The above figure shows a piston driving a crank OP pivoted at the end $O$. The piston slides in a straight cylinder and the crank is made to rotate with constant angular velocity $ \omega $. Find the distance $OQ$ in terms of the lengths $b,c$ and the angle $\theta$. Show that, when $b/c$ is small, $OQ$ is given approximately by $OQ = c + b\cos(\theta)-\frac{b^2}{2c}\sin^2(\theta)$
I have sketched a little diagram that goes as follows;
$ \cos \theta = \frac{x}{b} $ this implies that $ b\cos\theta = x$
$ \sin \theta = \frac{h}{b} $ this implies that $ b\sin\theta = h$
now $ c^2 = h^2 + y^2 $ so $ c^2 - h^2 = y^2 $
now i am letting the length $OQ = z$.
$z = x + y = b\cos\theta + \sqrt{c^2 - b^2(\sin\theta)^2}$
Now I know I can manipulate this more. but I feel as though i am getting further and further away. I may have made a mistake, but it is basic trig?
The key thing to remember is that the linking rods are rigid and must maintain their length throughout.
Put the origin at O. The point P is instantaneously $b\cos\theta, b\sin\theta$ which automatically satisfies $|OP|=b$. If $Q=(z,0)$, one must have $|PQ|=c$. This gives you
$$ (z-b\cos\theta)^2 + (b\sin\theta)^2 = c^2 \\ z^2 - 2bz \cos\theta + b^2-c^2 = 0 $$ Solving this quadratic for $z$ you have
$$ z = b\cos\theta \pm \sqrt{c^2-b^2\sin^2\theta} \\ = b\cos\theta \pm c\sqrt{1-\frac{b^2}{c^2}\sin^2\theta} $$ You have to take the positive root since the length is b+c at $\theta=0$.
For small $b/c$, you can expand the square root term in a Taylor series as $(1+x)^{(1/2)} \approx 1 + \frac{x}{2}$, giving you $$ z \approx b\cos\theta + c - \frac{b^2}{2c} \sin^2\theta $$