Engineering Mechanics

Chapter 22 : Combined Motion of Rotation and Translation „„„„„ 465

We also know that velocity of B,

vB=ω 1 × AB ...(iii)

Dividing equation (i) by (ii),

2

2

C

B

v OC OC AN

vOBOBAB

ω×

===

ω×

or CB 1

ANAN

vv AB

ABAB

=× =ω× ×

11

AN

rAN

r

=ω × × =ω × ...(iv)

Now from the geometry of the figure, we find that

CA=CM + MA = CB cos φ + AB cos θ = l cos φ + r cos θ

and AN=CA tan φ = (l cos φ + r cos θ ) tan φ

=l cos φ tan φ + r cos θ tan φ
=l sin φ + r cos θ tan φ
Substituting this value of AN in equation (iv) velocity of C,
vC=ω 1 (l sin φ + r cos θ tan φ)
Example 22.4. In a crank and connecting rod mechanism, the radius of crank and length of
the connecting rod are 300 mm and 1200 mm respectively. The crank is rotating at 180 r.p.m. Find
the velocity of the piston, when the crank is at an angle of 45°, with the horizontal.

Solution. Given : Radius of the crank (r) = 300 mm = 0.3 m ; Length of connecting rod (l) = 1200
mm = 1.2 m ; Angular rotation of crank (N) = 180 r.p.m and angle traversed by the crank (θ ) = 45°

Fig. 22.12.

We know that Angular velocity of the crank,

1

2180

6rad/s

60

π×

ω= = π

From the geometry of the figure, we find that

sin 45 0.3 0.707

sin

1.2

BM AB

BC BC

°×

φ= = =

= 0.1768 or φ = 10.18°

We also know that velocity of the piston,

vC=ω 1 (l sin φ + r cos θ tan φ)

=6π (1.2 sin 10.18° + 0.3 cos 45° tan 10.18°) m/s

=6π [(1.2 × 0.1768) + (0.3 × 0.707 × 0.1796)] m/s

= 4.72 m/s Ans.