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?
Best Answer
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 $$ z \approx b\cos\theta + c - \frac{b^2}{2c} \sin^2\theta $$