Find the velocity of piston, when the crank is at an angle of 45° with the horizontal. Direction of rotation of crank is clockwise.

Procedure (Do not write in exam)

-Construct the geometry.

-Draw the velocity vectors.

-Draw perpendiculars to the velocities.

-Point of intersection is the I-centre

Measure the lengths IB and IC.

Given that:

$N_{AB}=180rpm \\ \therefore \omega_{AB} = \dfrac {2π(180)}{60}=18.849 rad/s \\ \therefore V_B =5654.87 mm/s \\ Or\space \space V_B =5.6549 m/s $

Velocity of piston $= V_c \\ \omega_{IB}=\omega_{IC } \\ \therefore \dfrac {V_C}{IC}=\dfrac {V_B}{IB} \\ \therefore V_C= \dfrac {IC}{IB} ×V_B \\ =\dfrac {139}{197} × 5.6549 \\ \therefore V_C=3.9899 m/s \approx 4 m/s$

Ans: Thus the velocity of the piston is 4m/s

Note: Due to errors in measurements, the uncertainty in the answer is more, but is acceptable.