Derving piston position and instant velocity

  • Thread starter VooDoo
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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 [tex]\omega[/tex] gets into the answer. if I take the derivative with respect to time of both sides, i dont end up with an [tex]\omega[/tex].

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


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

[tex]\dot{x}[/tex] = R[tex]\omega[/tex](sin[tex]\theta[/tex] + [tex]\frac{\tau}{2}[/tex]sin2[tex]\theta[/tex])
 

Answers and Replies

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

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