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.