Rotating links

Let an inertial coordinate system be placed at point A (A is the origin) with the x-axis pointing rightward horizontal and y-axis pointing vertically up. Let $AB = L_1$ and $BC = L_2$ . Looking at the X-coordinate of point $C$

we have:

$x_c = -L_1\sin{\theta} - L_2\sin{\phi}$

Since the motion at point C is constrained by a translational joint, the x-coordinate of point C remains constant throughout, and therefore:

$\dot{x}_c = 0 = -L_1\cos{\theta}\dot{\theta} -L_1\cos{\phi}\dot{\phi}$

Solving yields:

$\dot{\phi} = -\frac{L_1}{L_2}\frac{\cos{\theta}}{\cos{\phi}}\dot{\theta} = -15\sqrt{2}$

Now at this stage, I have a reservation about the problem statement. The problem asks for the angular velocity which should be inclusive of a sign. This is because the sign is indicative of whether the rotation is clockwise or anticlockwise. Yet the answer is a positive number. I got my initial attempt wrong due to this. I request that the problem statement be updated to asking for the floor of the magnitude of angular velocity .

Coordinates and velocity of point $B$ , assuming that point $A$ is at the origin:

$x_B = -L_{AB} \, \sin \theta \\ y_B = L_{AB} \, \cos \theta \\ \dot{x}_B = -L_{AB} \, \cos \theta \, \dot{\theta} \\ \dot{y}_B =- L_{AB} \, \sin \theta \, \dot{\theta}$

Coordinates and velocity of point $C$ :

$x_C = x_B - L_{BC} \, \sin \phi \\ y_C = y_B - L_{BC} \, \cos \theta \\ \dot{x}_C = -L_{AB} \, \cos \theta \, \dot{\theta} - L_{BC} \, \cos \phi \, \dot{\phi} \\ \dot{y}_C = -L_{AB} \, \sin \theta \, \dot{\theta} + L_{BC} \, \sin \phi \, \dot{\phi}$

We know that $\dot{x}_C = 0$

$\dot{x}_C = -L_{AB} \, \cos \theta \, \dot{\theta} - L_{BC} \, \cos \phi \, \dot{\phi} = 0 \\ \dot{\phi} = \frac{L_{AB} \, \cos \theta \, \dot{\theta}}{- L_{BC} \, \cos \phi}$

Plugging in numbers yields:

$|\dot{\phi}| = 15 \sqrt{2} \approx 21.213$

Rotation About Fixed Axis. For link AB

$V_{B} = \boldsymbol{\omega}_{AB} \times \textbf{r}_{AB}$

= $(60\textbf{k}) \times (-0.3 \sin 60°\textbf{i} + 0.3 \cos 60°\textbf{j})$

={ $-9\textbf{i} - 9\sqrt{3}\textbf{j}$ } m/s

General Plane Motion for rod BC ,so use the relative velocity equation:

$\textbf{V}_{C} = \textbf{V}_{B} + \boldsymbol{\omega}_{C/B} \times \textbf{r}_{C/B}$

$-v_{C} \textbf{j} = (-9\textbf{i} - 9\sqrt{3} \textbf{j} ) + (\omega_{C/B} \textbf{k}) \times (-0.6 \sin 45°\textbf{i} - 0.6 \cos 45°\textbf{j})$

$-v_{C} \textbf{j} = (0.3\sqrt{2} \omega_{C/B} - 9)\textbf{i} + ( -0.3\sqrt{2}\omega_{C/B} - 9\sqrt{3})\textbf{j}$

Equating i components,

$(0.3\sqrt{2} \omega_{C/B} - 9)=0$

$\omega_{C/B} =21.2 rad/s$

round down to 21 rad/s