Using the loop closure method calculate the velocity of the

Using the loop closure method, calculate the velocity of the piston when the crank is at 51 degree (measured from the positive X-axis). Shown your work in the engineering problem solving format.

Solution

y = L2*sin theta + L3*cos phi...........where theta = angle of crank with x-axis and phi = angle of rod with y-axis

y = 63*sin theta + 130*cos phi

Also, L3*sin phi = 52 + L2*cos theta

130*sin phi = 52 + 63*cos theta

sin phi = (52 + 63*cos theta)/130

cos phi = sqrt (1 - sin^2 phi)

= sqrt(1 - ((52 + 63*cos theta)/130)^2)

= sqrt (130^2 - (52 + 63*cos theta)^2) / 130

y = 63*sin theta + 130*sqrt (130^2 - (52 + 63*cos theta)^2) / 130

y = 63*sin theta + sqrt (130^2 - (52 + 63*cos theta)^2)

dy/dtheta = 63*cos theta + (1/2) * (130^2 - (52 + 63*cos theta)^2)^(-1/2) * (126*(52 + 63*cos theta)*sin theta)

dy/dtheta = 63*cos theta + (130^2 - (52 + 63*cos theta)^2)^(-1/2) * (63*(52 + 63*cos theta)*sin theta)

Velocity of piston = dy/dt = dy/dtheta * dtheta/dt

But dtheta/dt = w

So, Velocity of piston = 63*w*cos theta + (130^2 - (52 + 63*cos theta)^2)^(-1/2) * (63*w*(52 + 63*cos theta)*sin theta)

Putting values,

Velocity of piston =  63*25*cos 51 + (130^2 - (52 + 63*cos 51)^2)^(-1/2) * (63*25*(52 + 63*cos 51)*sin 51)

= 2207.8 mm/s

= 2.207 m/s