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This question already has an answer here:

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...

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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):

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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

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marked as duplicate by user147263, user99914, Harish Chandra Rajpoot, user228113, Chris Godsil Dec 24 '15 at 2:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I don't see how this is a duplicate. The answer to this question was posted before the "original" question was even asked. $\endgroup$ – robjohn Dec 24 '15 at 4:16
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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:

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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.

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$$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}D &=50\text{ mm}\left(0,\cos(\phi)-\frac12\sqrt{14+2\cos(\phi)}\cos(\phi/2)+\frac{\sin(\phi/2)\sin(\phi)}{\sqrt{14+2\cos(\phi)}}\right)\,\phi'\\ &=0.05\text{ m}\left(0,\cos(\phi)-\frac12\sqrt{14+2\cos(\phi)}\cos(\phi/2)+\frac{\sin(\phi/2)\sin(\phi)}{\sqrt{14+2\cos(\phi)}}\right)\,\left(-\frac{2000\cdot2\pi}{60\text{ s}}\right)\\ &=\left(0,\cos(\phi)-\frac12\sqrt{14+2\cos(\phi)}\cos(\phi/2)+\frac{\sin(\phi/2)\sin(\phi)}{\sqrt{14+2\cos(\phi)}}\right)\,\left(-\frac{10\pi}{3}\right)\text{ m/s}\tag{13} \end{align} $$ Plug $\phi=\theta+\frac\pi2=\frac{\pi}3+\frac\pi2=\frac{5\pi}{6}$ into $(13)$ and we get $$ \bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{d}}{\mathrm{d}t}D=12.3716\text{ m/s}}\tag{14} $$

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  • $\begingroup$ I see what you are doing, unfortunately i don't see how knowing the height of point D would help me further... $\endgroup$ – privetDruzia Dec 23 '15 at 11:20
  • $\begingroup$ @privetDruzia: are you not trying to find the velocity of $D$? $\endgroup$ – robjohn Dec 23 '15 at 11:26
  • $\begingroup$ Well this is how I am trying to solve it: By using basic geometry i can find the position of the velocity vectors and their angles amongs each other. Once I find that I might be able to form a (velocity) triangle, where the length of the edges is the magnitude of the vectors. Example: imgur.com/IpeeGSD My current desperate situation: imgur.com/ADEp1bE This is why I don't see how the solution u offered can help me. $\endgroup$ – privetDruzia Dec 23 '15 at 11:42
  • $\begingroup$ Or one could simply compute $y'$ from my equation and scale for $2000$ rpm. $\endgroup$ – robjohn Dec 23 '15 at 11:46
  • $\begingroup$ sorry, i really don't want to be rude. But I am quite new to mechanics. And the method I showed seems to work well on quite a lot of problems. So I'd like to stick with that method untill I feel comfortable enough. Could you please show me how it should be done using the method I posted? I even think it is a relatively easy method because it is very visual. $\endgroup$ – privetDruzia Dec 23 '15 at 11:48

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