Chapter 8.3, Problem 54AYU

To determine

Find the angle of rotation of a piston?

Answer to Problem 54AYU

  $\boxed{x=r\cos \theta -\sqrt{r^2{\cos }^2\theta +L^2-r^2}}$

Explanation of Solution

Given information:

Road $OA$ rotates about the fixed point $O$ so that point $A$ travels on a circle radius r. Connected to point $A$ is another rod $AB$ of length $L>2r$ , and point $B$ is connected to a piston. See the figure. Show that the distance $x$ between point $O$ and point $B$ is given by

  $x=r\cos \theta +\sqrt{r^2{\cos }^2\theta +L^2-r^2}$

Where $\theta$ is the angle of rotation of rod $OA$ .

  

Calculation:

Let us consider the following figure of Roads and Piston:

The triangle $\Delta 0AB$ is SAS and SSS triangle, apply Law of Cosines in the triangle,we get

  ${\left(AB\right)}^2={\left(OA\right)}^2{\left(OB\right)}^2-2\left(OA\right)\left(OB\right)\cos \theta$

  $L^2=r^2+x^2-2rx\cos \theta$

  ${\left(AB\right)}^2={\left(OA\right)}^2+{\left(OB\right)}^2-2\left(OA\right)\left(OB\right)\cos \theta$

  $L^2=r^2+x^2-2rx\cos \theta$

  $x^2-2rs\cos \theta +\left(r^2-L^2\right)=0$

This is quadratic in $x$ , we get

  $x=\frac{2r\cos \theta \pm \sqrt{{\left(2r\cos \theta \right)}^2-4\left(1\right)\left(r^2-L^2\right)}}{2}$

  $x=r\cos \theta \pm \sqrt{r^2{\cos }^2\theta +L^2-r^2}\mathrm{....}(1)$

Now ,we have

  $0<\theta <\pi$

  $-1<\cos \theta <1$

  $-r<r\cos \theta <r\mathrm{........}\left(2\right)$

And $l>2r$

  $L^2>4r^2$

  $L^2-r^2>3r^2\mathrm{.....}(3)$

Also, $0<{\cos }^2\theta <1$

  $0<r^2{\cos }^2\theta <r^2$

  $L^2-r^2<r^2{\cos }^2\theta +L^2-r^2<L^2$

  $3r^2<r^2{\cos }^2\theta +L^2-r^2<L^2$

  $\sqrt{3r}<\sqrt{r^2{\cos }^2\theta +L^2-r^2}<L\mathrm{......}(4)$

From (2) and (4), we get

  $r\cos \theta <\sqrt{r^2{\cos }^2\theta +L^2-r^2}$ …….(5)

Thus, $x=r\cos \theta -\sqrt{r^2{\cos }^2\theta +L^2-r^2}$ is not possible as the distance can not be negative (5)).

Hence, $x=r\cos \theta -\sqrt{r^2{\cos }^2\theta +L^2-r^2}$ .