Question #281319

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.

1
Expert's answer
2021-12-24T10:19:02-0500

N=200rpm

OB=50mm=0.05m

AB=225mm=0.225m

GB=75mm=0.075m

wOB=2πN60=2π20060=20.94rad/sw_{OB}=\frac{2πN}{60}=\frac{2π*200}{60}=20.94rad/s

VOB=OBwOB=0.0520.94=1.047m/sV_{OB}=OB*w_{OB}=0.05*20.94=1.047m/s

From velocity diagram,

VA=0.975m/sV_{A}=0.975m/s

bgba=BGBA\frac{bg}{ba}=\frac{BG}{BA}

bg0.75=75225\frac{bg}{0.75}=\frac{75}{225}

Hence, bg=0.25m/sbg=0.25m/s

Plot in velocity diagram to find VOG

VOG=0.975m/sV_{OG}=0.975m/s


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