# Derving piston position and instant velocity

1. Mar 28, 2008

### VooDoo

hey guys,

Im doing a question where it asks to derive the instantaneous velocity of the piston. We are given the potion of the piston as a function of instantaneous angular displacement of the crank.
Now I checked the answer...and i m not sure how the $$\omega$$ gets into the answer. if I take the derivative with respect to time of both sides, i dont end up with an $$\omega$$.

tau = ratio of crank radius to connecting rod lenght R/L

x = R[(1-cos$$\theta$$) + $$\frac{\tau}{4}$$(1-cos2$$\theta$$)]

$$\dot{x}$$ = R$$\omega$$(sin$$\theta$$ + $$\frac{\tau}{2}$$sin2$$\theta$$)

2. Mar 28, 2008

### Integral

Staff Emeritus
You have x( $\theta$ ) and you are finding $\frac {dx} {dt}$ so you must use the chain rule, what will show up is $\frac {d \theta } {dt} = \omega$. So you need to finish your derivitve.