How to calculate the velocity of a piston-like linkage?

I ve tried to solve this problem in so many ways but still didn't manage to do it... What would be the correct way to solve it please?

This arm of this mechanism has a length of 0,2m. The piston has an angular velocity of 2000 tours/min. What would be the velocity of point D for an angle theta of 60 degrees?

I think that what I am missing is the angle formed by the arm and the line, which is 50mm long. Example like here (different exercise): I am trying to look for this angle beta which could help me solve the problem

• First you build a kinematic chain between A, B and D with radius $r$ between AB and the length $\ell$ between BD. The orientation angles from vertical are $\theta$ and $\phi$ respectively.
1. $(x_B,y_B) = (x_A,y_A) + (-r \sin\theta,r \cos \theta)$
2. $(x_D,y_D) = (x_B,y_B) + (-\ell \sin\phi, - \ell \cos \phi)$
• You find the angle $\phi$ from the constraint that $x_A-x_D = r \sin\theta + \ell \sin \phi = x_{AD}$ $$\sin \phi = \frac{x_{AD}}{\ell}-\frac{r}{\ell} \sin \theta$$
• Next you differentiate with respect to time using the chain rule to get the velocities
1. $(\dot{x}_B,\dot{y}_B) = (-r \dot{\theta} \cos\theta, -r \dot{\theta}\sin\theta)$
2. $(\dot{x}_D,\dot{y}_D) = (\dot{x}_B,\dot{y}_B) + (-\ell \dot{\phi} \cos\phi, \ell \dot{\phi} \sin \phi)$
• You find the rotational velocity of the connecting rod from the constraint $\dot{x}_D=-r\dot{\theta}\cos\theta -\ell \dot{\phi} \cos \phi=0$ $$\dot{\phi} = - \frac{r \dot{\theta} \cos\theta}{\ell \cos{\phi}}$$
• The collar speed is

$$v = \dot{y}_D = \ell \dot{\phi} \sin \phi - r \dot{\theta} \sin\theta = -\frac{r \sin(\theta+\phi)}{\cos\phi} \dot{\theta}$$ where $\dot{\theta}$ is the rotation of the disk in radians per second.

The angle $\beta$ can be obtained from the geometry

$$0.2*1000 \sin (\beta )+50 \sin (60 {}^{\circ})=150$$

This gives $\beta = 0.5627 rad$.

Assuming the angular velocity of the arm relative to B is $w$, the velocity of the point D can be computed as

$$v_D=\frac{2000}{60} (2 \pi ) \{0,0,1\}\times \frac{50}{1000} \{-\sin (60 {}^{\circ}),\cos (60 {}^{\circ}),0\}+\{0,0,w\}\times 0.2 \{-\sin (\beta ),-\cos (\beta ),0\}$$

$$v_D= \{0.169161 w-5.23599,-0.106699 w-9.069,0.\}$$

The $x$ component of the velocity at D is zero, so setting the first element to zero we get $w = 30.9527$. Substitute this in the second element to get the vertical velocity as $-12.3716$. It is moving downwards at 12.3716 m/s.