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1) The bent rod $ABCD$ rotates about the line $AD$ whit a constant angular velocity of $90 rad / s$. Determine the velocity and acceleration of the vertex $B$ when the rod is in the position shown in Figure.

2) Determine the velocity and acceleration of the vertex $B$ assuming now that the angular velocity is of $95rad / s$ decreasing the rate of $380rad / s^2$.

enter image description here

My attempt was to apply the formula $\vec{v}=\vec{r}\times \vec{\omega}$ but I think it is not valid if the rotation axis is not aligned with any of the axes in the $xyz$ space. Maybe I'm talking nonsense. I was thinking about making a change coordinatas to the rotation axis coincide with the $z$ axis. But it seems very counterproductive and does not seem to be the purpose of the exercise.

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I think $\vec{v}=\vec{r}\times \vec{\omega}$ is perfectly valid. It doesn't depend on whether $\omega$ is aligned to a Cartesian axis. What you do need to find, is the proper $\vec{r}$. – Evert May 17 '13 at 14:43
@Evert In this case would be the vector $\vec{r}$ is $\vec{OB}$? And the vector $\vec{\omega}$? – MathOverview May 17 '13 at 14:48
$\vec{\omega}$ would have the same direction as the rotation, so $\vec{AD}$. And $\vec{r}$ would be the distance relative to that axis, so $\vec{AB}$ (or $\vec{DC}+\vec{CB}$). However you could also calculate the projected distance from $B$ to the axis of rotation, in which case the angle between $\vec{\omega}$ and $\vec{r}$ would be zero so $\vec{r}\times\vec{\omega}$ would equal $\|\vec{r}\|\|\vec{\omega}\|$. – fibonatic May 17 '13 at 15:09
I usually have it as $\vec{v}_B = \vec{v}_A + \vec{ \omega} \times (\vec{r}_B-\vec{r}_A)$ so I do not get the sign wrong. – ja72 May 17 '13 at 15:23
up vote 2 down vote accepted

Your angular velocity vector is

$$ \vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|} $$

where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12) $ in meters and $\Omega = 90\;{\rm rad/s}$.

Your velocity kinematics is

$$ \vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A ) $$

And acceleration kinematics

$$ \vec{a}_B = \dot{\vec{\omega}} \times ( \vec{r}_B - \vec{r}_A ) + \vec{\omega} \times \vec{\omega} \times ( \vec{r}_B - \vec{r}_A ) $$ $$ = \dot{\vec{\omega}} \times ( \vec{r}_B - \vec{r}_A ) + \vec{\omega} \times \vec{v}_B $$

From here you plug-and-chug.

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I'm almost convinced of his answer. But because the direction is $r_D-r_A$ not the $r_A-r_D$? – MathOverview May 18 '13 at 1:14
OP says line is $A$ to $D$, therefore it is along $\vec{r}_D-\vec{r}_A$. – ja72 May 18 '13 at 3:33

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