1-1

1. Kinematics of Particles 1.1 Introduction to Dynamics

Dynamics

- Kinematics: the study of the geometry of motion; relate displacement, velocity, acceleration, and

time, without reference to the cause of the motion.

- Kinetics: the study of the relation existing between the forces acting on a body, the mass of the

body, and the motion of the body; predict the motion caused by given forces or determine the

forces required to produce a given motion.

1.2 Position, Velocity, and Acceleration Rectilinear motion: motion along a straight line

Position coordinate: x

SI Units: m, sec

Average velocity =

Instantaneous velocity (m/s) =

Speed: magnitude of v

Average acceleration (m/s2) =

Instantaneous acceleration =

or

tx

ΔΔ

dtdx

txv

t=

ΔΔ

=→Δ

lim0

tv

ΔΔ

2

2

0lim dt

xddtdv

tva

t==

ΔΔ

=→Δ

dxdvv

vdxdv

dtdva ===

1-2

1.3 Determination of the Motion of a Particle In general, the acceleration of a particle can be expressed as a function of one or more of variables x, v,

and t. Let us consider three common classes of motion:

( )tfa =

( )xfa =

( )vfa =

( )tfdtdva ==

( )( )

( )( )tvv

dttfvv

dttfdv

dttfdv

t

t

t

t

v

v

=

=−

=

=

∫

∫∫

0

00

0

( )( )∫=−

=t

tdttvxx

dttvdx

00

( )

( )( )

( )

( )xvv

dxxfvv

dxxfvdv

dxxfvdv

xfdxvdv

dtdva

x

x

x

x

v

v

=

=−

=

=

===

∫

∫∫

0

00

20

2

21

21

( )

( )∫=−

=

x

x xvdxtt

dtxv

dx

00

( )

( )

( )vfvdvdx

vfdvdt

vfdxvdv

dtdva

=

=

===

1-3

Example 1) The position of a particle which moves along a straight line is defined by the

relation , where x is expressed in meters 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 the

time, (c) the acceleration of the particle at that time, (d) the distance traveled by the particle from t = 4s to

t = 6s.

Ans) The equations of motion are

(1)

(2)

(3)

a) Time at which v = 0.

t = 5s

b) Position and Distance Traveled when v = 0.

Position at t = 5s:

Distance at t = 5s: 100m

c) Acceleration when v = 0.

d) Distance traveled from t = 4s to t = 6s.

The particle moves in the negative direction from t = 4s to t = 5s and in the positive direction from t = 5s

to t = 6s.

Example 2) A ball is tossed with a velocity of 10m/s directed vertically upward from a window located

20m above the ground. Knowing that the acceleration of the ball is constant and equal to 9.81m/s2

40156 23 +−−= tttx

40156 23 +−−= tttx

15123 2 −−== ttdtdxv

126 −== tdtdva

( )( ) 015542 =+−=−− tttt

)(6040515565 235 mx −=+×−×−=

)(400 mx =

100406005 −=−−=− xx

( )25 /181256 sma =−×=

( ) ( ) ( )mxxdisxxdiscedistotal 18108tan 5645 =+−=−+−=

1-4

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.

Ans)

a) velocity and elevation

b) highest elevation

c) ball hits the ground

( )2/81.9, smggadtdv

=−==

( )smvgtvv 10, 00 =−=−

tv 81.910 −=

tdtdy 81.910 −=

( )2

02

0

905.41020

20,905.410

tty

myttyy

−+=

=−=−

( )myst

tdtdy

highest 1.25019.1905.4019.11020019.1

081.910

2 =×−×+=

=

=−=

( )smv

sorsttty

t 2.2228.381.910

28.3243.10905.41020

28.3

2

−=×−=

−=

=−+=

=

1-5

1.4 Uniform Rectilinear Motion When the velocity of a particle is constant,

1.5 Uniformly Accelerated Rectilinear Motion When the acceleration is constant,

or

1.6 Motion of Several Particles

Relative Motion of Two Particles

The relative position coordinate of B with respect to A is denoted by

The relative velocity of B with respect to A is denoted by

vtxx

dtvdx

tconsvdtdx

t

t

x

x

+=

=

==

∫∫0

00

tan

200

0

0

21

tan

00

attvxx

atvvdtdx

atvv

dtadv

tconsadtdv

t

t

v

v

++=

+==

+=

=

==

∫∫

( )02

02 2

tan

00

xxavv

dxavdv

tconsadxvdv

dtdv

x

x

v

v

−+=

=

===

∫∫

ABAB

ABAB

xxxor

xxx

+=

−=

ABAB

ABAB

vvvor

vvv

+=

−=

1-6

The relative acceleration of B with respect to A is denoted by

Dependent Motions

In the case of three blocks of Fig. 11.9,

then

Example 4)

A ball is thrown vertically 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 5-m level, moving upward with a constant velocity

of 2m/s. Determine (a) when and where the ball will hit the elevator, (b) the relative velocity of the ball

with respect to the elevator when the ball hits the elevator.

Ans)

Motion of the Ball

: uniformly accelerated motion

Motion of the Elevator

: uniform motion

ABAB

ABAB

aaaor

aaa

+=

−=

variablesstate3

equationdependentlinearly1

2DOF:constant22 =++ CBA xxx

022022

=++=++

CBA

CBA

aaavvv

2200

0

905.4181221

81.918

ttattvyy

tatvv

B

B

−+=++=

−=+=

ttvyysmv

EE

E

252

0 +=+=+=

1-7

Ball Hits the Elevator

When the ball hits the elevator,

Example 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 constant velocity of 75m/s. At t=0,

collar A starts moving downward form position K with a

constant acceleration and no initial velocity. Knowing that

the velocity of collar A is 300 mm/s as it passes through

point L,

( )myt

tttt

yy

E

BE

30.1265.32565.3

or39.0905.4181225 2

=×+==

−=

−+=+

=

( )sm

ttvvv EBEB

81.1965.381.916

81.916281.918

−=×−=

−=−−=−=

1-8

determine the change in elevation, the velocity, and the acceleration of block B when collar A passes

through L.

Ans)

Motion of Collar A

We place the origin O at the upper horizontal surface

and choose the positive direction downward.

At t = 0

When

Motion of Pulley D

At t = 1.333s

Motion of Block B

The total length of the cable is constant. Considering

the times t = 0 and t = 1.333s,

( )( )0

0 0

AA

A

xxv

=

=

LxA =

smmvA /300=( ) ( )[ ]

2

20

20

2

/225

20020300

2

smma

a

xxavv

A

A

AAAAA

=

×+=

−+=

( )

stt

tavv AAA

333.12250300

0

=+=

+=

smmva

D

D

/750

==

( ) ( ) txtvxx DDDD 7500 +=+=

( ) ( )mmxx DD 100333.1750 =×=−

( ) ( ) ( )( )[ ] ( )[ ] ( )[ ] 02

22

000

000

=−+−+−

++=++

BBDDAA

BDABDA

xxxxxxxxxxxx

( ) ( ) ( )( )[ ] ( )[ ] ( )[ ]

( )[ ]( ) mmxx

xxxxxxxx

xxxxxx

BB

BB

BBDDAA

BDABDA

40001002200

0222

0

0

000

000

−=−

=−+×+

=−+−+−

++=++

1-9

Homework 1

Problems 11.34, 11.41, 11.45, 11.47, 11.57

1.9 Position Vector, Velocity, and Acceleration

Curviliear motion: motion along a curve other than a straight

line

Position vector: position of the particle w.r.t. the reference

frame, r

Average velocity:

Instantaneous velocity:

Instantaneous acceleration:

1.10 Derivatives of Vector Functions Let P(u) and Q(u) be vector functions of the scalar variable u.

Then

Sum of two vector functions of the same scalar variable u:

Product of a scalar function f(u) and a vector function P(u):

2225

00222502

4500752300

02

smma

aaaa

smmvv

vvv

B

B

BDA

B

B

BDA

−=

=+×+=++

−==+×+

=++

( )tt

ΔΔr

( ) ( ) ( )dt

tdttt

t

rrv =Δ

Δ=

→Δlim

0

( ) ( ) ( )dt

tdttt

t

vva =Δ

Δ=

→Δlim

0

( )dud

dud

dud QPQP

+=+

( )dudf

dudf

dufd PPP

+=

1-10

Scalar product and vector product:

Rectangular components of the derivative of a vector function P(u):

where i, j, k are unit vectors corresponding to the x, y, z axes.

Rate of Change of a Vector:

The rate of change of a vector is the same with respect to a fixed frame and with respect to a frame

in translation.

1.11 Rectangular Components of Velocity and Acceleration The position vector r of a particle in rectangular axes can be written

where the coordinates x, y, z are functions of t. Differentiating twice, we obtain

1.12 Motion Relative to a Frame in Translation Fixed frame of Reference: the reference frame

attached to the earth

Moving Frames of Reference: the reference frame in

either translation or rotation

Any frame can be designated as “fixed; all other

frames not rigidly attached to this frame will be

described as “moving”.

( )dud

dud

dud QPQPQP

•+•=•

( )dud

dud

dud QPQPQP

×+×=×

kjiP ZYx PPP ++=

kjiPdudP

dudP

dudP

dud zyx ++=

kjiPdt

dPdt

dPdt

dPdtd zyx ++=

kjir zyx ++=

kjiva

kjirv

2

2

2

2

2

2

dtzd

dtyd

dtxd

dtd

dtdz

dtdy

dtdx

dtd

++==

++==

1-11

: The absolute motion of B can be obtained by combining the motion of A and the relative motion of B

with respect to the moving frame attached to A.

Example 1.8)

A projectile is fired with an initial velocity of 240m/s at a target B located 600m above the gun A and at a

horizontal distance of 3600m. Neglecting air resistance, determine the value of the firing angle α.

Ans)

Horizontal Motion

At x = 3600m

Vertical Motion

Projectile Hits Target

ABAB

ABAB

ABAB

aaa

vvv

rrr

+=

+=

+=

( )( )

α

αα

α

cos15

cos2403600cos240

cos240

0

0

=

=

==

=

t

tttvx

v

x

x

( )

( ) 220

20

90.4sin24021

81.9

sin240

ttattvy

sma

v

y

y

−=+=

−=

=

α

α

oo 6.69and9.29

69.2and574.0tan

600cos

1590.4sincos

152402

=

=

=⎟⎠⎞

⎜⎝⎛−=

α

αα

αα

y

1-12

Example1.9)

Automobile A is traveling east at the constant speed of

36km/h. As automobile A crosses the intersection shown,

automobile B starts from rest 35m north of the intersection

and moves south with a constant acceleration of 1.2m/s2.

Determine the position, velocity, and acceleration of B

relative to A 4s after A crosses the intersection.

Ans) We choose x and y axes with origin at the intersection

of the two streets and with positive senses directed

respectively east and north.

Motion of Automobile A

For any time t

For t = 5s

Motion of Automobile B

For any time t

For t = 5s

smvA 10=

( ) ttvxxsmv

a

AAA

A

A

100100

0 +=+===

( )mxsmv

a

A

A

A

50510100

=×===

( )

( ) ( ) 2200

0

2

2.121035

21

2.102.1

ttatvyy

ttavvsma

BBBB

BBB

B

××−+=++=

−=+=−=

( )

( )my

smvsma

B

B

B

2052.12135

0.652.12.1

2

2

=××−=

−=×−=−=

1-13

Motion of B Relative to A

We draw the triangle corresponding to the vector equation rB = rA+ rB/A and obtain the magnitude and

direction of the position vector of B relative to A.

rB/A= 53.9m α = 21.8°

vB/A = 11.66m/s β = 31.0°

Homework

Problems 11.97, 11.103, 11.120, 11.131

1-14

1.13 Tangential and Normal Components

Plane Motion of a Particle

See Fig. 11.21 and Fig. 11.22

1-15

Let us consider a particle which moves along a curve contained in the plane of the figure. Let P be the

position of the particle at a given instant. We attach at P a unit vector et tangent to the path of the

particle and pointing in the direction of motion. Let et / be the unit vector corresponding to the position

P/ of the particle at a later instant. Drawing both vectors from the same origin O/, we define the vector

Δet = et / - et . Since et / and et are of unit length, their tips lie on a circle of radius 1. Denoting by Δθ

the angle formed by et / and et , we find that the magnitude of Δet is 2sin(Δθ/2). Considering now the

vector Δet / Δθ, we note that as Δθ approaches zero, this vector becomes tangent to the unit circle,

perpendicular to et , and that its magnitude approaches

Denoting this vector by en , we wirte

Since the velocity v of the particle is tangent to the path, it can be expressed as the product of the scalar v

and the unit vector et :

But

where ρ is the radius of curvature of the path at P. Thus,

The scalar components of the acceleration are

( ) ( ) 12

2sin2sin2 limlim00

=Δ

Δ=

ΔΔ

→Δ→Δ θθ

θθ

θθ

θ

θθ

dd t

n

tn

ee

ee

=

ΔΔ

=→Δ

lim0

dtd

vdtdv

dtdv

tt

t

eeva

ev

+==

=

nnvv

dtds

dsd

dd

dtd

ee

ee tt

ρρ

θθ

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

1

nv

dtdv

dtd eeva t ρ

2

+==

ρ

2va

dtdva

n

t

=

=

1-16

1.14. Radial and transverse Components

See Fig. 11.25

In certain problems of motion, the position of the particle P is defined by its polar coordinates r and θ. It

is then convenient to resolve the velocity and acceleration of the particle into components parallel and

perpendicular, respectively, to the line OP. These components are called radial and transverse

components. A derivation similar to the one we used to determine the derivative of the unit vector et

leads to the relations

We express the time derivatives of the unit vectors as follows:

Using dots to indicate differentiation with respect to t,

The position vector r of P is expressed as rer. Thus,

Differentiating again wrt t,

rθ

θr

ee

ee

−=

=

θ

θ

dddd

dtd

dtd

dd

dtd

dtd

dtd

dd

dtd

θθθ

θθθ

rθθ

θrr

eee

eee

−==

==

rθ

θr

ee

ee

θ

θ&&

&&

−=

=

( )

θr

rrr

eev

eeev

θ&&

&&

rr

rrrdtd

+=

+==

( )θθθ

θr

eee

eeva

&&&&&&&&&&

&&

θθθ

θ

rrrerer

rrdtd

dtd

rr ++++=

+==

1-17

In the case of a particle moving along a circle of center O, we have r = constant:

Extension to the Motion of a Particle in Space: Cylindrical Coordinates See fig. 11.26

The position of a particle P in space is sometimes defined by

its cylindrical coordinates R, θ, and z; unit vectors er, eθ, and k.

Then the position vector r of the particle P can be written as

Thus

Example 1.10)

A motorist is traveling on a curved section of highway of

radius 2500ft at the speed of 60 mi/h. The motorist suddenly

applies the brakes, causing the automobile to slow down at a

constant rate. Knowing that after 8s the speed has been

reduced to 45mi/h, determine the acceleration of the

automobile immediately after the brakes have been applied.

Ans)

Tangential Component of Acceleration

( ) ( ) θr eea θθθ &&&&&&& rrrr 22 ++−=

θev θ&r=

θr eea θθ &&& rr +−= 2

ker R zR +=

( ) ( ) keeva

keerv

θR

θR

zRRRRdtd

zRRdtd

&&&&&&&&&

&&&

+++−==

++==

θθθ

θ

22

sfthmi

sftsfthmi

/66/45

/883600528060/60

=

=⎟⎠⎞

⎜⎝⎛×=

2/75.28

/88/66 sfts

sftsfttvat −=

−=

ΔΔ

=

1-18

Normal Component of Acceleration

Magnitude and Direction of Acceleration

Homework

Problems 11.141, 11.142, 11.169, 11.177, 11.C3

( ) 222

/10.32500

/88 sftftsftvan ===

ρ

2

2

2

/14.4sin

4.48/75.2/10.3tan

sfta

a

sftsft

aa

n

t

n

==

=

==

α

α

α

o