BEF 25903 - Kinematics of Translating and Rotating Bodies

8/19/2019 BEF 25903 - Kinematics of Translating and Rotating Bodies

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Kinematics of

Translating and

Rotating Bodies


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• Kinematics studies the motion of particles or bodies,

without considering the mass and without regard to the

cause of the motion.

• Kinetics describes the speed and acceleration of a

particle or a body and the relations between them.

Introduction


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The kinematics of a particle is characterised by specifying at

any given instant the particle’s

. !osition

". #elocity

$ %cceleration.


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Rectilinear Kinematics

Rectilinear kinematics is concerned with the motion of a particle in a

straight line.


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&onsider a particle ! moving on a straight line along the s'a(is.

The position of the particle at any instant is specified by the

position vector r , measured from ).

)

!r

s s

Note

. The direction of r is always along the s a(is

2. s is a scalar function representing the magnitude and

sense of r as a function of time.

POSITION


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The displacement of particle ! is defined as the change in its position.

)

!r

s

s

*r

*s

r’

s’

+n vector form -isplacement *r r ’' r

+n scalar form -isplacement *s s’ ' s

DISPLACEMENT


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The distance travelled is a positive scalar which represents the total

length of the path over which the particle travels.

)

!

r

s

s

*r

*s

r’

s’

!’

!ath travelledby particle t i m e

DISTANCE


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. The minimum distance between two points is called

displacement while the actual path covered is called

distance.

". The displacement is a vector term and distance is scalar

term.

$. -istance and displacement both have /+ unit as meter.

NOTES


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+f the particle moves through a displacement *r from ! to !’ during

the time interval *t, the average velocity of the particle during this

time interval is

t∆∆= rv

O

!

r

s

*r

r’

!’

∆t

time

AVERAE VELOCIT!


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The instantaneous velocity of particle ! is defined as the velocity at

the limit where *t01, i.e.

)

!r s

*r

r’

!’

dt

d

tlim

0t

rrv =

∆

∆=

→∆

INSTANTANEO"S VELOCIT! #2&T)R -23+4+T+)4


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Instantaneous #elocit$ /calar -efinition

)!

s

s

*s

s’

!’

dt

ds

t

slimv

0t=

∆

∆==

→∆v

t i m e


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Speed o% a particle

) !’s

!

The average speed of a particle is defined as the total distance travelled

by the particle, sT, divided by the elapsed time, *t5 i.e.

sT

t

sv T

av

∆=

t i m e

∆ t


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A#era&e acceleration vector definition

The average acceleration during the time interval *t is defined as

t∆

∆=

v

a

)

! s!’

# #’

where *# #’ ' #


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Instantaneous acceleration vector definition

The instantaneous acceleration during the time interval *t is defined as

dt

d

t

lim0t

vv

a =∆

∆=

→∆

)

!s

!’

# #’


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Instantaneous acceleration scalar definition

The instantaneous acceleration during the time interval *t is defined as

dt

dv

t

vlima

0t

=

∆

∆=

→∆ )

!s

!’

v v’

Alternate %orm

6sing v ds7dt, we can also e(press the acceleration as

2

2

dt

sda =


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E'uations o% motion 8eneral velocity' speed relationship

3rom the e9uation v ds7dt, we can write

vdtds =

+ntegrating both sides of the e9uation, we obtain

∫ ∫ =t

t

s

s 00

vdtds

Therefore,

0

t

t

svdts

0

+= ∫

where s( displacement at initial time t1

s displacement at time t.


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29uations of motion constant velocity


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3or the special case where v is constant, we get

0

t

t

sdtvs

0

+= ∫

00 s)tt(vs +−=

or

0svts +=

Note

. :hen the initial time t1 is 1, the e9uation reduces to

". +n the constant velocity condition, the acceleration a is ;ero, since

0dt

dva ==


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E'uations o% motion 8eneral acceleration'velocity relationship

3rom the e9uation a dv7dt, we can write

adtdv =

+ntegrating both sides of the e9uation, we obtain

∫ ∫ =t

t

v

v 00

adtdv

Therefore,

where t( initial time t1

v1 velocity at initial time t1

v velocity at time t

∫ =−t

t

0

0

adtvv


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E'uations o% motion constant acceleration

3or the special case where a is a constant, we get

0

t

t

vdtav

0

+= ∫

00 )( vt t av +−=

or

0vat v +=

Note

. :hen the initial time t1 is 1, the e9uation reduces to


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%n electron moving along the ( a(is has a position given by (


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To find the velocity of the electron as a function of time, take the first

derivative of x(t)

again where t is in seconds, so that the units for v are m7s.

4ow the electron momentarily stops when the velocity is ;ero. 3rom

our e(pression for v we see that this occurs at t s. %t this

particular time we can find the value of x

The electron was A.C m from the origin when the velocity was ;ero.

mms 89.5)1(16)1( 1 == −e x

m/s)1(161616 t eteedt

dxv

t t t −=−== −−−

SOL"TION


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/mall blocks are being dropped into a $'foot'wide chute by means

of a conveyor belt from a height h as shown in the figure. +f h = ft,

determine the range of velocity v o of the conveyor belt for which the

blocks will fall into the chute. 4eglect air resistance. Det g $"."

ft7s".

E*ERCISE


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+f the conveyor belt in 2(ample , is moving with a velocity of "1 ms7s,

determine the range of the height h of the conveyor for which the blocks

will fall into the chute.

E*ERCISE


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rap+ical Solution o% Rectilinear/Motion

Pro,lems

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• 8iven the x-t curve, the v-t curve is e9ualto the x-t curve slope.

• 8iven the v-t curve, the a-t curve is

e9ual to the v-t curve slope.


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• Given the a-t curve, the change in velocit !et"een t 1

and t 2

is

e#ual to the area under the a-t curve !et"een t 1 and t

2.

• Given the v-t curve, the change in $osition !et"een t 1 and t 2 is

e#ual to the area under the v-t curve !et"een t 1 and t

2.

RAP3ICAL SOL"TION O4 RECTILINEAR/MOTION

PRO5LEMS


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• Moment-area method to determine $article $ositionat time t directl %rom the a-t curve&

( )∫ −+=

−=−

1

0

110

01 curve underarea

v

v

dvt t t v

t v x x

using dv ' a dt ,

( )∫ −+=−1

0

11001

v

v

dt at t t v x x

( ) =−∫ 1

0

1

v

v

dt at t %irst moment o% area under a-t curve

"ith res$ect to t ' t 1 line.

( ) ( )

C t

t t a-t t v x x

centroido% a!scissa

curveunderarea 11001

=

−++=

OT3ER RAP3ICAL MET3ODS


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• ethod to determine $article acceleration

%rom v-x curve&

==

=

=

BC

AB

dx

dvva

θ tan

subnormal to v-x curve

OT3ER RAP3ICAL MET3ODS


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Kinematics o%Circular Motion


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CIRC"LAR MOTION

&onsider the figure on the right which

shows a particle ! moving in a circle ofradius r.

r

)

θ

ω

s

!The angle θ made by theradius vector r and the

reference line )% can be

defined in terms of thelength of r and the arc

length s as follows

r

s

=

%

The unit of θ is the radian and isdimensionless.


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/ince the circumference of a circle of

radius r is "E, angle subtended by ! inmoving a full circumference is then

r

)

θ

ω

s

!

π=π

= 2r

r 2

%Thus we obtain the

proportionality relation

*602

degrad θ

=π

θ


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An&ular #elocit$

/uppose that the particle !

rotates in the anticlockwise

direction and sweeps through

an angle *θ during the timeinterval *t. The average

angular velocity made by ! isdefined as

r

)

θ

ω

s

!

t∆

θ∆=ω %

!’

*θ

*s

Note

. The unit of angular velocity is

the radians7second.

". The dimension of angular velocity is T '.


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An&ular #elocit$

The instantaneous angular

velocity ω is obtained in thelimit *t01, i.e.

r

)

θ

ω

s

!

dtd

tlim

0tθ=

∆θ∆=ω

→∆ %


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An&ular Acceleration

The instantaneous angular

acceleration α is defined asthe time diffential of the

angular velocity, i.e.r

)

θ

ω

s

!

dt

d

tlim

0t

ω=

∆

ω∆=α

→∆ %


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Tan&ential #elocit$

+f *s is the tangential distance

travelled by the point ! in the

time interval *t, then the

average tangential velocity of

particle ! is

r

)

θ

ω

s

!

t

sv

∆

∆=

%


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r

)

θ

ω

s

!

%

The instantaneous tangential

velocity v is defined in the limit

*t01, i.e.

dtds

tslimv

0t=∆∆= →∆


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r

)

θ

ω

s

!

%

3or small increments, we can

write ds r dθ. ?ence,

ω=θ

== r dt

dr

dt

dsv

Relations+ip ,et-een an&ular #elocit$ and linear #elocit$


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Tan&ential acceleration

The average tangential

acceleration of particle ! is

defined by

t

va

∆

∆=

The instantaneous tangential

acceleration a is defined in

the limit *t01, i.e.

r

)θ

ω

s

!

%

dt

dv

t

vlima0t

=∆∆=

→∆


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r

)θ

ω

s

!

%[ ]α=

ω=

ω== r

dt

dr

dt

r d

dt

dva

6sing v rF, we can write

Relations+ip ,et-een linear acceleration and an&ular

acceleration


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r

)θ

ω

s

!

%

dtd α=ωdt

dω=α

dtd

0 0

t

t

∫ ∫ ω

ω

α=ω

3rom we can write

Therefore

?ence

∫ α=ω−ωt

0

0 dt

or

∫ ω+α=ωt

0

0dt


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r

)θ

ω

s

!

%

3or the special case of linear angular

acceleration, we obtain

∫ ω+α=ω

t

0

0dt

0t ω+α=


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Relation ,et-een an&ular #elocit$ and speed o% rotation

+f F is the angular velocity, then.

?ence

dt

dω=α

α

ω=

ddt

+ntegrating both sides we obtain

∫ ∫ ω

ω α

ω=

0

t

to

ddt

or

∫ ω

ω αω=−

0

dtt 0

where Fo is angular velocity at time to and F angular velocity at time t.


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3or the case of constant angular acceleration we can bring α to theleft of the integral sign and write

∫ ω

ω

+ωα

=0

0td1

t

?ence

( ) 00 t1

t +ω−ωα

=

%ngular motion ,

dt

dn2π=ω

or

dt2

dnπ

ω=

?

ence

τωπ

= ∫ d2

1n

t

0

3or the special case of w constant, we have t n ω π 2

1=


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3or the special case of w constant, we have

t n ω π 2

1=


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E6ample

% direct'connected pump and motor accelerate uniformly from rest to

GA1 rpm in 1.$ second. -etermine the angular acceleration.

Solution

8iven that the acceleration F is constant, we have

t n ω π 2

1=

?ence,

rad+s86.610*.0

60

1,502

2=

⋅⋅==

π π

ω t

N


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E6ample

% rotating drive shaft decelerates uniformly from C11 rpm to =A1 rpm in =

seconds. -etermine the angular deceleration and the total number ofrevolutions in the ='second interval.

Solution

complete rev "E rad

4 rev "E4 rad

4 rev7s "E4 rad7s

4 rev7min "E47=1 rad7s

3or constant deceleration case, we have

( ) 00 t1t +ω−ωα

=

?ence,

( )( )

rad+s*6*.-

6

60

6509002

0 =

−

=

−

−=

π ω ω

α

ot t


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E6ercise

% table fan, rotating at a speed of "H11 rpm, is switched off and the

resulting variation of the rpm with time is shown in 3igure . -raw the a't and I't curves,

-etermine the total number of revolutions the fan has turned.

JAns. "C rev

600

2-00

ω (r$m)

8 2-

t (s)


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E6ercise

Block B in 3igure moves downward from rest with a linear acceleration of "

m7s

"

. &alculate the linear velocity v and displacement s of the block, and theangular velocity and displacement I of the pulley at the end of A s.

r

r ' 0.5 m

O /

/


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%ngular acceleration L d"I7dt" is given by

Solution

2rad.s-5.0

2

r

a===α

where a dv7dt is linear acceleration, and a " m7s.

Dinear velocity v is related to the linear acceleration a via the e(pression

o

t

0

vadtv += ∫

where v1 is the initial velocity, and v1 1. 3or the case of linear acceleration

starting from rest,

atdtav

t

0

==

∫


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Therefore,

v at " ( A 1 m.s'

%ngular velocity

atd

t

0

=τα=ω

∫ 3or the case of constant angular acceleration,

td

t

0

α=τα=ω

∫ Therefore,

1rad.s205-t ==α=ω


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%ngular displacement is related to angular velocity via the relation

∫ τω=θt

0

d

3or the case of constant angular velocity,

td

t

0

ω=τω=θ ∫

Therefore,

1rad.s100520t ==ω=θ


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E6ercise

+n the manufacture of steel sheet, the sheet, which has a velocity of "m7s, is drawn between two rollers. +f the rollers are 1 mm in diameter,

determine their speed of rotation in rpm.

180 mm

180 mm

v ' 12 m+s

BEF 25903 - Kinematics of Translating and Rotating Bodies

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Transcript of BEF 25903 - Kinematics of Translating and Rotating Bodies