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

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  • I've updated the first part of the calculation for the acceleration. You can see that the values are different than in your case. Could you verify my calculation? Also could you update the expected values for the acceleration?
– NMech
Feb 17, 2021 at 16:04

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: $$\alpha_{AC} = -3.07806, \qquad a_{Rel} =-0.4$$

• could you please try to format with latex the equations. Its very difficult to look the equations in ascii format.
– NMech
Feb 17, 2021 at 11:11