Implicit differentiation and piston engine speed

Question

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:

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.

Answer 1

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 $\omega t=\dfrac \pi2$,

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

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More generally,

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

Answer 2

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.

Answer 3

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

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.