Angular acceleration of a solid bar in a specific example of a dynamic riveted link arm configuration

Question

Determine the angular acceleration of the $CD$ bar for the situation illustrated in the figure. The solution is $474 \; {}^c/s^2$ $counterclockwise$.

I've done the following.

$v_B = v_A + w \cdot r_{AB} \iff v_B= -3j$

$a_B = a_A + α_{AD} \cdot r_{AD} - w^2 \cdot r_{AB} \iff a_B = -6j - 9 i $

But then, I dont know how to complete the analysis. My main lack of clarity is regarding the relation between acceleration and velocities of points $B$ and $C$.

Answer 1

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:

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