The crank is rotating at 180 rpm. Find the velocity of piston, when the crank is at an angle of 45° with the horizontal.

Direction of rotation of crank is clockwise.

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 06

Years : DEC 2012

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

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

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.