Relative Motion

Lecture VI

Relative Motion

In previous lectures, the particles motion have been described using coordinates referred to fixed reference axes. This kind of motion analysis is called absolute motion analysis.

Not always easy to describe or measure motion by using fixed set of axes.

The motion analysis of many engineering problems is sometime simplified by using measurements made with respect to moving reference system.

Combining these measurements with the absolute motion of the moving coordinate system, enable us to determine the absolute motion required. This approach is called relative motion analysis.

vB

vA

vA/B

Relative Motion (Cont.)-The motion of the moving coordinate system is specified w.r.t. a fixed coordinate system.

-The moving coordinate system should be nonrotating (translating or parallel to the fixed system).

-A/B is read as the motion of A relative to B (or w.r.t. B).

-The relative motion terms can be expressed in whatever coordinate system (rectangular, polar, n-t).

Fixed system

Moving system

Moving system

Fixed system

BABA

BABA

BABA

aaa

vvv

rrr

ABAB

ABAB

ABAB

aaa

vvv

rrr

Note: rB/A = -rA/B

vB/A = -vA/B

aB/A = -aA/B

Path

Path

Path

Path

Note: rA & rB are measured from the origin of the fixed axes X-Y.

Note: In relative motion analysis, acceleration of a particle observed in a translating system x-y is the same as observed in a fixed system X-Y, when the moving system has a constant velocity.

Relative Motion Exercises

Exercise # 1

2/186: The passenger aircraft B is flying east with a velocity vB = 800 km/h. A military jet traveling south with a velocity vA = 1200 km/h passes under B at a slightly lower altitude. What velocity does A appear to have to a passenger in B, and what is the direction of that apparent velocity?.

Exercise # 2

At the instant shown, race car A is passing race car Bwith a relative velocity of 1 m/s. Knowing that the speeds of both cars are constant and that the relative acceleration of car A with respect to car B is 0.25 m/s2 directed toward the center of curvature, determine (a) the speed of car A, (b) the speed of car B.

Exercise # 3

At the instant shown, cars A and B are traveling with speeds

of 18 m/s and 12 m/s, respectively. Also at this instant, car A

has a decrease in speed of 2 m/s2, and B has an increase in

speed of 3 m/s2. Determine the velocity and acceleration of

car B with respect to car A.

Exercise # 4

Instruments in airplane A indicate that with respect to the air the plane is headed north of east with an airspeed of 480 km/h. At the same time radar on ship B indicates that the relative velocity of the plane with respect to the ship is 416 km/h in the direction north of east. Knowing that the ship is steaming due south at 20 km/h, determine (a) the velocity of the airplane, (b) the wind speed and direction.

Constrained Motion

Lecture VII

Constrained Motion

Here, motions of more than one particle are interrelated because of the constraints imposed by the interconnecting members.

In such problems, it is necessary to account for these constraints in order to determine the respective motions of the particles.

Constrained Motion (Cont.)

Notes:

-Horizontal motion of A is twice the vertical motion of B.

-The motion of B is the same as that of the center of its pulley, so we establish position coordinates x and y measured from a convenient fixed datum.

-The system is one degree of freedom, since only one variable, either x or y, is needed to specify the positions of all parts of the system.

bryr

xL 12 2

2

One Degree of Freedom System

L, r1, r2, and b are constants

Differentiating once and twice gives:

BA

BA

aayx

vvyx

20or 20

20or 20

Datum

+

+

Constrained Motion (Cont.)

Note:

-The positions of the lower pulley C depend on the separate specifications of the two coordinates yA & yB.

-It is impossible for the signs of all three terms to be +ve simultaneously.

constant

constant2

DCCBB

DAA

yyyyL

yyL

Two Degree of Freedom System

Differentiating once gives:

DCBDA

DCBDA

yyyyy

yyyyy

20 and 20

20 and 20

Differentiating once gives:

042 and 042

042 and 042

CBACBA

CBACBA

aaayyy

vvvyyy

Eliminating the terms in gives:

DD yy and

24

BAC

dydydy

Datum Datum++

++

Constrained Motion Exercises

Exercise # 1

2/208: Cylinder B has a downward velocity of 0.6 m/s and an upward acceleration of 0.15 m/s2. Calculate the velocity and acceleration of block A .

Exercise # 2

2/210: Cylinder B has a downward velocity in meters per second

given by vB = t2/2 + t3/6, where t is in seconds. Calculate the

acceleration of A when t = 2 s.

Exercise # 3

2/211: Determine the vertical rise h of the load W during 5

seconds if the hoisting drum wraps cable around it at the constant

rate of 320 mm/s.

Exercise # 4

Block C moves downward with a constant velocity of 2 ft/s. Determine (a) the velocity of block A, (b) the velocity of block