Lecture VI Relative Motion
Relative Motion vB vA
vA/B
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.
Relative Motion (Cont.) Path
Path
-The motion of the moving coordinate system is specified w.r.t. a fixed coordinate system. Moving system
-A/B is read as the motion of A relative to B (or w.r.t. B).
Moving system
Path Path Fixed system
-The moving coordinate system should be nonrotating (translating or parallel to the fixed system).
-The relative motion terms can be expressed in whatever coordinate system (rectangular, polar, n-t).
Fixed system
r A rB r A B
rB rA rB A
v A vB v A B
vB v A vB A
aA aB aA B
aB aA aB A
Note: rA & rB are measured from the origin of the fixed axes X-Y. Note: rB/A = -rA/B vB/A = -vA/B aB/A = -aA/B
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 B with 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.
Lecture VII Constrained Motion
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.) +
Datum +
One Degree of Freedom System
Notes:
L x
r2 2
2 y r1 b L, r1, r2, and b are constants
Differentiating once and twice gives:
0 x 2 y 0 x 2 y
or
0 v A 2vB
or
0 a A 2a B
-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.
Constrained Motion (Cont.) Datum
Datum
+
+ +
+
Two Degree of Freedom System Note:
LA y A 2 y D constant
LB y B yC yC y D constant Differentiating once gives:
0 y A 2 y D 0 yA 2 yD
and and
0 y B 2 y C y D 0 yB 2 yC yD
Eliminating the terms in y D and yD gives:
y A 2 y B 4 y C 0 yA 2 yB 4 yC 0
and
v A 2vB 4vC 0
and
a A 2aB 4aC 0
-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.
dyC
dy A dy B 4 2
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
