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

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 $$C$$.

• 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?
– NMech
Apr 14, 2021 at 14:25
• yes, it seems to high to me too. The levers are not that great to justify several hundred rad/s
– NMech
Apr 16, 2021 at 11:36
• I just realised that I haven't replied to your other question. Are you still interested?
– NMech
Apr 16, 2021 at 11:37
• I’m voting to close this question because the OP has vandalized their own question & closed their account.
– Fred
May 20, 2021 at 10:52

The relationship between points B and C

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

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

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

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

$$$$\vec{v}_D = \vec{v}_B + \vec{W}_{BC} \times \vec{r}_{BC} + \vec{W}_{CD} \times \vec{r}_{CD}$$$$

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:

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

• 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.
– NMech
Apr 16, 2021 at 11:39
• 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
– NMech
Apr 16, 2021 at 23:28