a}\cdot \mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$ is used above also.

** NOTE: In 3D, there isn't a unique instant centre point, but rather an axis of rotation. It is allowed to have linear velocity parallel to the axis of rotation and the location still is considered the instant centre.

I encourage the reader to read more about the motion properties of a rigid body from this answer and more importantly from my full explanation of the terms twist and wrench which describe the geometry of rigid body mechanics.

John Alexiou
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