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I see that $W_a(1) = \dot U_a(1)=\ddot{X_a}(1) = 0.3 $

Since $U_{O'}=0 $ then O' is Instant centre of rotation. Then $U_b = 2U_a = 0.6$

I tried a lot, about a week, i find the speed, but there is a problem with acceleration. I know that we can find the acceleration using projections. But i don't understand what to do with $\varepsilon$. (Angular acceleration). Help me please.

I mean, that for points B, C i can find ${W_b}_n = OB*\omega_{ob}^2 , {W_c}_n = OC*\omega_{oc}^2$, because it simple to find $\omega_{ob},\omega_{ob}$ when you have $U_C, U_B,U_A$. But it hard for me to find ${W_b}_\tau = OB*\varepsilon_{ob} , {W_c}_\tau = OC*\varepsilon_{oc}$. I also don't know their directions.

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I have 4 A4 papers with attempts but i very slow in LaTeX and English too.

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closed as off-topic by Bernhard, user10851, John Rennie, Danu, ACuriousMind Dec 7 '14 at 13:02

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What do you mean you don't know what to do with epsilon(angular acceleration) ? Remember the crucial equation that:

translational acceleration = angular acceleration * distance to rotational axis

a = epsilon*r. Otherwise I am not sure what the problem is.

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  • $\begingroup$ I added more information, thank you for responding $\endgroup$ – Simankov Dec 7 '14 at 11:33
  • $\begingroup$ hmmm... look you have the force at point B, great news, force carries information about the tangential acceleration. F = ma, a = OBepsilon. so if you have Wbt you have the answer. (I am too lazy for latex too :) ) $\endgroup$ – Engin007 Dec 7 '14 at 11:45
  • $\begingroup$ There isn't m. We look at B as a point. $\endgroup$ – Simankov Dec 7 '14 at 11:48
  • $\begingroup$ Full $A = Wb = \sqrt{W_{b\tau}^2+W_{bn}^2}$ $\endgroup$ – Simankov Dec 7 '14 at 11:50
  • $\begingroup$ oh ok A is acceleration, (normally small a). can you tell me what is given? is Wbr or Wbn or Wc given ? or did you calculate any of them ? $\endgroup$ – Engin007 Dec 7 '14 at 11:56

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