Relative Motion / Piston Motion

Hello

I'm trying to find the velocity of the piston in the following position, but I cant seem to get the right answer? Can you please help me? I really want to learn how to do this!

Thanks Heaps

Edward

Using the triangle in the figure, by Law of Cosines,
L^2 = r^2 +x^2 -2(L)(r)cos(theta)
(68)^2 = (17)^2 +x^2 -2(17)(x)cos(theta)
x^2 -34x*cos(theta) -4335 = 0 -----------------(i)

We are to find dx/dt when theta = 30degrees and d(theta)/dt = 206 rpm.

Differentiate both sides of (i) with respect to time t,
2x(dx/dt) -34[x(-sin(theta))(d(theta)/dt) +cos(theta)(dx/dt)] = 0
Divide both sides by 2, then expand,
x(dx/dt) +17x*sin(theta)(d(theta)/dt) -17cos(theta)(dx/dt) = 0
[x -17cos(theta)](dx/dt) +[17x*sin(theta)](d(theta)/dt) = 0 ----------(ii)

When theta = 30degrees,
x^2 -34x*cos(theta) -4335 = 0 -----------------(i)
x^2 -29.445x -4335 = 0
x = {29.445 +,-sqrt[(29.445)^2 -4(1)(-4335)]} / 2(1)
x = 82.189 or -52.744 mm
Or,
x = 82.189 mm

d(theta)/dt = 206 rev/min = (206rev/1min)(2pi radians /1rev) = 412pi radians/min.

Substitute those into (ii),
[82.189 -14.722](dx/dt) +[17(82.189)(0.5)(412pi)] = 0
dx/dt = -904,231.664 / 67.467
dx/dt = -13,402.577 mm/min

Therefore, at that instant, the piston is moving towards O at the rate of 13,403 mm/min. ----------------answer.