I am not sure whether this is the correct sub stackexchange to ask my question but I ll have a try.
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
expected answer:2,88m/s
EDIT: This is what I have so far: imgur.com/ADEp1bE
Method 1: Let $y$ be the vertical position of $D$: $$ (150-50\sin(\theta))^2+(y-50\cos(\theta))^2=200^2\tag{1} $$ Implicit differentiation yields $$ y'=\frac{7500\cos(\theta)-50y\sin(\theta)}{y-50\cos(\theta)}\,\theta'\tag{2} $$ For $\theta=\frac\pi3$, we have $y=25-\sqrt{200^2-\left(150-25\sqrt3\right)^2}$. If we plug this into $(2)$, we get $$ y'=-59.0701\text{ mm}\,\theta'\tag{3} $$ $\theta'=2000\text{ rpm}=\frac{4000\pi}{60\text{ s}}$. If we plug this into $(3)$, we get $$ \bbox[5px,border:2px solid #C0A000]{y'=-12.3716\text{ m/s}}\tag{4} $$
Method 2:
Let $A=(0,0)$. Then $B=(-\sin(\theta),\cos(\theta))\,50\text{ mm}$. Since $\theta'=2000\text{ rpm}=\frac{4000\pi}{60\text{ s}}$, we get the velocity of $B$ to be $$ \begin{align} B' &=(-\cos(\theta),-\sin(\theta))\,\theta'\,50\text{ mm}\\ &=-(\cos(\theta),\sin(\theta))\,\frac{4000\pi}{60\text{ s}}\cdot50\text{ mm}\\ &=-(\cos(\theta),\sin(\theta))\,\frac{10\pi}3\text{ m/s}\tag{5} \end{align} $$ Furthermore, since $x^2+y^2=40000\text{ mm}^2$, we have $xx'+yy'=0$, therefore, $$ y'=-\frac xyx'\tag{6} $$ Since $x=B_x+150\text{ mm}$, at $\theta=60^\circ$, we get $$ (x,y)=\left(6-\sqrt3,\sqrt{25+12\sqrt3}\right)25\text{ mm}\tag{7} $$ Since $x'=B_x'$, at $\theta=60^\circ$, $(5)$, $(6)$, and $(7)$ yield $$ \left(x',y'\right)=\left(-\frac{5\pi}3,\frac{6-\sqrt3}{\sqrt{25+12\sqrt3}}\frac{5\pi}3\right)\text{ m/s}\tag{8} $$ The downward speed of $D$ is the sum of the downward speed of $B$, which is $\frac{5\pi}{\sqrt3}\text{ m/s}$, plus $y'$. That is, $$ \bbox[5px,border:2px solid #C0A000]{\left[\frac{5\pi}{\sqrt3}+\frac{6-\sqrt3}{\sqrt{25+12\sqrt3}}\frac{5\pi}3\right]\text{ m/s}=12.3716\text{ m/s}}\tag{9} $$
Method 3: Parametrize the mechanics. Using $\phi=\theta+\frac\pi2$, let $$ B=50\,(\cos(\phi),\sin(\phi))\tag{10} $$ and $$ D=50\left(-3,\sin\left(\phi\right)-\sqrt{14+2\cos(\phi)}\sin(\phi/2)\right)\tag{11} $$ then $$ \begin{align} \left|B-D\right| &=50\sqrt{(\cos(\phi)+3)^2+(14+2\cos(\phi))\sin^2(\phi/2)}\\ &=50\sqrt{\left(\cos^2(\phi)+6\cos(\phi)+9\right)+(7+\cos(\phi))(1-\cos(\phi))}\\ &=50\sqrt{16}\\[4pt] &=200\tag{12} \end{align} $$ Here is a plot of the circle of radius $50$, the line $x=-150$, and the segment between $B$ and $D$. The vertical speed of $D$ in $\text{m/s}$ is included.
$$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}D &=50\text{ mm}\left(0,\cos(\phi)-\frac12\sqrt{14+2\cos(\phi)}\cos(\phi/2)+\frac{\sin(\phi/2)\sin(\phi)}{\sqrt{14+2\cos(\phi)}}\right)\,\phi'\\ &=0.05\text{ m}\left(0,\cos(\phi)-\frac12\sqrt{14+2\cos(\phi)}\cos(\phi/2)+\frac{\sin(\phi/2)\sin(\phi)}{\sqrt{14+2\cos(\phi)}}\right)\,\left(-\frac{2000\cdot2\pi}{60\text{ s}}\right)\\ &=\left(0,\cos(\phi)-\frac12\sqrt{14+2\cos(\phi)}\cos(\phi/2)+\frac{\sin(\phi/2)\sin(\phi)}{\sqrt{14+2\cos(\phi)}}\right)\,\left(-\frac{10\pi}{3}\right)\text{ m/s}\tag{13} \end{align} $$ Plug $\phi=\theta+\frac\pi2=\frac{\pi}3+\frac\pi2=\frac{5\pi}{6}$ into $(13)$ and we get $$ \bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{d}}{\mathrm{d}t}D=12.3716\text{ m/s}}\tag{14} $$