1

Kinematics of Particles(Part 2)

Chapter 11

2

11.7 Graphical Solution of Rectilinear Motion Problem

• Given the x-t curve, the v-t curve is equal to the x-t curve slope.

• Given the v-t curve, the a-t curve is equal to the v-t curve slope.

3

• Given the a-t curve, the change in velocity between t1 and t2 is equal to the area under the a-t curve between t1 and t2.

• Given the v-t curve, the change in position between t1 and t2 is equal to the area under the v-t curve between t1 and t2.

4

11.8 Other Graphical Methods

• Moment-area method to determine particle position at time t directly from the a-t curve:

1

0

110

01 curve under areav

v

dvtttv

tvxx

using dv = a dt ,

1

0

11001

v

v

dtatttvxx

1

0

1

v

v

dtatt first moment of area under a-t curve with respect to t = t1 line.

Ct

tta-ttvxx

centroid of abscissa

curve under area 11001

5

• Method to determine particle acceleration from v-x curve:

BC

ABdx

dvva

tan

subnormal to v-x curve

6

Sample Problem 11.6

A subway car leaves station A; it gains speed at the rate of 1.2 m/s for 6 s and then at the rate of 1.8 m/s until it has reached the speed of 14.6 m/s.The car maintains the same speed until it approaches station B ; brakes are then applied, giving the car a constant deceleration and bringing it to a stop in 6 s.The total running time for A to B is 40 s.Draw the a-t, v-t , and x-t curves, and determine the distance stations A and B.

2

2

7

Solution

Acceleration Time Curve Since the acceleration is either constant or zero, the a-t curve is made of horizontal straight-linesegments. The values of t2 and a4 are determine as follows :

The acceleration being negative, the corresponding area is below the t-axisthis area represents a decrease in velocity.

8

Velocity Time Curve Since the acceleration is either constant or zero, the v-t curveis made of straight line segments connecting the points determined above.

Solution

Position Time Curve The points determined above should be joined by three arcs ofvparabola and one straight line segment. In constructing the x-t curve is equal to the value of v at that instant.

9

PROBLEMS

10

11

Solution

A1= -12m/s

A2= 8m/s

a=( m/s)

t(s)

2

6

10 14

-2

0

2

12

v=( m/s)

t(s)

4

-4

8

0

A3

A4

A5

A6

A7

12

6 104

14

16

12

4-8

-4

10 12 14

t(s)

x(m)

13

11.9 Position Vector, Velocity, and Acceleration

CURVILINEAR MOTION OF PARTICLES

• Particle moving along a curve other than a straight line is in curvilinear motion.

• Position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle.

• Consider particle which occupies position P defined by at time t and P’ defined by at t + t, r

r

dt

ds

t

sv

dt

rd

t

rv

t

t

0

0

lim

lim

instantaneous velocity (vector)

instantaneous speed (scalar)

14

dt

vd

t

va

t

0

lim

instantaneous acceleration (vector)

• Consider velocity of particle at time t and velocity at t + t,

v

v

• In general, acceleration vector is not tangent to particle path and velocity vector.

15

11.10 Derivatives of Vector Functions

uP

• Let be a vector function of scalar variable u,

u

uPuuP

u

P

du

Pd

uu

00limlim

• Derivative of vector sum,

du

Qd

du

Pd

du

QPd

du

PdfP

du

df

du

Pfd

• Derivative of product of scalar and vector functions,

• Derivative of scalar product and vector product,

du

QdPQ

du

Pd

du

QPd

du

QdPQ

du

Pd

du

QPd

16

• When position vector of particle P is given by its rectangular components,

kzjyixr

• Velocity vector,

kvjviv

kzjyixkdt

dzj

dt

dyi

dt

dxv

zyx

• Acceleration vector,

kajaia

kzjyixkdt

zdj

dt

ydi

dt

xda

zyx

2

2

2

2

2

2

11.11 Rectangular Components of Velocity And Acceleration

17

• Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile,

00 zagyaxa zyx

with initial conditions, 0,,0 000000 zyx vvvzyx

Integrating twice yields

0

02

21

00

00

zgtyvytvx

vgtvvvv

yx

zyyxx

• Motion in horizontal direction is uniform.

• Motion in vertical direction is uniformly accelerated.

• Motion of projectile could be replaced by two independent rectilinear motions.

18

11.12 Motion Relative to a Frame in Translation

• Designate one frame as the fixed frame of reference. All other frames not rigidly attached to the fixed reference frame are moving frames of reference.

• Position vectors for particles A and B with respect to the fixed frame of reference Oxyz are . and BA rr

• Vector joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and

ABr

ABAB rrr

• Differentiating twice,ABv

velocity of B relative to A.ABAB vvv

ABa

acceleration of B relative to A.

ABAB aaa

• Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A.

19

Sample Problem 11.7

A projectile is fired from the edge of a 150m cliff with an initial velocity of 180 m/s at an angle of 30ºwith the horizontal. Neglecting air resistance, find

(a) the horizontal distance from the gun to the point where the projectile strikes the ground.(b) the greatest elevation above the ground reached by the projectile.

20

Solution

The vertical and the horizontal motion will be considered separately.

Vertical Motion . Uniformly Acceleration MotionChoosing the positive sence of the y axis upward and placing the origin Oat the gun, we have

Substituting into the equations of uniformly accelerated motion, we have

Horizontal Motion. Uniform MotionChoosing the positive sense of the x axis to the right, we have

Substituting into the equation of uniform motion, we obtain

21

a. Horizontal Distance When the projectile strikes the ground, we have

Carrying this value into Eq. (2) for the vertical motion, we write

t = 19.91 s

Carrying t=19.91 s into Eq. (4) for the horizontal motion, we obtain

x = 3100 m

b. Greatest ElevationWhen the projectile reaches its greatest elevation, we have Vy=0 ;carrying this value into Eq. (3) for the vertical motion, we write

Greatest Elevation above ground = 150 m + 143 m = 563 m

22

Sample Problem 11.8

A projectile is fired with an initial velocity of 224 m/s at a target B located 610 m above the gun A and at a horizontal distance of 3658 m. Neglecting air resistance, determine the value of the firing angle α

23

Solution

The horizontal and the vertical motion will be considered separately.Horizontal motion . Placing the origin of the coordinate axes at the gun,

we have

Substituting into the equation of uniform horizontal motion, we obtain

The time required for the projectile to move through a horizontal distanceof 3658 m is obtained by setting x equal to 3658 m

Vertical Motion

Substituting into the equation of uniformly accelerated vertical motion,we obtain

t2

24

Solution

25

PROBLEMS

26

30º

27

28

29

50 m

3 m

13 m

h

Vo

30

50 m

3 m

13 m

h

Vo

31

50 m

3 m

13 m

h

Vo

32

33

34

35

11.13 Tangential and Normal Components

• Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components.

• are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, and

is the angle between them.

tt ee and

ttt eee

d

ede

eee

e

tn

nnt

t

2

2sinlimlim

2sin2

00

36

tevv• With the velocity vector expressed as

the particle acceleration may be written as

dt

ds

ds

d

d

edve

dt

dv

dt

edve

dt

dv

dt

vda tt

but

vdt

dsdsde

d

edn

t

After substituting,

22 va

dt

dvae

ve

dt

dva ntnt

• Tangential component of acceleration reflects change of speed and normal component reflects change of direction.

• Tangential component may be positive or negative. Normal component always points toward center of path curvature.

37

22 va

dt

dvae

ve

dt

dva ntnt

• Relations for tangential and normal acceleration also apply for particle moving along space curve.

• Plane containing tangential and normal unit vectors is called the osculating plane.

ntb eee

• Normal to the osculating plane is found from

binormale

normalprincipal e

b

n

• Acceleration has no component along binormal.

38

• When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP.

rr e

d

ede

d

ed

dt

de

dt

d

d

ed

dt

ed rr

dt

de

dt

d

d

ed

dt

edr

erer

edt

dre

dt

dr

dt

edre

dt

drer

dt

dv

r

rr

rr

• The particle velocity vector is

• Similarly, the particle acceleration vector is

errerr

dt

ed

dt

dre

dt

dre

dt

d

dt

dr

dt

ed

dt

dre

dt

rd

edt

dre

dt

dr

dt

da

r

rr

r

22

2

2

2

2

rerr

11.14 Radial and Transverse Components

39

• When particle position is given in cylindrical coordinates, it is convenient to express the velocity and acceleration vectors using the unit vectors . and ,, keeR

• Position vector,

kzeRr R

• Velocity vector,

kzeReRdt

rdv R

• Acceleration vector,

kzeRReRRdt

vda R

22

40

Sample Problem 11.10

A motorist is traveling on a curved section of highway of radius 762 m at the speed of 96 km/h.The motorist suddenly applies the brake, causing the automobile to slow down at a constant rate.knowing that after 8s the speed has been reduced to 72 km/h, determine the acceleration of the automobile immediately after the brakes have been applied.

VA = 96 km/h

762 m

41

Tangential Component of Acceleration. First the speeds are expressed in m/s

VA = 96 km/h

762 m

m/s7.26s

m

3600

1000x96

h

km96

m/s20s

m

3600

1000x72

h

km72

Since the automobile slows down at a constant rate, we have

2/8375.08

/7.26/20sm

s

smsm

t

vaaveragea tt

Normal component of acceleration. Immediately after the brakes have beenapplied, the speed is still 26.7 m/s, and we have

22

m/s94.0m762

m/s7.26

v

an

2m/s8375.0ta

2m/s94.0na

42

VA = 96 km/h

762 m

2m/s8375.0ta

2m/s94.0na

Magnitude and Direction of Acceleration.The magnitude and direction of the resultant a of the components an and at are

1223.1/8375.0

/94.0tan

2

2

sm

sm

a

a

t

n

3.48

22

/26.13.48sin

/94.0

sinsm

smaa n

43

Sample Problem 11.12

The rotation of the 0.9m arm about OA about O is defined by = 0.15t2 where is expressed in radians and t in seconds. Collar B slides along the arm in such a way that its distance from O is r = 0.9 - 0.12t2 where r is expressed in meters and t in seconds.

After the arm OA has rotated through 30o, determine

(a) the total velocity of the collar,

(b) the total acceleration of the collar,

(c) the relative acceleration of the collar with respect to the arm.

44

Time t at which θ=30º. Substituting θ=30º =0.524 rad into the expression for θ,we obtain

Equation of Motion. Substituting t=1.869 s in the expressions for r,θ,and their first and second derivatives, we have

sttt 869.115.0524.015.0 22

2

2

/240.024.0

/449.024.0

481.012.09.0

smr

smtr

mtr

2

2

/300.030.0

/561.030.0

524.015.0

srad

sradt

radt

a. Velocity of B We obtain the values of vr and vθwhen t = 1.869 s

smrv

smrvr

/270.0)561.0(481.0

/449.0

22 )()( vvv r

0.31/524.0 smv

45

b. Acceleration of B. We obtain

2

rrar22 /391.0)561.0(481.0240.0 sm

rra 22/359.0)561.0)(449.0(2)300.0(481.0 sm

c. Acceleration of B with Respect to Arm OA.. We note that the motionof the collar with respect to the arm is rectilinear and defined by the coordinate r. We write

Otowardssma

smra

OAB

OAB

2/

2/

/240.0

/240.0

46

PROBLEMS

47

Problem 11.136

At the instant shown, race car A is passing race car B with a relative velocity of 1m/s. Knowing that the speedsof both cars are constant and that the relative acceleration of car A with respect to car B is 0.25 m/s directedtoward the center of curvature, determine

(a) the speed of car A(b) the speed of car B

2

48

Bv AvBa

Aa

49

50

51

Problem 11.142

137.2 m

91.4 m

52

53

54

55

56

57

58

59

THE END