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I ve tried to solve this problem in so many ways but still didn't manage to do it...

enter image description here

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):

enter image description here

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

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closed as off-topic by Yagna Patel, user228113, Harish Chandra Rajpoot, Leucippus, Alex Provost Dec 24 '15 at 3:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Community, Harish Chandra Rajpoot, Alex Provost
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Your image is not at all readable? It is very difficult to help you as is with this image. $\endgroup$ – Satish Ramanathan Dec 23 '15 at 20:38
  • $\begingroup$ @satishramanathan: Which image and what parts of it are not readable? (the original image of the first one is not very clear neither) $\endgroup$ – privetDruzia Dec 23 '15 at 20:44
  • $\begingroup$ Possible duplicate of How to calculate the velocity for such a situation $\endgroup$ – amd Dec 23 '15 at 21:52
  • $\begingroup$ I just noticed that you’ve already asked this same question before and got a fairly detailed answer. Flagging as a duplicate. $\endgroup$ – amd Dec 23 '15 at 21:53
  • $\begingroup$ @amd unfortunately his answer is completely wrong $\endgroup$ – privetDruzia Dec 23 '15 at 21:56
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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}


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 $$ v_B=R_B\Omega,\,v_D=R_D\Omega\implies v_D=\frac{R_D}{R_B}v_B=\frac{DK}{BK}v_B=\frac{DK}{BK}R\omega $$

enter image description here


With reference to the following image

enter image description here

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 $\theta$.

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  • $\begingroup$ Why would you need a reference system and can't you just work with the absolute lengths of the shafts, like on the example? $\endgroup$ – privetDruzia Dec 23 '15 at 21:13
  • $\begingroup$ @privetDruzia: is only a matter of convenience, and the result is independent of the reference system (excluding moving reference systems). $\endgroup$ – enzotib Dec 23 '15 at 21:17
  • $\begingroup$ But there is no rod between AD... What doe omega and Rd stand for? i have not tested it yet, but are you sure this would word? It looks to me like you switched two terms from the vormula $v = r*\omega$ $\endgroup$ – privetDruzia Dec 23 '15 at 21:29
  • $\begingroup$ @privetDruzia: yes, it was $BD$, not $AD$. Moreover $R_D$ stands for the distance $DK$, the radius of the circle through $D$ and centered at the center of rotation $K$, and $\Omega$ is the angular velocity of the rod, $\omega$ is the angular velocity of the disc. $\endgroup$ – enzotib Dec 23 '15 at 21:38
  • $\begingroup$ I am going to have a look at it now. This is what I had in mind:imgur.com/xFpcbpX (similar to the example given) $\endgroup$ – privetDruzia Dec 23 '15 at 21:45
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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 $\sin\beta = {S-\sin\theta\over L}$.

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  • $\begingroup$ How would you find the angle the vector Vb makes with the vertival vector Vd: imgur.com/xFpcbpX ? $\endgroup$ – privetDruzia Dec 23 '15 at 21:49

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