0
$\begingroup$

I know when the body rotatates as well as translates IAR or ICR shouldn't be used but I am not able to understand why?

$\endgroup$

closed as unclear what you're asking by ja72, Jim, Ryan Unger, ACuriousMind, John Rennie Aug 12 '15 at 9:49

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ There is always an instantaneous axis of rotation defined. You heard wrong. $\endgroup$ – ja72 Aug 11 '15 at 18:06
2
$\begingroup$

Any rigid body in motion can be described as rotating about in instantaneous axis of rotation (IAR) and translating along the same axis at the same time.

Example/Proof

A rigid body in moving and at time instant a point A riding on the rigid body has position vector $\vec{r}_A$ and instantaneous linear velocity $\vec{v}_A$ at A. The whole body is rotating with $\vec{\omega}$. We are given 6 motion parameters and one 3 location parameters.

The motion is decomposed as:

  1. The direction of the IAR is given by $$\vec{e} = \frac{\vec{\omega}}{| \vec{\omega} |}$$
  2. The rotational speed (magnitude of rotation) is $$\omega = |\vec{\omega}|$$
  3. The IAR passes through a point $$ \vec{r}_C = \vec{r}_A + \frac{\vec{\omega}\times \vec{v}_A}{|\vec{\omega}|^2}$$
  4. The motion parallel to the IAR is described by the pitch $h$ which is the ratio of the linear speed along the axis to angular speed $$ h = \frac{\vec{\omega} \cdot \vec{v}_A}{|\vec{\omega}|^2}$$
  5. If the above is true then the linear velocity at C is zero and thus $$\vec{v}_A = (\vec{r}_C-\vec{r}_A) \times \vec{\omega} + h \vec{\omega}$$ Lets back substitute and see if we can go full circle: $$ \begin{align} \vec{v}_A & = \frac{\vec{\omega}\times \vec{v}_A}{|\vec{\omega}|^2} \times \vec{\omega} + \frac{\vec{\omega} \cdot \vec{v}_A}{|\vec{\omega}|^2} \vec{\omega} \\ & = \frac{\left(\vec{\omega}\times\vec{v}_{A}\right)\times\vec{\omega}+\left(\vec{\omega}\cdot\vec{v}_{A}\right)\vec{\omega}}{\left|\vec{\omega}\right|^{2}} \\ & = \frac{\vec{v}_{A}\left(\vec{\omega}\cdot\vec{\omega}\right)-\vec{\omega}\left(\vec{v}_{A}\cdot\vec{\omega}\right)+\left(\vec{\omega}\cdot\vec{v}_{A}\right)\vec{\omega}}{\left|\vec{\omega}\right|^{2}} \\ & = \frac{\vec{v}_{A}\left|\vec{\omega}\right|^{2}}{\left|\vec{\omega}\right|^{2}}=\vec{v}_{A} \end{align} $$

Summary

  • Given the motion parameters $\vec{\omega}$, $\vec{v}_A$ and of any point A on a moving rigid body, the instant axis of rotation and parallel velocity ratio are uniquely defined by the unit direction vector $\vec{e}$ the closest point on the axis $\vec{r}_C$, the scalar pitch $h$ and the rotation magnitude $\omega$.

  • Given the axis of rotation position $\vec{r}_C$, direction $\vec{e}$, the pitch $h$ and the magnitude $\omega$ the motion of any point A located at $\vec{r}_A$ is fully defined by $\vec{\omega} = \omega \vec{e}$ and $\vec{v}_A =\omega \left((\vec{r}_C-\vec{r}_A) \times \vec{e} + h \vec{e}\right)$

Appendix

In robotics there is talk about a joint screw axis $\mathbf{s}_A$ such that

$$ \mathbf{v}_A =\mathbf{s}_A \, \omega $$ $$ \begin{bmatrix} \vec{v}_A \\ \vec{\omega} \end{bmatrix} = \begin{vmatrix} (\vec{r}_C-\vec{r}_A) \times \vec{e} + h \vec{e} \\ \vec{e} \end{vmatrix} \,\omega $$ The coordinates of $\mathbf{s}_A$ are the pluecker coordinates of a line (or screw) in space.

NOTE: $\times$ is the vector cross product and $\cdot$ the vector dot product.

$\endgroup$
  • $\begingroup$ This is one of many of the consequences of Euler's rotation theorem. $\endgroup$ – David Hammen Aug 12 '15 at 0:36
  • $\begingroup$ That, plus the parallel translation which is not included in Euler's theorem. Rotation is the $\vec{r}\times \vec{e}$ part and translation the $h \vec{e}$ part of any linear velocity vector. $\endgroup$ – ja72 Aug 12 '15 at 1:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.