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Suppose $r$ and $r'$ are the distances of a point P from points A and B in a rotating rigid body ( all points A, B, P are inside the rigid body). Now I can choose the body coordinate system whose origin lies at A or B ( or any point inside the body).

Let the distance between A and B be $a$, then,

$r=r' +a$. Now what will be the derivative of $r$ will it be $\dot{r} = \dot{r'} $ that is will the time derivative of $a$ be 0 or not.

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you have three points on a rigid body.

$$\vec{R}_P\,,\vec{R}_A\,,\vec{R}_B$$

thus

$$\vec{r}=\vec{R}_A-\vec{R}_P$$ $$\vec{r}'=\vec{R}_B-\vec{R}_P$$ and $$\vec{a}=\vec{r}-\vec{r}'=\vec{R}_A-\vec{R}_B$$

If only point P is changing with the time t; $\vec{R}_P=\vec{R}_p(t)$ thus $\vec{\dot{a}}=0$ , but if the body is rotate about axis that goes throw point P thus $\quad \vec{\dot{a}}=\vec{\dot{\varphi}}\times \vec{a}=\dot{\varphi}\,\hat{\vec{u}}$ , where $\varphi=\varphi(t)$ is the rotation angle about the axis $\hat{\vec{u}}=\frac{\vec{u}}{\sqrt{\vec{u}\cdot\vec{u}}}$

so any rotation axis $\hat{\vec{u}}$ that not parallel to vector $\vec{a}$ , the components of $\vec{\dot{a}}$ are not zero

edit

first you transform the components of vector a from B frame to O frame

$$\vec{a}_O=S\,\vec{a}_B$$

where S is the rotation matrix.

take the time derivative

$$\vec{\dot{a}}_O= \left[\frac{d}{dt}\,S\right]\,\vec{a}_B$$

For the rotation matrix I use the Rodriguez matrix

$$S=I_3+\sin(\varphi)\,\tilde{u}+(1-\cos(\varphi))\,\tilde{u}\,\tilde{u}$$

where

$$\tilde{u}=\left[ \begin {array}{ccc} 0&-u_{{z}}&u_{{y}}\\ u_{ {z}}&0&-u_{{x}}\\ -u_{{y}}&u_{{x}}&0\end {array} \right] $$

and the rotation axis : $$\vec{u}\mapsto \frac{\vec{u}}{\sqrt{\vec{u}\cdot\vec{u}}}=\begin{bmatrix} u_x \\ u_y \\ u_z \\ \end{bmatrix}_B$$

thus

$$\dot{S}=\dot{\varphi}\cos(\varphi)\,\tilde{u}+\dot{\varphi}\sin(\varphi)\,\tilde{u}\,\tilde{u}$$

thus:

$$\vec{\dot{a}}_O= \dot{S}\,\vec{a}_B= \dot{\varphi}\left[\cos(\varphi)\,\tilde{u}+\sin(\varphi)\,\tilde{u}\,\tilde{u}\right]\,\vec{a}_B$$

so if $\vec{{u}}\parallel \vec{a}\quad$ the components of $\vec{\dot{a}}_O$ are equal zero.

with $\tilde{u}\,\vec{a}=\vec{\hat{u}}\times \vec{a}$

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  • $\begingroup$ I don't know if can agree with your last line. $\dot{a}$ should always be 0 if a body is rotating otherwise then the angular velocity will not remain constant... See this question of mine to see why - physics.stackexchange.com/q/559207/113699 $\endgroup$
    – Shashaank
    Jun 25 '20 at 19:46
  • $\begingroup$ What if the rotation axis passes through A or B... Will then $\dot{a}$ be 0 or not $\endgroup$
    – Shashaank
    Jun 25 '20 at 19:48
  • $\begingroup$ @Shashaank in this case you are right $\dot a=0$ But only in this case $\endgroup$
    – Eli
    Jun 25 '20 at 19:59
  • $\begingroup$ yes precisely in this case. But I just know it should be 0 but I cannot intuitively imagine as to why it should be 0. It was precisely in this case I wanted to know why it should be 0. Could you please add a few lines as to what reason will you give for it being 0 in this case. I will accept the answer. $\endgroup$
    – Shashaank
    Jun 25 '20 at 20:08
  • $\begingroup$ @Shashaank please look new comment $\endgroup$
    – Eli
    Jun 25 '20 at 20:39

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