Determine the angular velocity and angular acceleration of the AB bar at the instant shown

Question

Consider a sliding cursor C that is fixed to the disk which has a constant angular velocity counterclockwise at 4 rad/s. Determine the angular velocity and angular acceleration of the AB bar at the instant shown. [0.667 rad/s (counterclockwise); 3.84 rad/s2 (Clockwise)]

to solve this exercise I draw the two axes

I've done

$$\begin{align} v_C &= v_O + \Omega_{OC} \times r_{OC} \\ &= 0 + 4k * (-0.06\sin(30)i+0.06\cos(30)j) \\ &= -0.12j-0.2078i \\ v_C &= v_A + \Omega_{AC}r_{AC} + v_{rel} \\ &= 0 + \Omega_{AC}k\cdot 0.180\cos(60)i + 0.18\sin(60)j) + v_{rel}\cos(60)i + v_{rel}\sin(60)j \\ &= (-0.156\Omega_{AC} + 0.5v_{rel})i + (0.09\Omega_{AC} + 0.866v_{rel})j \end{align}$$

and therefore i got $v_{rel} = -0.2079\text{ m/s}$ and $\Omega_{AC}=0.667\text{ rad/s}$ which are the correct results

$$\begin{align} a_C &= a_0 + \alpha_{OC}\times r_{OC} + \Omega_{OC}\times (\Omega_{OC}\times r_{OC}) \\ &= 0 + 0 + 4k(4k(-0.06\sin(30)i + 0.06\cos(30)j)) \\ &= 0.48i-0.8312j \\ a_C &= a_A + \alpha_{AC}\times r_{AC} + \Omega_{AC}\times (\Omega_{AC}\times r_{AC}) + a_{coriolis} + a_{rel} \\ &= 0 + 0 + \alpha_{AC}k(0.18\cos(60)i + 0.18\sin(60)j) + (-0.44i - 0.069j) +(-0.139j + 0.24i) + a_{rel}(\cos(60)i + \sin(60j)) \\ &= (-0.156\alpha_{AC} - 0.2 + 0.5a_{rel})i + (0.99\alpha_{AC} - 0.208 + 0.866a_{rel})j \end{align}$$

and therefore i got $a_{rel}=0.56\text{ m/s}^2$ and $\alpha_{AC} = -6.156\text{ rad/s}^2$, which is the wrong value for $α_{AC}$.

is it possible to solve the exercise using the 2 axes that I draw? because I can't understand where my resolution is wrong

Answer 1

The problem is with

$$v_C = v_A + \Omega_{AC} \times r_{AC} + v_{rel}$$

More specifically, you are getting confused with what it means to have a different coordinate system. In your image you have $XY$ (which is at an angle) and $\color{red}{XY}$ which is horizontal and vertical. That means that you have $\vec{i}$ and $\color{red}{\vec{i}}$ respectively.

So for example when you are substituting $\vec{v}_{rel} \rightarrow v_{rel} \vec{i}$ and eventually you come up with

$$\vec{v}_{C} =v(rel)\vec{i} + 0.180\Omega_{AC}\vec{j} \qquad\text{ and } \qquad\vec{v}_{C} = -0.12\color{red}{\vec{j}}-0.2078\color{red}{\vec{j}}$$

you have to remember that $\vec{i}$ and $\color{red}{\vec{i}}$ are different, and they are related with the angle:

$$\vec{i} = \cos\phi \color{red}{\vec{i}} + \sin\phi \color{red}{\vec{i}}$$ $$\vec{j} = -\sin\phi \color{red}{\vec{i}} + \cos\phi \color{red}{\vec{j}}$$

Substitute that in your relationships and you should get the correct values.

Similarly with the acceleration.

$$a_C = a_0 + \alpha_{OC}\times r_{OC} + \Omega_{OC}\times (\Omega_{OC}\times r_{OC}) $$

and because $\alpha_{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ and $\alpha_{OC}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ the above simplifies to

$$a_C = \Omega_{OC}\times (\Omega_{OC}\times r_{OC}) $$

where

• $a_C =\begin{bmatrix}a_{C,x}\\a_{C,y}\\0\end{bmatrix}$,
• $r_{OC}=0.06\begin{bmatrix}-\sin(30^o)\\\cos(30^o)\\0\end{bmatrix}$,
• $\Omega_{OC}=\begin{bmatrix}0\\ 0 \\ 4\end{bmatrix}$

Therefore:

$$\begin{bmatrix}a_{C,x}\\a_{C,y}\\0\end{bmatrix} = \begin{bmatrix}0\\ 0 \\ 4\end{bmatrix}\times \left(\begin{bmatrix}0\\ 0 \\ 4\end{bmatrix}\times 0.06\begin{bmatrix}-\sin(30^o)\\\cos(30^o)\\0\end{bmatrix}\right) \Rightarrow $$

$$\begin{bmatrix}a_{C,x}\\a_{C,y}\\0\end{bmatrix} = \begin{bmatrix}0.48\\ -0.831384 \\ 0\end{bmatrix}$$

if you do the same with this equation you will get

$$a_C = a_A + \alpha_{AC}\times r_{AC} + \Omega_{AC}\times (\Omega_{AC}\times r_{AC}) + a_{coriolis} + a_{rel} $$

$$\begin{bmatrix}a_{C,x}\\a_{C,y}\\0\end{bmatrix} = \begin{bmatrix} 0.200178 - 0.155885 \alpha_{AC} + \frac{a_{Rel}}{2}\\ -0.207949 + 0.09 \alpha_{AC} + \frac{\sqrt{3}}{2} a_{Rel}\\ 0\end{bmatrix}$$

By equating the x, and y coordinates for $a_C$ you will get: $$\alpha_{AC} = -3.07806, \qquad a_{Rel} =-0.4$$