Dynamics chapter 1

7/26/2019 Dynamics chapter 1

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Dynamics MDB 2043

Rectilinear Kinematics: Continuous Motion

May 2016 Semester

Lesson Outcomes

At the end of this lecture you should be able

to:

Determine the kinematic quantities (position,

displacement, velocity, and acceleration) of a

particle traveling along a straight path.


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Rectilinear Kinematics: Continious Motion

A particle travels along a straight-line

path defined by the coordinate axis 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.

Theposition of the particle at any

instant, relative to the origin, O, is

defined by the position vectorr, or the

scalar s.

Scalar s can be positive or negative.

Typical units for rand s is meters (m).

The displacement of the particle is

defined as its change in position.

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

Position, Displacement & Distance:Example

0 1 2 5 64 73-1-2-3

Time, t (sec) 0 2 4 6 8 10 12

Position, x (m) 2 3 6 7 3 0 -2

From time t=0 to t =12 sec,

The object moved from positionx= 2 m to another positionx= -2 sec.

The displacement of the object is -4 m. [ s = -4 m] (must state direction)

The distance travelled by the object is 14 m. (No information on direction but

must know the path taken)


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Velocity

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

It is a vectorquantity (it hasboth magnitude and direction). The magnitude of the velocity is called speed, with unit of m/s .

The average velocity of a particle during a

time interval t isvavg

= 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

1

Acceleration

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

It is a vectorquantity.

Typical unit is m/s2 .

Note that, the derivative equations for velocity and acceleration can be

manipulated to get a ds = v dv

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).

2

3


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Summary of kinematic relations: Rectilinear motion

Differentiateposition to get velocity and acceleration.

v = ds/dt

a = dv/dt

a = v dv/ds

Integrate acceleration for velocity and position.

Note that so and vo represent the initial position and

velocity of the particle at t = 0.

Velocity:

t

o

v

vo

dtadv s

s

v

v oo

dsadvvor t

o

s

so

dtvds

Position:

1

2

3

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 downward.

These equations are:

tavv coyields t

o

c

v

v

dtadvo

2coo

s

t(1/2) atvss yields

t

osdtvds

o

)s-(s2a)(vv oc2

o

2 yields s

sc

v

v oo

dsadvv


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Velocity as a Function of Time (v=v(t))

adt

dv2

t

t

t

t

v

v

dtaadtdv000

t

t

v

v tav

00

)( 00 ttavv

Position as a Function of Time (s=s(t))

vdt

ds1 dtttavvdtds

t

t

t

t

s

s

000

)( 00

2

0000 )(2

1)( ttattvss

Velocity as a Function of Position (v=v(s))

adsvdv3

s

s

v

v

adsvdv00

)(20

2

0

2ssavv

In special cases

atv 2

2

1ats asv 2

2

t0=0, s0=0 and v0=0

t0=0 atvv

02

002

1attvss )(2 0

2

0

2ssavv

Variable Acceleration

Depending on the nature of a problem, acceleration (a) may also be

known in different forms including

(a) a is a given function of time a=a(t)

(b) a is a given function of velocity a=a(v)

(c) a is a given function of displacement a=a(s)

(a) Given a=a(t), develop v-t and s-t relationships

)(tadt

dv

2

t

t

v

v

dttadv00

)(

t

t

dttavv0 )(0

Tip: Velocity v(t) as a function of

time can be found by integrating a(t)

)(tvdt

ds1

t

t

dttvss0

)(0Tip: Distance s(t) as a function oftime can be found by integrating v(t)

t

t

s

s

dttvds00

)(


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(b) Given a=a(v), develop v-t and s-v relationships

)(vadt

dv2

t

t

v

v

dtva

dv

00 )(

00 )(

ttva

dvv

v

dsvavdv )(3

s

s

v

v

dsdvva

v

00 )(

00 )(

ssdvva

vv

v

This gives a relationship between velocityvand time taken t.

This gives the distance travelled s before

the velocity vis reached.

(c) Given a=a(s), develop v-s and s-t relationships

dssavdv )(3

s

s

v

v

dssadvv00

)(

s

s

dssavv0

)(2202

This gives velocity v(s) as a function of distance s.

)(svdt

ds1

s

s

t

t sv

dsdt

00 )(

s

s sv

dstt

0 )(0

This gives a relationship between distance s and time taken t.


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References:

R.C. Hibbeler, Engineering Mechanics: Dynamics,

SI 13th Edition, Prentice-Hall, 2012.

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Dynamics chapter 1