11 Lecture ppt

VECTOR MECHANICS FOR ENGINEERS: DYNAMICSDYNAMICS

Tenth Tenth Edition in Edition in SI UnitsSI Units

Ferdinand P. BeerFerdinand P. BeerE. Russell Johnston, Jr.E. Russell Johnston, Jr.Phillip J. CornwellPhillip J. CornwellLecture Notes:Lecture Notes:Brian P. SelfBrian P. SelfCalifornia Polytechnic State UniversityCalifornia Polytechnic State University

Sanjeev SanghiIndian Institute of Technology, Delhi

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

11 Kinematics of

Particles


Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics

Tenth Editionin SI Units Contents

11 - 2

IntroductionRectilinear Motion: Position, Velocity & Acceleration

Determination of the Motion of a Particle

Sample Problem 11.2Sample Problem 11.3Uniform Rectilinear-Motion

Uniformly Accelerated Rectilinear-Motion

Motion of Several Particles: Relative Motion

Sample Problem 11.4Motion of Several Particles: Dependent Motion

Sample Problem 11.5Graphical Solution of Rectilinear-Motion Problems

Other Graphical MethodsCurvilinear Motion: Position, Velocity & Acceleration

Derivatives of Vector Functions

Rectangular Components of Velocity and Acceleration

Motion Relative to a Frame in Translation

Tangential and Normal Components

Radial and Transverse Components

Sample Problem 11.10Sample Problem 11.12



Tenth Editionin SI Units

11 - 3

Kinematic relationships are used to help us determine the trajectory of a golf ball, the orbital speed of a satellite, and the accelerations during acrobatic flying.



Tenth Editionin SI Units Introduction

11 - 4

•Dynamics includes:

Kinetics: study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion.

Kinematics: study of the geometry of motion. Relates displacement, velocity, acceleration, and time without reference to the cause of motion.Fthrust

Flift

Fdrag



Tenth Editionin SI Units Introduction

11 - 5

•Particle kinetics includes:

•Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line.

•Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions.




Rectilinear Motion: Position, Velocity & Acceleration

11 - 6

•Rectilinear motion: particle moving along a straight line•Position coordinate: defined by positive or negative distance from a fixed origin on the line.

•The motion of a particle is known if the position coordinate for particle is known for every value of time t.

•May be expressed in the form of a function, e.g.,

326 ttx or in the form of a graph x vs. t.





11 - 7

• Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed.

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

txv

tx

t

0lim

Average velocity

Instantaneous velocity

• From the definition of a derivative,

dtdx

txv

t

0lim

e.g.,

2

32

312

6

ttdtdxv

ttx





11 - 8

• Consider particle with velocity v at time t and v’ at t+t, Instantaneous

acceleration tva

t

0lim

tdtdva

ttvdt

xddtdv

tva

t

612

312e.g.

lim

222

0

• From the definition of a derivative,

• Instantaneous acceleration may be:- positive: increasing positive velocity

or decreasing negative velocity- negative: decreasing

positive velocityor increasing negative velocity.



Tenth Editionin SI Units Concept Quiz

11 - 9

What is true about the kinematics of a particle?

a) The velocity of a particle is always positive

b) The velocity of a particle is equal to the slope of the position-time graph

c) If the position of a particle is zero, then the velocity must zero

d) If the velocity of a particle is zero, then its acceleration must be zero





11 - 10

• From our example,326 ttx

2312 ttdtdxv

tdt

xddtdva 6122

2

- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0

- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2

• What are x, v, and a at t = 2 s ?

• Note that vmax occurs when a=0, and that the slope of the velocity curve is zero at this point.

• What are x, v, and a at t = 4 s ?




Determination of the Motion of a Particle

11 - 11

•We often determine accelerations from the forces applied (kinetics will be covered later)

•Generally have three classes of motion- acceleration given as a function of time,

a = f(t)- acceleration given as a function of

position, a = f(x)- acceleration given as a function of

velocity, a = f(v)

• Can you think of a physical example of when force is a function of position?When force is a function

of velocity?

a spring drag




Acceleration as a function of time, position, or velocity

11 - 12

a a t 0 0

v t

v

dv a t dt ()dv a tdt

vdv a x dx

0 0

v x

v x

vdv a x dx a a x and dx dvdt a

v dt

dvv a vdx

0 0

v t

v

dv dta v

0 0

x v

x v

vdvdxa v

a a v

( )dv a vdt

If…. Kinematic relationship Integrate



Tenth Editionin SI Units Sample Problem 11.2

11 - 13

Determine:• velocity and elevation above ground at time t,

• highest elevation reached by ball and corresponding time, and

• time when ball will hit the ground and corresponding velocity.

Ball tossed with 10 m/s vertical velocity from window 20 m above ground.

SOLUTION:• Integrate twice to find

v(t) and y(t).• Solve for t when velocity equals zero (time for maximum elevation) and evaluate corresponding altitude. • Solve for t when altitude equals zero (time for ground impact) and evaluate corresponding velocity.




11 - 14

tvtvdtdv

adtdv

ttv

v81.981.9

sm81.9

00

2

0

ttv

2s

m81.9sm10

2

210

081.91081.910

81.910

0

ttytydttdy

tvdtdy

tty

y

22sm905.4s

m10m20 ttty

SOLUTION:• Integrate twice to find v(t) and y(t).




11 - 15

• Solve for t when velocity equals zero and evaluate corresponding altitude. 0

sm81.9s

m10 2

ttv

s019.1t

• Solve for t when altitude equals zero and evaluate corresponding velocity.

22

22

s019.1sm905.4s019.1s

m10m20

sm905.4s

m10m20

y

ttty

m1.25y




11 - 16

• Solve for t when altitude equals zero and evaluate corresponding velocity.

0sm905.4s

m10m20 22

ttty

s28.3

smeaningles s243.1

tt

s28.3sm81.9s

m10s28.3

sm81.9s

m10

2

2

v

ttv

sm2.22v




11 - 17

Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil. As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity. Determine v(t), x(t), and v(x).

kva

SOLUTION:• Integrate a = dv/dt = -kv to find v(t).• Integrate v(t) = dx/dt to find x(t).• Integrate a = v dv/dx = -kv to find v(x).




11 - 18

SOLUTION:• Integrate a = dv/dt = -kv to find v(t).

0 00

lnv t

v

v tdv dva kv k dt ktdt v v

ktevtv 0

• Integrate v(t) = dx/dt to find x(t).

0

0 000 0

1

kt

tx tkt kt

dxv t v edt

dx v e dt x t v ek

ktekvtx 10




11 - 19

• Integrate a = v dv/dx = -kv to find v(x).

kxvv

dxkdvdxkdvkvdxdvva

xv

v

0

00

kxvv 0

• Alternatively,

00 1

vtv

kvtx

kxvv 0

0

0 or v

tveevtv ktkt

ktekvtx 10wit

handthen



Tenth Editionin SI Units Group Problem Solving

11 - 20

A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 5 m/s. Assuming the ball experiences a downward acceleration of a =10 - 0.01v2 when in the water, determine the velocity of the ball when it strikes the bottom of the lake.

Which integral should you choose?

0 0

v t

v

dv a t dt 0 0

v x

v x

vdv a x dx

0 0

v t

v

dv dta v

0 0

x v

x v

vdvdxa v

(a)

(b)

(c)

(d)

+y



Tenth Editionin SI Units Concept Question

11 - 21

When will the bowling ball start slowing down?

A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 5 m/s. Assuming the ball experiences a downward acceleration of a =10 - 0.01v2 when in the water, determine the velocity of the ball when it strikes the bottom of the lake. The velocity would have to be high enough for the 0.01 v2 term to be bigger than 10

+y




11 - 22

SOLUTION:• Determine the proper kinematic relationship to apply (is acceleration a function of time, velocity, or position?• Determine the total distance the car travels in one-half lap• Integrate to determine the velocity after one-half lap

The car starts from rest and accelerates according to the relationship

23 0.001a v It travels around a circular track that has a radius of 200 meters. Calculate the velocity of the car after it has travelled halfway around the track. What is the car’s maximum possible speed?




11 - 23

200 m

dvv a vdx

0 0

x v

x v

vdvdxa v

Choose the proper kinematic relationship

Given: 23 0.001a v vo = 0, r = 200 m

Find: v after ½ lap

Maximum speed

Acceleration is a function of velocity, and we also can determine distance. Time is not involved in the problem, so we choose:

Determine total distance travelled3.14(200) 628.32 mx r




11 - 24

0 0

x v

x v

vdvdxa v

Determine the full integral, including limits

628.32

20 0 3 0.001

v vdx dvv

Evaluate the interval and solve for v2

01628.32 ln 3 0.0010.002

vv

2628.32( 0.002) ln 3 0.001 ln 3 0.001(0)v

2ln 3 0.001 1.2566 1.0986= 0.15802v

2 0.158023 0.001v e Take the exponential of each side




11 - 25

2 0.158023 0.001v e Solve for v

0.158022 3 2146.20.001

ev

46.3268 m/sv

How do you determine the maximum speed the car can reach?23 0.001a v Velocity is a maximum

when acceleration is zeroThis occurs when 20.001 3v

30.001maxv max 54.772 m/sv



Tenth Editionin SI Units Uniform Rectilinear Motion

11 - 26

For a particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant.

vtxxvtxx

dtvdx

vdtdx

tx

x

00

00

constant

During free-fall, a parachutist reaches terminal velocity when her weight equals the drag force. If motion is in a straight line, this is uniform rectilinear motion.

Careful – these only apply to uniform rectilinear motion!



Tenth Editionin SI Units Uniformly Accelerated Rectilinear Motion

11 - 27

If forces applied to a body are constant (and in a constant direction), then you have uniformly accelerated rectilinear motion.

Another example is free-fall when drag is negligible



Tenth Editionin SI Units Uniformly Accelerated Rectilinear Motion

11 - 28

For a particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. You may recognize these constant acceleration equations from your physics courses.

0

00

constantv t

v

dv a dv a dt v v atdt

0

210 0 0 0 2

0

x t

x

dx v at dx v at dt x x v t atdt

0 0

2 20 0constant 2

v x

v x

dvv a vdv a dx v v a x xdx

Careful – these only apply to uniformly accelerated rectilinear motion!



Tenth Editionin SI Units Motion of Several Particles

11 - 29

We may be interested in the motion of several different particles, whose motion may be independent or linked together.




Motion of Several Particles: Relative Motion

11 - 30

• For particles moving along the same line, time should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction. ABAB xxx relative

position of B with respect to A

ABAB xxx

ABAB vvv relative velocity of B with respect to A

ABAB vvv

ABAB aaa relative acceleration of B with respect to A

ABAB aaa




11 - 31

Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s. Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact.

SOLUTION:• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.

• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact.

• Substitute impact time into equation for position of elevator and relative velocity of ball with respect to elevator.




11 - 32

SOLUTION:• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.

22

22100

20

sm905.4s

m18m12

sm81.9s

m18

ttattvyy

tatvv

B

B

• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.

ttvyy

v

EE

E

sm2m5

sm2

0




11 - 33

• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. 025905.41812 2 ttty EB

s65.3

smeaningles s39.0

tt

• Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator. 65.325Ey

m3.12Ey

65.381.916

281.918

tv EB

sm81.19EBv




Motion of Several Particles: Dependent Motion

11 - 34

• Position of a particle may depend on position of one or more other particles.• Position of block B depends on position of block A. Since rope is of constant length, it follows that sum of lengths of segments must be constant.

BA xx 2 constant (one degree of freedom)• Positions of three blocks are

dependent. CBA xxx 22 constant (two degrees of freedom)

• For linearly related positions, similar relations hold between velocities and accelerations.

022or022

022or022

CBACBA

CBACBA

aaadt

dvdt

dvdt

dv

vvvdt

dxdt

dxdt

dx




11 - 35

Pulley D is attached to a collar which is pulled down at 75 mm/s. At t = 0, collar A starts moving down from K with constant acceleration and zero initial velocity. Knowing that velocity of collar A is 300 mm/s as it passes L, determine the change in elevation, velocity, and acceleration of block B when block A is at L.

SOLUTION:• Define origin at upper horizontal surface with positive displacement downward.• Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L.• Pulley D has uniform rectilinear motion. Calculate change of position at time t.• Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t. • Differentiate motion relation twice to develop equations for velocity and acceleration of block B.




11 - 36

SOLUTION:• Define origin at upper horizontal surface with positive displacement downward.

• Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L.

( )02

mm mm300 225 1.333 ss s

= +

= =

A A Av v a t

t t

220 02

2

2

mm mm300 2 200mm 225s s

A A A A A

A A

v v a x x

a a




11 - 37

• Pulley D has uniform rectilinear motion. Calculate change of position at time t.

• Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t. Total length of cable remains constant,

0 0 0

0 0 0

0

2 22 0

200mm 2 100mm 0

A D B A D B

A A D D B B

B B

x x x x x x

x x x x x x

x x

( )0 400mmB Bx x- = -




11 - 38

• Differentiate motion relation twice to develop equations for velocity and acceleration of block B.

2 constant2 0mm mm300 2 75 0s s

A D B

A D B

B

x x xv v v

v

mm mm450 450s sBv = - =

2

2 0mm225 2(0) 0s

A D B

B

a a a

a

2 2m m mm225 225s sBa = - =




11 - 39

Slider block A moves to the left with a constant velocity of 6 m/s. Determine the velocity of block B.

Solution steps

• Sketch your system and choose coordinate system

• Write out constraint equation

• Differentiate the constraint equation to get velocity




11 - 40

Given: vA= 6 m/s left Find: vBxA

yB

This length is constant no matter how the blocks move

Sketch your system and choose coordinates

Differentiate the constraint equation to get velocity

constnts3 aA Bx y L Define your constraint equation(s)

6 m/s + 3 0Bv

2 m/sB v

Note that as xA gets bigger, yB gets smaller.




Graphical Solution of Rectilinear-Motion Problems

11 - 41

Engineers often collect position, velocity, and acceleration data. Graphical solutions are often useful in analyzing these data.

0

20

40

60

80

100

120

140

160

180

47.76 47.77 47.78 47.79 47.8 47.81Tim e (s)

Acceleration (g)

Acceleration data from a head impact during a round of boxing.





11 - 42

•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.





11 - 43

• 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.



Tenth Editionin SI Units Other Graphical Methods

11 - 44

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

1

0110

01 curve under areav

vdvtttv

tvxx

using dv = a dt ,

1

011001

v

vdtatttvxx

1

01

v

vdtatt first moment of area

under a-t curve with respect to t = t1 line.

Cttta-ttvxx

centroid of abscissacurve under area 11001



Tenth Editionin SI Units Other Graphical Methods

11 - 45

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

BCAB

dxdvva

tansubnormal to v-x curve




Curvilinear Motion: Position, Velocity & Acceleration

11 - 46

The softball and the car both undergo curvilinear motion.

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





11 - 47

• The 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 a particle which occupies position P defined by at time t and P’ defined by at t + t,

rr





11 - 48

0lim

t

s dsvt dt

Instantaneous velocity (vector)

Instantaneous speed (scalar)

0lim

t

r drvt dt





11 - 49

0lim

t

v dvat dt

instantaneous acceleration (vector)

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

v v

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



Tenth Editionin SI Units Derivatives of Vector Functions

11 - 50

uP

• Let be a vector function of scalar variable u,

uuPuuP

uP

duPd

uu

00limlim

• Derivative of vector sum,

duQd

duPd

duQPd

duPdfP

dudf

duPfd

• Derivative of product of scalar and vector functions,

• Derivative of scalar product and vector product,

duQdPQ

duPd

duQPd

duQdPQ

duPd

duQPd

Delete or put in “bonus” slides




Rectangular Components of Velocity & Acceleration

11 - 51

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

• Velocity vector,

kvjviv

kzjyixkdtdzj

dtdyi

dtdxv

zyx

• Acceleration vector,

kajaia

kzjyixkdt

zdjdt

ydidt

xda

zyx

22

22

22




Rectangular Components of Velocity & Acceleration

11 - 52

• 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

00

221

0000

zgtyvytvxvgtvvvv

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.




11 - 53

A projectile is fired from the edge of a 150-m cliff with an initial velocity of 180 m/s at an angle of 30°with 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.

SOLUTION:• Consider the vertical and horizontal motion separately (they are independent)• Apply equations of motion in y-direction

• Apply equations of motion in x-direction

• Determine time t for projectile to hit the ground, use this to find the horizontal distance

• Maximum elevation occurs when vy=0




11 - 54

SOLUTION:

Given: (v)o =180 m/s (y)o =150 m(a)y = - 9.81 m/s2(a)x = 0 m/s2

Vertical motion – uniformly accelerated:

Horizontal motion – uniformly accelerated: Choose positive x to the right as shown




11 - 55

SOLUTION:Horizontal distanceProjectile strikes the ground at:

Solving for t, we take the positive root

Maximum elevation occurs when vy=0

Substitute into equation (1) above

Substitute t into equation (4)

Maximum elevation above the ground =




11 - 56

If you fire a projectile from 150 meters above the ground (see Ex Problem 11.7), what launch angle will give you the greatest horizontal distance x?a)A launch angle of 45b)A launch angle less than 45c)A launch angle greater than

45d)It depends on the launch

velocity




11 - 57

A baseball pitching machine “throws” baseballs with a horizontal velocity v0. If you want the height h to be 1 m, determine the value of v0.

SOLUTION:• Consider the vertical and horizontal motion separately (they are independent)• Apply equations of motion in y-direction

• Apply equations of motion in x-direction• Determine time t for projectile to fall to 1 m.• Calculate v0=0




11 - 58

Analyze the motion in the y-direction

Given: x= 12 m, yo = 1.5 m, yf= 1 m. Find: vo

20

1(0) 2fy y t gt

2 210.5 m (9.81 m/s )2 t

213.5 5 2gt

0.3193 st

Analyze the motion in the x-direction

0 00 ( )xx v t v t

012 m ( )(0.3193 s)v

0 37.6 m/s 135.3 km/hv





11 - 59

A soccer player must consider the relative motion of the ball and her teammates when making a pass.

It is critical for a pilot to know the relative motion of his aircraft with respect to the aircraft carrier to make a safe landing.





11 - 60

• 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.




11 - 61

Automobile A is traveling east at the constant speed of 36 km/h. As automobile A crosses the intersection shown, automobile B starts from rest 35 m north of the intersection and moves south with a constant acceleration of 1.2 m/s2. Determine the position, velocity, and acceleration of B relative to A 5 s after A crosses the intersection.

SOLUTION:

• Define inertial axes for the system

• Determine the position, speed, and acceleration of car A at t = 5 s

• Using vectors (Eqs 11.31, 11.33, and 11.34) or a graphical approach, determine the relative position, velocity, and acceleration

• Determine the position, speed, and acceleration of car B at t = 5 s




11 - 62

SOLUTION: • Define axes along the road

Given: vA=36 km/h, aA= 0, (xA)0 = 0(vB)0= 0, aB= - 1.2 m/s2, (yA)0 = 35 mDetermine motion of

Automobile A:

We have uniform motion for A so:

At t = 5 s




11 - 63

SOLUTION:

Determine motion of Automobile B:We have uniform acceleration for B so:

At t = 5 s




11 - 64

SOLUTION:

We can solve the problems geometrically, and apply the arctangent relationship:

Or we can solve the problems using vectors to obtain equivalent results: B A B/Ar r r B A B/Av v v B A B/Aa a a

20 5020 50 (m)

B/A

B/A

j i rr j i

6 106 10 (m/s)

B/A

B/A

j i vv j i 2

1.2 01.2 (m/s )

B/A

B/A

j i aa j

Physically, a rider in car A would “see” car B travelling south and west.




11 - 65

If you are sitting in train B looking out the window, it which direction does it appear that train A is moving?

a)

b)

c)

d)25o

25o



Tenth Editionin SI Units Tangential and Normal Components

11 - 66

If we have an idea of the path of a vehicle, it is often convenient to analyze the motion using tangential and normal components (sometimes called path coordinates).




11 - 67

• The tangential direction (et) is tangent to the path of the particle. This velocity vector of a particle is in this direction

x

y

et

en

• The normal direction (en) is perpendicular to et and points towards the inside of the curve.

v= vt et

= the instantaneous radius of curvature

2dv vdt

t na e e

v tv e

• The acceleration can have components in both the en and et directions




11 - 68

• To derive the acceleration vector in tangential and normal components, define the motion of a particle as shown in the figure.

• 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

dede

eee

e

tn

nnt

t

22sinlimlim

2sin2

00




11 - 69

tevv • With the velocity vector expressed asthe particle acceleration may be written as dt

dsdsd

dedve

dtdv

dtedve

dtdv

dtvda tt

butv

dtdsdsde

ded

nt

After substituting,

22 vadtdvaeve

dtdva ntnt

• The tangential component of acceleration reflects change of speed and the normal component reflects change of direction.• The tangential component may be positive or negative. Normal component always points toward center of path curvature.




11 - 70

22 vadtdvaeve

dtdva ntnt

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

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

ntb eee

• The normal to the osculating plane is found from

binormale

normalprincipal e

b

n

• Acceleration has no component along the binormal.




11 - 71

A motorist is traveling on a curved section of highway of radius 750 m at the speed of 90 km/h. The motorist suddenly applies the brakes, causing the automobile to slow down at a constant rate. Knowing that after 8 s the speed has been reduced to 72 km/h, determine the acceleration of the automobile immediately after the brakes have been applied.

SOLUTION:

• Define your coordinate system

• Calculate the tangential velocity and tangential acceleration

• Determine overall acceleration magnitude after the brakes have been applied

• Calculate the normal acceleration




21.041m/s=a 53.1a =

2

20.833m /stan0.625m /s

n

t

aa

a = =

2 22(25m/s) 0.833m /s750mn

var

= = =

220m/s 25m /saverage 0.625m /s8st tva at

D -= = = = -D

km 1000m 1 h90km/h = 90 25m/sh 1km 3600s72km/h = 20m/s

Sample Problem 11.8

11 - 72

SOLUTION: • Define your coordinate system

et en

• Determine velocity and acceleration in the tangential direction

• The deceleration constant, therefore

• Immediately after the brakes are applied, the speed is still 25 m/s

2 2 2 2( 0.625) (0.833)n ta a a




11 - 73

In 2001, a race scheduled at the Texas Motor Speedway was cancelled because the normal accelerations were too high and caused some drivers to experience excessive g-loads (similar to fighter pilots) and possibly pass out. What are some things that could be done to solve this problem?

Some possibilities:

Reduce the allowed speed Increase the turn radius (difficult and costly)Have the racers wear g-suits




11 - 74

SOLUTION:

• Define your coordinate system• Calculate the tangential velocity and tangential acceleration

• Determine overall acceleration magnitude

• Calculate the normal acceleration

The tangential acceleration of the centrifuge cab is given bywhere t is in seconds and at is in m/s2. If the centrifuge starts from fest, determine the total acceleration magnitude of the cab after 10 seconds.

20.5 (m/s )ta t




11 - 75

In the side view, the tangential direction points into the “page”

Define your coordinate system

8 met

en

en

Top View

Determine the tangential velocity0.5ta t

2 200

0.5 0.25 0.25t ttv t dt t t

20.25 10 25 m/stv

Determine the normal acceleration 2 2

225 78.125 m/s8t

nvar

Determine the total acceleration magnitude

22 2 278.125 + (0.5)(10) mag n ta a a 278.285 m/smaga




11 - 76

a)The accelerations would remain the same

b)The an would increase and the at would decrease

c)The an and at would both increased)The an would decrease and the at

would increase

Notice that the normal acceleration is much higher than the tangential acceleration. What would happen if, for a given tangential velocity and acceleration, the arm radius was doubled?



Tenth Editionin SI Units Radial and Transverse Components

11 - 77

By knowing the distance to the aircraft and the angle of the radar, air traffic controllers can track aircraft. Fire truck ladders can rotate as well as extend; the motion of the end of the ladder can be analyzed using radial and transverse components.




11 - 78

• The position of a particle P is expressed as a distance r from the origin O to P – this defines the radial direction er. The transverse direction e is perpendicular to er • The particle velocity vector is

• The particle acceleration vector is

rerr




11 - 79

• We can derive the velocity and acceleration relationships by recognizing that the unit vectors change direction.

rr e

dede

ded

dtde

dtd

ded

dted rr

dtde

dtd

ded

dted

r

erer

edtdre

dtdr

dtedre

dtdrer

dtdv

r

rr

rr

• The particle velocity vector is

• Similarly, the particle acceleration vector is

errerrdted

dtdre

dtdre

dtd

dtdr

dted

dtdre

dtrd

edtdre

dtdr

dtda

r

rr

r

2222

22

rerr




11 - 80

If you are travelling in a perfect circle, what is always true about radial/transverse coordinates and normal/tangential coordinates?a) The er direction is identical to the

en direction.b) The e direction is perpendicular to

the en direction.c) The e direction is parallel to the

er direction.




11 - 81

• 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,kzeReR

dtrdv R

• Acceleration vector,

kzeRReRRdtvda R

22




11 - 82

Rotation of the arm about O is defined by = 0.15t2 where is in radians and t in seconds. Collar B slides along the arm such that r = 0.9 - 0.12t2 where r is in meters.After the arm has rotated through 30o, determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration of the collar with respect to the arm.

SOLUTION:• Evaluate time t for = 30o.• Evaluate radial and angular positions, and first and second derivatives at time t.• Calculate velocity and acceleration in cylindrical coordinates.• Evaluate acceleration with respect to arm.




11 - 83

SOLUTION:• Evaluate time t for = 30o.

s 869.1rad524.0300.152

tt

• Evaluate radial and angular positions, and first and second derivatives at time t.

2

2

sm24.0sm449.024.0m 481.012.09.0

rtr

tr

2

2

srad30.0srad561.030.0

rad524.015.0

tt




11 - 84

• Calculate velocity and acceleration.

rr

r

vvvvv

rvsrv

122 tan

sm270.0srad561.0m481.0m449.0

0.31sm524.0 v

rr

r

aaaaa

rra

rra

122

2

2

2

22

2

tan

sm359.0srad561.0sm449.02srad3.0m481.0

2sm391.0

srad561.0m481.0sm240.0

6.42sm531.0 a




11 - 85

• Evaluate acceleration with respect to arm.Motion of collar with respect to arm is rectilinear and defined by coordinate r. 2sm240.0ra OAB




11 - 86

SOLUTION:

• Define your coordinate system

• Calculate the angular velocity after three revolutions

• Determine overall acceleration magnitude

• Calculate the radial and transverse accelerations

The angular acceleration of the centrifuge arm varies according to

where is measured in radians. If the centrifuge starts from rest, determine the acceleration magnitude after the gondola has travelled two full rotations.




11 - 87

In the side view, the transverse direction points into the “page”

Define your coordinate system

8 me

er

er

Top View

Determine the angular velocity

Evaluate the integral

Acceleration is a function of position, so use:




11 - 88

er

Determine the angular velocity

Determine the angular acceleration

Find the radial and transverse accelerations

22 2 2( 63.166)+ 5.0265 mag ra a a 2 63.365 m/smaga

Magnitude:




11 - 89

You could now have additional acceleration terms. This might give you more control over how quickly the acceleration of the gondola changes (this is known as the G-onset rate).

What would happen if you designed the centrifuge so that the arm could extend from 6 to 10 meters?

r

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