Determine the angular acceleration of the $CD$ bar for the situation illustrated in the figure.

The solution is $474 rad/s^2$ $counterclockwise$

enter image description here

I've done:

$v_B = v_A + w *r_{AB} <=> v_B= -3j$

$a_B = a_A + α_{AD}*r_{AD} - w^2*r_{AB} <=> a_B = -6j - 9 i $

But then i dont know how to finish it

My main doubt is what is the relation between acceleration and velocities of points $B$ and $C$ ?

  • $\begingroup$ I didn't get your result. However, I am fairly confident that my approach is correct, do you have interdmediate values e.g. for angular velocity of BC, and BD? $\endgroup$
    – NMech
    Apr 14 at 14:25
  • $\begingroup$ yes, it seems to high to me too. The levers are not that great to justify several hundred rad/s $\endgroup$
    – NMech
    Apr 16 at 11:36
  • $\begingroup$ I just realised that I haven't replied to your other question. Are you still interested? $\endgroup$
    – NMech
    Apr 16 at 11:37
  • 3
    $\begingroup$ I’m voting to close this question because the OP has vandalized their own question & closed their account. $\endgroup$
    – Fred
    May 20 at 10:52

The relationship between points B and C

$$\vec{v}_C = \vec{v}_B + \vec{W}_{BC} \times \vec{r}_{BC} \tag{eq.1}$$

Additionally you know that

$$\vec{v}_D = \vec{v}_C + \vec{W}_{CD} \times \vec{r}_{CD} \tag{eq.2}$$

However you already know that:

  • $v_B = \begin{bmatrix}0 \\-3j\\0 \end{bmatrix}$
  • $v_D = \begin{bmatrix}0 \\0\\0 \end{bmatrix}$

If you do the calcs you get from the equations (1, 2):

\begin{equation} \vec{v}_D = \vec{v}_B + \vec{W}_{BC} \times \vec{r}_{BC} + \vec{W}_{CD} \times \vec{r}_{CD} \end{equation}

if you do the math you should get: $$ \begin{bmatrix}0 \\0\\0 \end{bmatrix} = \begin{bmatrix}-w_{BC} - 0.5*w_{CD} \\-0.5*w_{CD} - 3\\0 \end{bmatrix} $$

From which you can obtain:

  • $w_{CD}= -6 [\frac{rad}{s}]$
  • $w_{BC}= +3 [\frac{rad}{s}]$

Similarly the (chain) relationships for the acceleration of point B, C, D are:

$$a_B = a_A + α_{AB}\times r_{AB} + w_{AB}\times(w_{AB}\times r_{AB}) \Leftrightarrow a_B = -6j - 9 i$$ $$a_C = a_B + α_{BC}\times r_{BC} + w_{BC}\times(w_{BC}\times r_{BC}) $$ $$a_D = a_C + α_{CD}\times r_{CD} + w_{CD}\times(w_{CD}\times r_{CD}) $$

However as before, $a_D$ should be equal to zero. So if you do the math you should get to:

$$\begin{bmatrix}0 \\0\\0 \end{bmatrix} = \begin{bmatrix} -a_{BC} - 0.5*a_{CD} - 4.5 \\ -0.5*a_{CD} - 19.5\\ 0 \end{bmatrix}$$

The result I am getting is:

$$a_{CD}= -39[rad/s^2]\qquad a_{BC}= 15[rad/s^2] $$

  • $\begingroup$ The best thing to do, is to draw up the instantaneous center of rotation. which coincides with the midsection of the AB bar. The BC rotates around that point - CW- for the instant given. $\endgroup$
    – NMech
    Apr 16 at 11:39
  • $\begingroup$ yes. point B only has a velocity component on Y because its rotating about point A, while point C only has a velocity component on X because its rotating about point D $\endgroup$
    – NMech
    Apr 16 at 23:28

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