Dynamics Lecture5 Dependent Motion of Two Bodies and Relative Motion Using Translating Axes

©2007 Pearson Education South Asia Pte Ltd

Absolute Dependent Motion Analysis of Two Particles

• Motion of one particle will depend on the corresponding motion of another particle• Dependency occurs when particles are interconnected by the inextensible cords which are wrapped around pulleys• For example, the movement of block A downward along the inclined plane will cause a corresponding movement of block B up the other incline• Specify the locations of the blocks using position coordinate sA and sB


• Note each of the coordinate axes is (1) referenced from a fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of block A and block B and (3) has a positive sense from C to A and D to B• If total cord length is lT, the position coordinate are related by the equation

TBCDA lsls



• Here lCD is the length passing over arc CD• Taking time derivative of this expression, realizing that lCD and lT remain constant, while sA and sB measure the lengths of the changing segments of the cord

ABBA vvdtds

dtds 0 or

• The negative sign indicates that when block A has a velocity downward in the direction of position sA, it causes a corresponding upward velocity of block B; B moving in the negative sB direction



• Time differentiation of the velocities yields the relation between accelerations

aB = - aA

• For example involving dependent motion of two blocks• Position of block A is specified by sA, and the position of the end of the cord which block B is suspended is defined by sB



• Choose coordinate axes which are (1) referenced from fixed points and datums, (2) measured in the direction of motion of each block, (3) positive to the right (sA) and positive downward (sB)• During the motion, the red colored segments of the cord remain constant• If l represents the total length of the cord minus these segments, then the position coordinates can be related by

lshs AB 2



Since l and h are constant during the motion, the two time derivatives yields

ABAB aavv 22

• When B moves downward (+sB), A moves to left (-sA) with two times the motion• This example can also be worked by defining the position of block B from the center of the bottom pulley ( a fixed point)



ABAB

AB

aavv

lshsh

22

)(2

Time differentiation yields



PROCEDURE FOR ANALYSISPosition-Coordinate Equation• Establish position coordinates which have their origin located at a fixed point or datum• The coordinates are directed along the path of motion and extend to a point having the same motion as each of the particles• It is not necessary that the origin be the same for each of the coordinates; however, it is important that each coordinate axis selected be directed along the path of motion of the particle



• Using geometry or trigonometry, relate the coordinates to the total length of the cord, lT, or to that portion of cord, l, which excludes the segments that do not change length as the particles move – such as arc segments wrapped over pulleys• For problem involving a system of two or more cords wrapped over pulleys, then the position of a point on one cord must be related to the position of a point on another cord using the above procedure• Separate equations must be written for a fixed length of each cord of the system.



Time Derivatives• Two successive time derivatives of the position-coordinates equations yield the required velocity and acceleration equations which relate motions of the particles• The signs of the terms in these equations will be consistent with those that specify the positive and negative sense of the position coordinates



EXAMPLE 12.21

Determine the speed of block A if block B has an upward speed of 2 m/s.


Position Coordinate System. There is one cord in this system having segments which are changing length. Position coordinates sA and sB will be used since each is measured from a fixed point (C or D) and extends along each block’s path of motion. In particular, sB is directed to point E since motion of B and E is the same. The red colored segments of the cords remain at a constant length and do not have to be considered as the block move.

EXAMPLE 12.21


The remaining length of the cord, l, is also considered and is related to the changing position coordinates sA and sB by the equation

lss BA 3

Time Derivative. Taking the time derivative yields

03 BA vv

so that when vB = -2m/s (upward)vA = 6m/s ↓

EXAMPLE 12.21


Determine the speed of block A if block B has an upward speed of 2m/s.

EXAMPLE 12.22


Position Coordinate Equation. Positions of A and B are defined using coordinates sA and sB.Since the system has two cords which change length, it is necessary to use a third coordinate sC in order to relate sA to sB. Length of the cords can be expressed in terms of sA and sC, and the length of the other cord can be expressed in terms of sB and sC. The red colored segments are not considered in this analysis.

EXAMPLE 12.22


For the remaining cord length,

21 )(2 lssslss CBBCA

Eliminating sC yields,

1224 llss BA

Time Derivative. The time derivative gives04 BA vv

so that vB = -2m/s (upward)

smsmvB /8/8

EXAMPLE 12.22


Determine the speed with which block B rises if the end of the cord at A is pulled down with a speed of 2m/s.

EXAMPLE 12.23


Position-Coordinate Equation. The position of A is defined by sA, and the position of block B is specified by sB since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fixed pin at pulley D. Since the system consists of two cords, the coordinates sA and sB cannot be related directly. By establishing a third position coordinate, sC, and the length of the other cord in terms of sA, sB and sC.

EXAMPLE 12.23


Excluding the red colored segments of the cords, the remaining constant cord lengths l1 and l2 (along the hook and link dimensions) can be expressed as

12 24 llss BC

2

1

lsssss

lss

BCBCA

BC

Eliminating sC yields

EXAMPLE 12.23


Time Derivative. The time derivative give

smsmv

vv

B

BA

/5.0/5.0

04

when vA = 2m/s (downward)

EXAMPLE 12.23


A man at A is hoisting a safe S by walking to the right with a constant velocity vA = 0.5m/s. Determine the velocity and acceleration of the safe when it reaches the elevation at E. The rope is 30m long and passes over a small pulley at D.

EXAMPLE 12.24


Position Coordinate System. Rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the position of S and A, are specified by means of the x and y coordinates measured from a fixed point and directed along the paths of motion of the ends of the rope. The x and y coordinates may be related since the rope has a fixed length l = 30m, which at all times is equal to the length of the segment DA plus CD.

EXAMPLE 12.24

View Free Body Diagram


Using Pythagorean Theorem and the equation for lCD to get the relationship between y and x:

)2.(15)1.(15 22 eqyleqxl CDDA

15225

151530

)3.(

2

22

xy

yx

eqlll CDDA

(eq.4)

EXAMPLE 12.24


Time Derivative. Taking time derivative, using the chain rule where, vS = dy/dt and vA = dx/dt

A

S

vx

x

dtdx

x

xdtdy

v

2

2

225

225

221

At y = 10 m, x = 20 m, vA = 0.5 m/s, vS = 400mm/s ↑

(2)

EXAMPLE 12.24


The acceleration is determined by taking the time derivative of eqn (2),

2/32

2

2

22/322

2

225

225

225

1

225

1

)225(

)/(

x

vdtdvx

x

vdtdx

xxv

x

dtdxx

dt

yda

AA

AAS

At x = 20 m, with vA = 0.5 m/s,

2/6.3 smmaS

EXAMPLE 12.24


Relative Motion Analysis of Two Particles Using Translating Axes

• There are many cases where the path of the motion for a particle is complicated, so that it may be feasible to analyze the motions in parts by using two or more frames of reference• For example, motion of an particle located at the tip of an airplane propeller while the plane is in flight, is more easily described if one observes first the motion of the airplane from a fixed reference and then superimposes (vectorially) the circular motion of the particle measured from a reference attached to the airplane


Position.• Consider particle A and B, which moves along the arbitrary paths aa and bb, respectively• The absolute position of each particle rA and rB, is measured from the common origin O of the fixed x, y, z reference frame



• Origin of the second frame of reference x’, y’ and z’ is attached to and moves with particle A• Axes of this frame only permitted to translate relative to fixed frame• Relative position of “B with respect to A” is designated by a relative-position vector rB/A

• Using vector addition

ABAB rrr /



Velocity.• By time derivatives,

• Here refer to absolute velocities, since they are observed from the fixed frame• Relative velocity is observed from the translating frame

ABAB vvv /

dtrdvanddtrdv AABB //

dtrdv ABAB ///



• Since the x’, y’ and z’ axes translate, the components of rB/A will not change direction and therefore time derivative o this vector components will only have to account for the change in the vector magnitude• Velocity of B is equal to the velocity of A plus (vectorially) the relative velocity of “B relative to A” as measured by the translating observer fixed in the x’, y’ and z’ reference



Acceleration.• The time derivative yields a similar relationship between the absolute and relative accelerations of the particles A and B

• Here aB/A is the acceleration of B as seen by the observer located at A and translating with the x’, y’ and z’ reference frame

ABAB aaa /



PROCEDURE FOR ANALYSIS• When applying the relative position equations, rB = rA + rB/A, it is necessary to specify the location of the fixed x, y , z and translating x’, y’ and z’• Usually, the origin A of the translating axes is located at a point having a known position rA

• A graphical representation of the vector addition can be shown, and both the known and unknown quantities labeled on this sketch



• Since vector addition forms a triangle, there can be at most two unknowns, represented by the magnitudes and/or directions of the vector quantities• These unknown can be solved for either graphically, using trigonometry, or resolving each of the three vectors rA, rB and rB/A into rectangular or Cartesian components, thereby generating a set of scalar equations



• The relative motion equations vB = vA + vB/A and aB = aA + aB/A are applied in the same manner as explained above, except in this case, origin O of the fixed axes x, y, z axes does not have to be specified



A train, traveling at a constant speed of 90km/h, crosses over a road. If automobile A is traveling t 67.5km/h along the road, determine the magnitude and direction of relative velocity of the train with respect to the automobile

EXAMPLE 12.25


Vector Analysis. The relative velocity is measured from the translating x’, y’ axes attached to the automobile. Since vT and vA are known in both magnitude and direction, the unknowns become the x and y components of vT/A. Using the x, y axes and a Cartesian vector analysis,

hkmjiv

vjii

vvv

AT

AT

ATAT

/)~7.47~3.42{

)~45sin5.67~45cos5.67(~90

/

/

/

EXAMPLE 12.25


The magnitude of vT/A is

hkmv AT /8.63)7.473.42( 22/

The direction of vT/A defined from the x axis is

40.48

3.427.47

tan/

/

xAT

yAT

v

v

EXAMPLE 12.25


Plane A is flying along a straight-line path, while plane B is flying along a circular path having a radius of curvature of ρB = 400 km. Determine the velocity and acceleration of B as measured by the pilot of A.

EXAMPLE 12.26


Velocity. The x, y axes are located at an arbitrary fixed point. Since the motion relative to plane A is to be determined, the translating frame of reference x’. y’ is attached to it. Applying the relative-velocity equation in scalar form since the velocity vectors of both plane are parallel at the instant shown,

hkmhkmv

v

vvv

AB

AB

ABAB

/100/100

700600

/

/

/)(

EXAMPLE 12.26


Acceleration. Plane B has both tangential and normal components of acceleration, since it is flying along a curved path. Magnitude of normal acceleration,

22

/900 hkmv

a BnB

Applying the relative-acceleration equation,

2/

/

/

/~150~900

~50~100~900

hkmjia

ajji

aaa

AB

AB

ABAB

EXAMPLE 12.26


From the figure shown, the magnitude and direction of ABa /

46.9900150

tan/912 12/ hkma AB

EXAMPLE 12.26


At the instant, car A and B are traveling with the speed of 18 m/s and 12 m/s respectively. Also at this instant, 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 B with respect to A.

EXAMPLE 12.27


Velocity. The fixed x, y axes are established at a point on the ground and the translating x’, y’ axes are attached to car A. Using Cartesian vector analysis,

smv

smjiv

vjij

vvv

AB

AB

AB

ABAB

/69.9588.39

/~588.3~9

~60sin18~60cos18~12

22/

/

/

/

Thus,

EXAMPLE 12.27

View Free Body Diagram


Its direction is

7.21

9588.3

tan/

/

xAB

yAB

v

v

Acceleration. The magnitude of the normal component is

22

/440.1 smv

a BnB

EXAMPLE 12.27


Applying the equation for relative acceleration yields

2

/

/

/

/~

732.4~

440.2

~60sin2

~60cos2

~3

~440.1

smjia

ajiji

aaa

AB

AB

ABAB

Magnitude and direction is

7.62

/32.5 2/

sma AB

EXAMPLE 12.27

Dynamics Lecture5 Dependent Motion of Two Bodies and Relative Motion Using Translating Axes

Documents

Transcript of Dynamics Lecture5 Dependent Motion of Two Bodies and Relative Motion Using Translating Axes