Question

In a slider crank mechanism, the crank is 480 mm long

and rotates at 20 rad/s in the counter-clockwise direction.

The length of the connecting rod is 1.6 m. When the crank

turns 60° from the inner dead center; determine the

velocity of the slider, angular velocity of connecting rod

and velocity of a point P located at a distance 450 mm on

the connecting rod extended.​

Answer 1

Answer:

The answer is given below

Answer 2

Given :

OA = 480mm

AB = 1600mm

ω = 20rad/sec

To find :

i) velocity of the slider B

ii) velocity of a point E located at a distance 450mm on the connecting rod extended.

Solution :

For any distance

x =  rcosθ + lcosφ

∴ velocity,

x' = -rω[sinθ + λ/2sin2θ]

where, λ = r / l

Therefore,

velocity of the B,

x'= -rω[sinθ +λ/2sin2θ]

x' = -0.48 × 20[sin60° + 480/1600sin120°]

x' = 9.56m/sec

(-ve sgin indicates the velocity is in left direction[-x])

lsinφ = rcosθ

∴ φ = 8.626°

From velocity triangle,

oa = 0.45 × 20 = 9.6m/s

The vector equation is

Vbo = Vba + Vao

Vbg = Vao + Vba

or, gb = oa + ab

Take the vector Vao to convenient scale in the proper direction and sense (velocity diagram)

Locate the point e on ba extended such that

ae / ba =  AE / BA

ba = 5.25m/sec (measuring from velocity diagram)

∴ ae = 5.25 × 0.45/1.60 = 1.48m/sec

∴ Ve = oe = 10.2m/sec (Measured from diagram)