Lecture 1 · 2020. 5. 2. · For particle in uniformly accelerated rectilinear motion, the...

Lecture 1

• Kinematics of a particle

Material covered

• Kinematics of a particle

- Introduction

- Rectilinear kinematics:

Continuous motion

- How to analyze problems

- Problems

Next lecture; Rectilinear

kinematics: Erratic motion

Kinematics of a particle: Introduction

Important contributors

Galileo Galilei, Newton, Euler

Mechanics

Statics Dynamics

Equilibrium of a

body that is at

rest/moves with

constant velocity

Accelerated

motion of a body

• Kinematics: geometric aspects of the motion

• Kinetics: Analysis of forces which cause the motion

Kinematics of a particle: Introduction

• 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.

Today’s Objectives

Find the kinematic quantities (position, displacement,

velocity, and acceleration) of a particle traveling

along a straight path

Next lecture; Determine position, velocity, and

acceleration of a particle using graphs

Cartesian coordinates

Cylindrical coordinates

(Polar coordinates in 3D)

Choice of Coordinates

Rectilinear kinematics: Continuous Motion

A particle travels along a straight-line

path defined by the coordinate axis ‘s’

The POSITION of the particle at any

instant, relative to the origin, O, is

defined by the position vector r, or the

scalars. Scalars can be positive or

negative. Typical units for r and s are

meters (m) or feet (ft).

The displacement of the particle is defined

as its change in position.

Vector form: r = r’ - r Scalar form: s = s’ - s

The total distance traveled by the particle, sT, is a positive scalar that represents

the total length of the path over which the particle travels.

Velocity

Velocity is a measure of the rate of change in the position of a

particle. It is a vector quantity (it has both magnitude and

direction). The magnitude of the velocity is called speed, with

units of m/s or ft/s.

The average velocity of a particle during a

time interval t is

vavg = r/t

The instantaneous velocity is the time-derivative of position.

v = dr/dt

Speed is the magnitude of velocity: v = ds/dt

Average speed is the total distance traveled divided by elapsed

time: (vsp)avg = sT/ t

Acceleration

Acceleration is the rate of change in the velocity of a particle. It is a

vector quantity. Typical units are m/s2 or ft/s2.

The instantaneous acceleration is the time derivative of velocity.

Vector form: a = dv/dt

Scalar form: a = dv/dt = d2s/dt2

Acceleration can be positive (speed increasing)

or negative (speed decreasing).

As the book indicates, the derivative equations for velocity and

acceleration can be manipulated to get a ds = v dv

Constant acceleration

The three kinematic equations can be integrated for the special case

when acceleration is constant (a = ac) to obtain very useful equations.

A common example of constant acceleration is gravity; i.e., a body

freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2

downward. These equations are:

tavv co+=yields=

t

o

c

v

v

dtadv

o

2coo

s

t(1/2)at vss ++=yields= t

os

dtvds

o

)s-(s2a)(vv oc2

o2 +=yields=

s

s

c

v

v oo

dsadvv

1.Velocity as a Function of Time.

2. Position as a Function of Time.

3. Velocity as a Function of Position.

11

Uniform Rectilinear Motion

• For particle in uniform rectilinear motion, the

acceleration is zero and the velocity is constant.

0 0

0

0

x t

x

dx v dt

x x vt

x x vt

=

=

= +

constantdx

vdt

= =

Let the initial condition v = vo when to = 0.

ONLY CAN BE USED WHEN THE VELOCITY IS

CONSTANT.

12

Uniformly Accelerated Rectilinear

Motion

constantdv

adt

= =

0 0

v t

v

dv a dt= 0v v at =

For particle in uniformly accelerated rectilinear motion, the acceleration of

the particle is constant. Let initial condition v = vo when t = 0

0v v at= +

210 0 2

x x v t at= + +

2 2

0 02v v a x x= +

Let initial condition x = xo when t = 0

0

dxv v at

dt= = +

0

0

0

x t

x

dx v at dt= + 21

0 0 2x x v t at = +

Let initial condition v=vo at x = xo

constantdv

v adx

= =

0 0

v x

v x

v dv a dx= 2 210 02

v v a x x =

ONLY CAN BE USED WHEN THE ACCELARATION IS CONSTANT.

Rectilinear Motion: Position, Velocity &

Acceleration

• 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.

Rectilinear Motion: Position, Velocity & Acceleration

• 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

==

=

Rectilinear Motion: Position, Velocity & Acceleration

• 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.

Rectilinear Motion: Position, Velocity &

Acceleration

• Consider particle with motion given by

326 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

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)

Important points

• Dynamics: Accelerated motion of

bodies

• Kinematics: Geometry of motion

• Average speed and average velocity

• Rectilinear kinematics or straight-

line motion

• Acceleration is negative when

particle is slowing down

• a ds = v dv; relation of acceleration,

velocity, displacement

Analyzing problems in dynamics

Coordinate system

• Establish a position coordinate S along the path and specify its fixed origin and positive direction

• Motion is along a straight line and therefore s, v and α can be represented as algebraic scalars

• Use an arrow alongside each kinematic equation in order to indicate positive sense of each scalar

Kinematic equations

• If any two of a, v, s and t are related, then a third variable can be obtained using one of the kinematic equations.

• When performing integration, position and velocity must be known at a given instant (…so the constants or limits can be evaluated)

• Some equations must be used only when ‘a is constant’

Problem solving MUSTS

1. Read the problem carefully (and read it again)

2. Physical situation and theory link

3. Draw diagrams and tabulate problem data

4. Coordinate system!!!

5. Solve equations and be careful with units

6. Be critical. A mass of an aeroplane can not be 50 g

7. Read the problem carefully

A car moves in a straight line such that for a short time

its velocity is defined by

v = (3t2 + 2t) ft/s, where t is in seconds.

Determine its position and acceleration when t = 3 s.

When t = 0, s = o.

Solution:

The car’s velocity is given as a function of time so its

position can be determined from v = ds/dt,

since it relates v, s and t. Note that s=0 when t=0

v= ds = (3t2 + 2t)

dt

Integration over limit t=0, s=0

∫ ds = ∫ (3t2 + 2t) dt

s = t3 + t2

When t= 3s, s= (3)3 + (3)2 = 36 ft

For acceleration

Knowing velocity as function of time, acceleration is

determined from a = dv/dt, since this equation relates

a, v and t

a= dv= d (3t2 + 2t)

dt dt

a= 6t + 2

When t=3s, a=6(3) + 2 = 20 ft/s2

Problem 2

Solution: For velocity using eqn:

a ds= v dv

Integrating over the limit for t=0, s=0 v=0

∫(6+0.02s) ds = ∫v dv

6∫ ds +0.02 ∫ s ds = ∫v dv

6s + 0.02 s2/2 = v2 / 2

v= (12s+0.02s2) 1/2

For s= 2km, s = 2000m,

v = √12(2000)+ 0.02(2000) 2

v = 287.054 m/s

2 2

0 02v v a x x= +

Solution;:

Using the equation of velocity as function of

position

(80)2 = 0 + 2ac ( 500 – 0 )

ac = 6400 / 2 (500)

ac= 6.4 ft/s 2

80

80

0

0 0

2

: 6.4 6.4 [ ]

8012.5 /

6.4

t

Use adt dv dt dv t v

t ft s

= = =

= =

b) For time =?

500 80 2500 80

0 0

0 0

2

: [ ] [ ]2

64006.4 /

2*500

vUse ads v dv ads v dv a S

a ft s

= = =

= =

80

80

0

0 0

2

: 6.4 6.4 [ ]

8012.5 /

6.4

t

Use adt dv dt dv t v

t ft s

= = =

= =

The position of a particle along a straight line is

given by s = ( t3 - 9t2 + 15t) ft, where t is in seconds.

Determine its maximum acceleration and maximum

velocity during the time interval 0< t < 10 sec.

• Solution:

242

18106

1823

151823

135

151018103

152923

15293

15293

max

max

max

max

ft/sa

)(a

t)(a

)t)t(dt

d

dt

dva

ft/sv

)()(v

t)(tv

t)t(tdt

d

dt

dsv

ttts

=

=

=

+==

=

+=

+=

+==

+=

2.5 0.50.333 /

6av

The avargae velocity is found from the beginning and

ending positions and the total travel time.

V m s

The average speed is found from the total distance of

travel devided by the total tr

= =

{(0.5 ( 1.5)) (2.5 ( 1.5))}1 /

6

avel time.

Average speed= m s +

=

AB

C

00.5-1.5 2.5

3 2(6 9*6 15*6) 18At t=6 seconds, S ft

To determine the total distance traveled in 6 s, we

must find when the particle changes its direction. This

occurs when the drivative of the position is zero.

ds

st

= + =

23 18 15 3( 5)( 1) 0

1, 7 5, 25 .

t t t t

at t s ft and t s ft

= + =

= = = =

7 ft

-18 ft

-25 ft

7 ft

-18 ft

-25 ft

Total distance=

7+(7-(-25))-(-25-(-18))=46 ft

Total distance, over time t=6sec

For t=1, s= ( 1-9 +15)=7

t=2, s= 2

t=3, s= -9

t=4, s= -20

t=5, s= -25

t=6, s= -18

Problem 1