Answer to Question #198985 in Mechanical Engineering for Muhammad Idrees

Question

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]

Answer 1

Given:

Angular velocity of crank "OB"

$W_{BO}=\frac{2\pi N}{60}=\frac{2\pi \times600}{60}=62.83rad/s$

Velocity of crank

$V_B=W_{BO} \times BO=62.83\times0.125$

$=7.25m/s$

$\therefore$ velocity of slider A

$VA=ao\times scale$

$=3.35\times 2=6.70m/s$

Location of point 'G' on velocity

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

$\therefore$ velocity of point 'G'

$V_G=og\times scale =3.4\times 2$

$=6.8 m/s$