How to calculate the velocity for this

Question

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 clockwise. 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. A little bit more than just a formula as a hint would be great!

I am new to mechanics so a manner to solve this problem, which is similar to the example given (gemetrically), would be nice.

expected answer:2,88m/s

Answer 1

If you choose a reference system with origin in $A$, then $B=(-R\sin\theta,R\cos\theta)$ with $R=50\text{mm}$. Moreover, considering the right triangle whose hypotenuse is $DB$, $$ (x_B-x_D)^2+(y_B-y_D)^2=L^2 $$ with $L=0.2\text{m}$, then \begin{align} y_D&=y_B-\sqrt{L^2-(x_B-x_D)^2}=\\ &=R\cos\theta-\sqrt{L^2-(d-R\sin\theta)^2} \end{align} Taking into account that $\theta=\omega t$ \begin{align} v_D=\frac{dy_D}{dt}=-R\omega\sin\omega t-\frac{(d-R\sin\omega t)R\omega\cos\omega t}{\sqrt{L^2-(d-R\sin\omega t)^2}}\tag1 \end{align}

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There is another method: the motion of the rod $BD$ is a plane rigid motion, so the center of rotation is at the intersection of the line containing $AB$ and the horizontal line through $D$. Let's call $K$ this point. Then $$ v_B=R_B\Omega,\,v_D=R_D\Omega\implies v_D=\frac{R_D}{R_B}v_B=\frac{DK}{BK}v_B=\frac{DK}{BK}R\omega $$

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With reference to the following image

we have \begin{align} DE&=d-R\sin\theta\\ DB&=L\\ BE&=\sqrt{L^2-(d-R\sin\theta)^2}\\ BK&=\frac{BE}{\cos\theta}=\frac{1}{\cos\theta}\sqrt{L^2-(d-R\sin\theta)^2}\\ EK&=BE\tan\theta=\tan\theta\sqrt{L^2-(d-R\sin\theta)^2}\\ DK&=DE+EK=d-R\sin\theta+\tan\theta\sqrt{L^2-(d-R\sin\theta)^2}\\ v_D&=\frac{DK}{BK}R\omega=R\omega\frac{d-R\sin\theta+\tan\theta\sqrt{L^2-(d-R\sin\theta)^2}}{\dfrac{1}{\cos\theta}\sqrt{L^2-(d-R\sin\theta)^2}}=\\ &=R\omega\left(\frac{(d-R\sin\theta)\cos\theta}{\sqrt{L^2-(d-R\sin\theta)^2}}+\sin\theta\right) \end{align}

Also note that, apart from the sign, this is the same result obtained in $(1)$ when $\omega t$ is substituted with $\theta$.

Answer 2

If you’re just looking for the angle $\beta$, you can calculate that pretty easily. Call the distance between the vertical rod and the center of the disk $S$, $L$ the length of the pushrod $BD$, and $R$ the length of $BA$, i.e., the radius of the circle that point $B$ follows. The distance between $B$ and the vertical rod is then $S-\sin\theta$. This is one side of a right triangle with hypotenuse $L$, and is opposite the angle $\beta$, so $\sin\beta = {S-\sin\theta\over L}$.