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I have tried many times to solve this problem in different ways but with nos success: The angular velocity of the shaft AB is 3rad/s counterclockwise. Calculate the velocity of the shafts BD and DE.

Could someone please show me how it should be done?

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expected answers: 0 rad/s, 2 rad/s

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  • $\begingroup$ Is it 0 or 0.66 for bd $\endgroup$ – Archis Welankar Dec 23 '15 at 14:15
  • $\begingroup$ What do you mean? The shafts length? $\endgroup$ – privetDruzia Dec 23 '15 at 14:26
  • $\begingroup$ Angular velocity $\endgroup$ – Archis Welankar Dec 23 '15 at 14:27
  • $\begingroup$ this angular velocity is not given. But according to the answers it is 0 (for some mysterious reason) $\endgroup$ – privetDruzia Dec 23 '15 at 15:01
  • $\begingroup$ I am talking about answer is it 0.6 for Ab so approx $0$ $\endgroup$ – Archis Welankar Dec 23 '15 at 15:21
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Hint $\omega=\frac{v}{r}$ you can calculate $r$ by pythaogoras theorem for BD same for DE and no need of pythagoras theorem there

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  • $\begingroup$ Yes indeed I did something similar to that, but didn't work... imgur.com/PsFg73Y And Vbd is already incorrect, so Vde will be too $\endgroup$ – privetDruzia Dec 23 '15 at 16:03
  • $\begingroup$ see lets talk in SI units i e (meters ) convert mm to meters and you get it . $\endgroup$ – Archis Welankar Dec 23 '15 at 16:49
  • $\begingroup$ I think that converting to SI units would still not do the trick. Could you please post your method to clarify your explanation? $\endgroup$ – privetDruzia Dec 23 '15 at 17:25
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In the given position, the velocity of $B$ and $D$ are vertical, the first one tangent to the circle of center $A$ and passing though $B$, the second tangent to the circle of center $E$ and passing through $D$.

The center of rotation of $BD$ is at the intersection of the lines perpendicular to this two velocities, but being the velocities parallel, also the two perpendicular lines are parallel, so there is no intersection, and this mean that $BD$ does not rotate, but translate, so its angular velocity is $0\text{rad/s}$ and $v_B=v_D$.

Furthermore, given that $v_B=r_{AB}\omega_{AB}$ and $v_D=r_{ED}\omega_{ED}$, then $$ r_{AB}\omega_{AB}=r_{ED}\omega_{ED}\implies \omega_{ED}=\frac{r_{AB}}{r_{ED}}\omega_{AB}=\frac{150\text{mm}}{225\text{mm}}3\text{rad/s}=2\text{rad/s} $$

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