BEF 25903 - Kinematics of Translating and Rotating Bodies
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Transcript of 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
00 / 12
• 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