Answer to Question #281319 in Mechanical Engineering for Dev

Question

The engine mechanism shown in Fig. 5 has crank OB = 50 mm and length of connecting rod

AB= 225 mm. The centre of gravity of the rod is at G which is 75 mm from B. The engine

speed is 200 r.p.m.

Fig. 5

For the position shown, in which OB is turned 45° from OA, Find 1. the velocity of G and the

angular velocity of AB, and 2. the acceleration of G and angular acceleration of AB.

Answer 1

N=200rpm

OB=50mm=0.05m

AB=225mm=0.225m

GB=75mm=0.075m

$w_{OB}=\frac{2πN}{60}=\frac{2π*200}{60}=20.94rad/s$

$V_{OB}=OB*w_{OB}=0.05*20.94=1.047m/s$

From velocity diagram,

$V_{A}=0.975m/s$

$\frac{bg}{ba}=\frac{BG}{BA}$

$\frac{bg}{0.75}=\frac{75}{225}$

Hence, $bg=0.25m/s$

Plot in velocity diagram to find V_OG

$V_{OG}=0.975m/s$