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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])

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# Derving piston position and instant velocity

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