Engineering Mechanics

Chapter 22 : Combined Motion of Rotation and Translation „„„„„ 459

1. First of all, draw the initial and final positions of the rigid link as AB and A 1 B 1 respectively.


2. Join AA 1 and BB 1.


3. Draw PQ right bisector of AA 1 ,and RS right bisector of BB 1.


4. Extend the two right bisectors PQ and RS to meet at O, which is the required centre of
   rotation or instantaneous centre as shown in Fig. 22.2.
   Let ω = Angular velocity of the rod AB about A.
   ∴ Linear velocity of point A.

vA = ω × OA or vA

OA

ω= ...(i)

Similarly, vB

OB

ω= ...(ii)

Equating equations (i) and (ii),

B

A

v OB

vOA

=

The direction of vA and vB will be at right angles to OA and OB respectively as shown in
Fig. 22.2.

Note. Sometimes, three or more members are hinged together to form a triangle, a quadrilateral
etc., In such a case , each member will have its own instantaneous centre. The instantaneous centre for
each member may be found out by drawing perpendiculars at the directions of motions at the two
ends of the member as usual.

Example 22.1. A link AB is moving in a vertical plane. At a certain instant, when the link is
inclined at 60° to the horizontal, the point A is moving horizontally at 2 m/s, while B is moving in a
vertical direction. Find the velocity of B.

Solution. Given : Inclination of the link with horizontal = 60° and
velocity of point A in horizontal direction (vA) = 2 m/s

Let vB = Velocity of B in the vertical direction.
First of all, let us locate the position of instantaneous centre O, graphically,
as shown in Fig. 22.3 and as discussed below :

1. First of all draw the position of the link AB, such that it is inclined at
   an angle of 60° with the horizontal.


2. Now draw the lines indicating the directions of motions of points A
   (in horizontal direction) and B (in vertical direction).


3. Now draw perpendiculars at A and B on the directions of motion vA
   and vB.


4. Let these perpendiculars meet at O, which is the instantaneous centre of the link AB.
   From the geometry of velocity of B, the triangle AOB, we find that

cot 60 0.577

OB

OA

=°=

We know that

0.577

B

A

v OB

vOA

==

∴ vB = vA × 0.577 = 2 × 0.577 = 1.15 m/s Ans.

Fig. 22.3.