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Question:

The crankshaft rotates with the constant angular velocity $\omega$ rad/s. Calculate the piston speed when

$$\omega t = \frac{\pi}{2}$$

Attempted solution:

Here is an image of the problem:

enter image description here

I have absolutely no knowledge whatsoever about engines or pistons. So I am far outside my comfort zone. But I have worked on a few implicit differentiation problems, so I get the overarching strategy: find relevant function, differentiate both sides, plug in the values for derivatives and variables/constants and solve for whatever you need to get.

From the image, I am guessing that finding $x(t)$ seems like a productive start since it is a straight distance from the center. Then taking the derivative and plugging in the expression given in the question.

The inner part seems like a triangle with sides $x$, $b$ and $R$, then the angle velocity $\omega t$. Perhaps the law of cosines will give us something:

$$x(t) = \sqrt{R^2 + b^2 -2Rb \cos \omega t}$$

Taking the derivative:

$$x'(t) = \frac{1}{2\sqrt{R^2 + b^2 -2Rb \cos \omega t}} 2Rb\omega \sin(\omega t)$$

Putting in the value for \omega t:

$$x'(t) = \frac{1}{2\sqrt{R^2 + b^2 -2Rb \cos \frac{\pi}{2}}} 2Rb\omega \sin \frac{\pi}{2}$$

Simplfying gives:

$$x'(t) = \frac{Rb\omega}{\sqrt{R^2 + b^2}}$$

However, I am not able to get it to go anywhere from here. The answer is $R \omega$ downwards. I am doubtful that I have even gotten the basic idea of the question right.

Edit:

Using Aretino's equation:

$$b^2 = x^2 + R^2 - 2Rx \cos{\omega t}$$

Implicit differentiation gives:

$$0 = 2x + 2Rx\omega \sin{\omega t}$$

Putting in $\omega t = \frac{\pi}{2}$:

$$0 = 2x + 2Rx\omega =2x(1+R\omega)$$

...but I think this is wrong and I cannot get any further.

I am looking to finish this question off using a calculus-based answer.

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  • $\begingroup$ I personally vote that this should be in Physics StackExchange. $\endgroup$ – S.C.B. Aug 1 '16 at 13:56
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    $\begingroup$ The question is from a general first-year single variable calculus textbook that does not involve any emphasis on physics or engineering. $\endgroup$ – MathInferno Aug 1 '16 at 14:01
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    $\begingroup$ Aretino’s answer points out where you went wrong in setting up the equation. However, for this particular problem, you can skip the calculus. When $\omega t=\frac\pi2$, the short arm of length $R$ is perpendicular to the direction of the piston’s motion (there’s no “wasted” sideways motion) so the piston’s velocity must be equal to the linear velocity of the end of the rotating arm. $\endgroup$ – amd Aug 1 '16 at 18:39
  • $\begingroup$ The bounty should be granted to @amd. $\endgroup$ – Yves Daoust Jul 11 '19 at 11:54
  • $\begingroup$ That answer is in a comment, was posted two years before the bounty and does not fulfill the stated bounty requirements (answer needed to use implicit differentiation, even if simpler answers were possible). I have awarded the bounty to the first answer that fulfilled the bounty, which was yours. $\endgroup$ – MathInferno Jul 13 '19 at 10:59
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From

$$b^2=x^2+R^2-2xR\cos(\omega t),$$ by differentiation on time $$0=2x\dot x-2\dot xR\cos(\omega t)+2xR\omega\sin(\omega t).$$

Then at $\omega t=\dfrac \pi2$,

$$0=\dot x+R\omega.$$


More generally,

$$\dot x=\frac{xR\omega\sin(\omega t)}{R\cos(\omega t)-x}=\frac{\left(R\cos(\omega t)-\sqrt{b^2-R^2\sin^2(\omega t)}\right)R\omega\sin(\omega t)}{\sqrt{b^2-R^2\sin^2(\omega t)}}.$$

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You must be careful with the angle: $\omega t$ is the angle between $R$ and $x$. You should then use: $b^2=x^2+R^2-2Rx\cos\omega t$ and differentiate this equation.

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Biella_manov_1

I would solve the crankshaft kinematics through the sines law $$ {{\sin \alpha } \over R} = {{\sin \omega \,t} \over b} = {{\sin \left( {\pi - \omega \,t - \alpha } \right)} \over x} = {{\sin \left( {\omega \,t + \alpha } \right)} \over x} = {{\sin \omega \,t\cos \alpha + \cos \omega \,t\sin \alpha } \over x} $$ which is valid also for $\pi < \omega \, t < 2 \pi$, that is for negative values of the sines.

From there it is easy to deduce $$ x = b{{\sin \omega \,t\cos \alpha + \cos \omega \,t\sin \alpha } \over {\sin \omega \,t}} = b\sqrt {1 - \left( {R/b} \right)^{\,2} \sin ^{\,2} \omega \,t} + R\cos \omega \,t $$ which for $R<<b$ correctly reduces to $b+R \cos \omega \, t$.

The derivate is then $$ x' = - \,\omega \,R\sin \omega \,t\left( {1 + {{R\cos \omega \,t} \over {b\sqrt {1 - \left( {R/b} \right)^{\,2} \sin ^{\,2} \omega \,t} }}} \right) $$

An example of the dispacement - speed pattern is sketched here

Biella_manov_2

Engineering-wise then, without any calculation, it is clear that at $\omega\,t = \pi / 2$, the speed of the piston will be same as the radial speed $\omega R$ (with the minus sign as giving a diminuishing $x$), because the two are parallel which means that the crank has only translatory speed with no rotation.

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