# How to calculate the velocity for such a situation [duplicate]

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

EDIT: This is what I have so far: imgur.com/ADEp1bE

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

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

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 $$\bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{d}}{\mathrm{d}t}D=12.3716\text{ m/s}}\tag{14}$$

• I see what you are doing, unfortunately i don't see how knowing the height of point D would help me further... – privetDruzia Dec 23 '15 at 11:20
• @privetDruzia: are you not trying to find the velocity of $D$? – robjohn Dec 23 '15 at 11:26
• 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. – privetDruzia Dec 23 '15 at 11:42
• Or one could simply compute $y'$ from my equation and scale for $2000$ rpm. – robjohn Dec 23 '15 at 11:46
• 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. – privetDruzia Dec 23 '15 at 11:48