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Kinematics of Particles(Part 1)

Chapter 11

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• Dynamics includes:

- Kinematics: study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion.

- 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.

• 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.

Introduction

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11.2 Position, Velocity and Acceleration

RECTILINEAR MOTION OF PARTICLES

• Particle moving along a straight line is said to be in rectilinear motion.

• Position coordinate of a particle is defined by positive or negative distance of particle 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. Motion of the particle may be expressed in the form of a function, e.g.,

326 ttx

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

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• 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,

t

xv

t

x

t

0lim

Average velocity

Instantaneous velocity

• From the definition of a derivative,

dt

dx

t

xv

t

0lim

e.g.,

2

32

312

6

ttdt

dxv

ttx

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

Instantaneous accelerationt

va

t

0lim

tdt

dva

ttv

dt

xd

dt

dv

t

va

t

612

312e.g.

lim

2

2

2

0

• From the definition of a derivative,

• Instantaneous acceleration may be:

- positive: increasing positive velocity

or decreasing negative velocity

- negative: decreasing positive velocity

or increasing negative velocity.

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• Consider particle with motion given by326 ttx

2312 ttdt

dxv

tdt

xd

dt

dva 612

2

2

• at t = 0, x = 0, v = 0, a = 12 m/s2

• 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

• at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2

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11.3 Determination of the motion of a particle

• Recall, motion of a particle is known if position is known for all time t.

• Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations.

• Three classes of motion may be defined for:

- 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)

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• Acceleration given as a function of time, a = f(t):

tttx

x

tttv

v

dttvxtxdttvdxdttvdxtvdt

dx

dttfvtvdttfdvdttfdvtfadt

dv

00

0

00

0

0

0

• Acceleration given as a function of position, a = f(x):

x

x

x

x

xv

v

dxxfvxvdxxfdvvdxxfdvv

xfdx

dvva

dt

dva

v

dxdt

dt

dxv

000

202

1221

or or

11.3 Determination of the motion of a particle

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tv

v

tv

v

tx

x

tv

v

ttv

v

vf

dvvxtx

vf

dvvdx

vf

dvvdxvfa

dx

dvv

tvf

dv

dtvf

dvdt

vf

dvvfa

dt

dv

0

00

0

0

0

0

• Acceleration given as a function of velocity, a = f(v):

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Sample problem 11.1

The position of a particle which moves along a straight line is defined by the relation ,

where x is expected in meter and t in seconds. Determine

(a) the time at which the velocity will be zero(b) the position and distance traveled by the particle at that time (c) the acceleration of the particle at that time(d) the distance traveled by the particle from t= 4s to t= 6s

40156 23 tttx

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Solution

x (m)

v (m/s)

a (m/s )2

t(s)

t(s)

t(s)

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x (m)

v (m/s)

a (m/s )2

t(s)

t(s)

t(s)

Solution

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Sample problem 11.2

A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20m above the ground. Knowing that the acceleration of the ball is constant and equal to 9.81 m/s downward,Determine (a) the velocity v and elevation y of the ball above the ground at any time t,(b) the highest elevation reached by the ball and the corresponding value of t,(c) the time when the ball will hit the ground and the corresponding velocity.

(d) Draw the v-t and y-t curves.

2

14

tvtvdtdv

adt

dv

ttv

v

81.981.9

sm81.9

00

2

0

ttv

2s

m81.9

s

m10

2

21

00

81.91081.910

81.910

0

ttytydttdy

tvdt

dy

tty

y

22s

m905.4

s

m10m20 ttty

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

Solutiona.Velocity and Elevation

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

0s

m81.9

s

m10

2

ttv

s019.1t

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

22

22

s019.1s

m905.4s019.1

s

m10m20

s

m905.4

s

m10m20

y

ttty

m1.25y

b. Highest Elevation

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

0s

m905.4

s

m10m20 2

2

ttty

s28.3

smeaningles s243.1

t

t

s28.3s

m81.9

s

m10s28.3

s

m81.9

s

m10

2

2

v

ttv

s

m2.22v

c. Balls Hits the Ground

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The brake mechanism used to reduce recoil in certain types of guns consists essentially of piston attached to the barrel and moving in a fixed cylinder filled with oil. As the barrel recoils with an initial velocity v0, the piston moves and oil is forced through orifices in the piston, causing the piston and the barrel to decelerate at a rate proportional to their velocity. that is, a = -kv

Determine v(t), x(t), and v(x).

Draw the corresponding motion curves.

kva

Sample problem 11.3

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

ktv

tvdtk

v

dvkv

dt

dva

ttv

v

00

ln0

ktevtv 0

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

tkt

tkt

tx

kt

ek

vtxdtevdx

evdt

dxtv

00

00

0

0

1

ktek

vtx 10

Solution

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

kxvv

dxkdvdxkdvkvdx

dvva

xv

v

0

00

kxvv 0

• Alternatively,

0

0 1v

tv

k

vtx

kxvv 0

0

0 or v

tveevtv ktkt

ktek

vtx 10with

and

then

Solution

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11.4 Uniform Rectilinear Motion

For particle in uniform rectilinear motion, the acceleration is zero and

the velocity is constant.

vtxx

vtxx

dtvdx

vdt

dx

tx

x

0

0

00

constant

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For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant.

atvv

atvvdtadvadt

dv tv

v

0

000

constant

221

00

221

000

000

attvxx

attvxxdtatvdxatvdt

dx tx

x

020

2

020

221

2

constant00

xxavv

xxavvdxadvvadx

dvv

x

x

v

v

11.5 Uniformly Acceleration Rectilinear Motion

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11.6 Motion of Several Particles

• 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

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• 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

dv

dt

dv

dt

dv

vvvdt

dx

dt

dx

dt

dx

Dependent Motions

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Sample Problem 11.4

A ball is thrown vertically upward from the 12 m level in an elevator shaft with an initial velocity of 18 m/s. At the same instant, an open-platform elevator passes the 5m level, moving upward with a constant velocity at 2 m/s.

Determine

(a) when and where the ball will hits the elevator

(b) the relative velocity of ball with respect to the elevator when the ball hits the elevator.

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

22

21

00

0

905.418m12

81.918

ttattvyy

tatvv

B

B

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

ttvyy

smv

EE

E

2m5

/2

0

(1)

(2)

(3)(4)

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Ball hits elevatorWe first note that the same time t and the same origin O were used in writingthe equations of motion of both the ball and the elevator. We see from the figure that when the ball hits the elevator,

BE yy (5)

Substituting for yE and yB from (2) and (4) into (5), we have

2905.4181225 ttt

st 39.0 st 65.3and

Only the root t=3.65s corresponds to a time after the motion has begun,Substituting this value into (4), we have

myE 30.12)65.3(25

Elevation fro ground = 12.30m

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The relative velocity of the ball with respect to the elevator is

ttvvv EBEB 81.9162)81.918(/

When the ball hits the elevator at time t= 3.65 s , we have

smv

v

EB

EB

/81.19

)65.3(81.916

/

/

The negative sign means that the ball is observed from the elevator to be moving in the negative sense (downward)

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200 mm

Sample Problem 11.5

Collar A and block B are connected by a cable passing over three pulleys C,D, and E as shown.Pulleys C and E are fixed, while D is attached to a collar which is pulled downward with a constantvelocity of 75mm/s. At t=0, collar A starts moving downward from position K with constant accelerationand no initial velocity.Knowing that the velocity of collar A is 300 mm/s as it passes through point L, determine the change in elevation, the velocity and the acceleration of block B when collar A passes through L.

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XA

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THE END