Question #198985

In a slider crank mechanism, the length of crank OB and connecting rod AB are 125 mm and 500 mm respectively. The centre of gravity G of the connecting rod is 275 mm from the slider A. The crank speed is 600 r.p.m. clockwise. When the crank has turned 45° from the inner dead centre position, determine:

1. velocity of the slider A,

2. velocity of the point G, and

3. angular velocity of the connecting rod AB. Use relative velocity method.

[Ans. 6.45 m/s ; 6.75 m/s ; 10.8 rad/s]



1
Expert's answer
2021-05-27T15:25:01-0400

Given:



Angular velocity of crank "OB"

WBO=2πN60=2π×60060=62.83rad/sW_{BO}=\frac{2\pi N}{60}=\frac{2\pi \times600}{60}=62.83rad/s

Velocity of crank

VB=WBO×BO=62.83×0.125V_B=W_{BO} \times BO=62.83\times0.125

=7.25m/s=7.25m/s

\therefore velocity of slider A

VA=ao×scaleVA=ao\times scale

=3.35×2=6.70m/s=3.35\times 2=6.70m/s

Location of point 'G' on velocity

ag=AGAB×ab=1.57cm=\frac{AG}{AB} \times ab=1.57cm

\therefore velocity of point 'G'

VG=og×scale=3.4×2V_G=og\times scale =3.4\times 2

=6.8m/s=6.8 m/s


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