Thursday, March 22, 2018

rotational dynamics - Instantaneous axis of rotation of a rigid body


For the description of rigid body motion, any point O of the rigid body could be taken as reference, since the velocity of a generic point P can be written in function of the angular velocity Ω and of the velocity of O, independently of the choice of O.


˙P=˙O+Ω(PO)


That means that the motion of P is seen as the composition of the traslation of O, plus a rotational motion about an axis passing through O: let's call this axis γ.


In which cases it is correct to say that γ is an instantaneous axis of rotation of the rigid body?


In order to do that must the point O (on the axis γ) have zero velocity (i.e. ˙O=0)? Or can I define γ as an instantaneous axis of rotation in any case when I write down (1)?



Answer




A body can have parallel velocity on the instantaneous axis of rotation. This parallel velocity is sometimes designated with a scalar pitch value h, such that vO=hω


Consider a point O on the IAR and a point P outside of it. You have


vP=vO+(rPrO)×ω


Now I can prove that given the motion vP of an arbitrary point P and the rotational velocity ω. The motion can always be decomposed into an axis of rotation point rO and a parallel velocity vO.


Lemma Given vP and ω there exists a unique (relative) location r such that vO=vP+ω×r=hω so the velocity vO=hω is parallel to ω only. This location is on the instantaneous axis of rotation closest to P such that rO=rP+r.


Proof


Take r=ω×vP in the transformation and expand out the terms


\vec{v}_O = \vec{v}_P + \vec{\omega}\times \left( \frac{\vec{\omega} \times \vec{v}_P}{\| \vec{\omega}\|^2} \right)


Now use the vector triple product identity a \times (b \times c) = b (a\cdot c) - c (a \cdot b)


\require{cancel} \vec{v}_O =\vec{v}_P +\frac{\vec{\omega}(\vec{\omega}\cdot \vec{v}_P) - \vec{v}_P (\vec{\omega}\cdot \vec{\omega})}{\| \vec{\omega}\|^2} = \cancel{\vec{v}_P} + \frac{\vec{\omega}(\vec{\omega}\cdot \vec{v}_P)}{\| \vec{\omega}\|^2}-\cancel{\vec{v}_P}



Now define the scalar pitch as h = \frac{\vec{\omega} \cdot \vec{v}_P}{\| \vec{\omega} \|^2} and the above becomes \vec{v}_O = h \vec{\omega}


So the velocity on O is parallel to the rotation only.


Example A body rotates by \vec{\omega} = (0,0,10) and a point P located at \vec{r}_P=(0.8,0.2,0) has velocity \vec{v}_P = (2,-3,1). Find the IAR point O and pitch value.h.


\vec{r}_O = \vec{r}_P + \frac{\vec{v}_P \times \vec{\omega}}{\| \vec{\omega} \|^2} = (0.8,0.2,0) + \frac{(-30,20,0)}{10^2} = (0.5,0,0)


h = \frac{\vec{\omega} \cdot \vec{v}_P}{\|\vec{\omega}\|^2} = \frac{10}{10^2} = 0.1


Confirmation


\vec{v}_P = h \vec{\omega} + (\vec{r}_P - \vec{r}_O) \times \vec{\omega} = (0,0,1) + (0.3,0.2,0) \times (0,0,10) = (2,3,-1) \; \checkmark


NOTES: See also related answer to question Why is angular velocity of any point about any other point of a rigid body always the same?


No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...