Kinematics of circular Motion

CDO
Angular Position O and AngularDisplacement

fparticle
Tr
Angle made by Position rector with a fixed
t ref reference line is called Angular Position O
fixed

AngularDisplacement Da Of Oi change in Angular

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Anglecovered Position

NOTE i Units SI unit radian nad

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A had 180 degrees

IS IIT
I revolution 21T had

B
ii
MR
A
Angdisplacement AngularPosition
depends on frame of reference as well
IT O
pathof as they are dependenton position
particle of observer in the same frame
H SF

DO Dez
to
ROSIC

0 Oz

Iii Angle covered in large is scaler but differential angular

interval
displacement is rector quantity The direction of dot is perpendicular
BYHY

to Plane containing do can be found

using Right hand Thumb rule
B
A
yr A
t
P

B qd0
todo
do
O O do

Angular velocity of a particle 8

Angularvelocity B is equal to rate of change of AngularPosition or
Angular displacement
w DOI
dt
NOTE
 d1
I 9pm 2 hadIsec
Gerolutionteeming

Laps 2T hadCsec
revolution peesec
dot
Ii Direction of w is
alongis hence it can also be found using Right
hand Thumb rule it L to the plane containing do

Ciii Angularvelocity Os in generalMotion

similar to DT Os is also observed position dependent

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i
ta nio v
Tdt

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vatsino

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do r
j a
t
ao

1 Jo
MR
0
IT O
do gaardcy.us Vdt Sino date VSino
r r
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W dd Y v1 is comp of T L to Positionvector et
rose
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d T
n
i
F I no Vt
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To

Angular acceleration T
P

It is defined as Rate ofchange of Angularvelocity

I did
dt
NOTE
Units i hadIsec2
Cii X ddwt dd.fz wddh.fr

Ciii Direction of 8
when 101 increasing then I is along w
 ng II 1 increasing
w
I
On yw F
I
Ydeceasing

eg ay
t A Particle starts moving from10,3
to Vo lmisec t TO with const velocity Imisec along
0,3 o
ol n ro
i the X direction
d u Prosino find angularvelocity angular
O
I w acceleration at secs about0
x At t 4See

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j tano 0 370
0 dy 2
w Vod ring VI Sin 13 239 233hadBec
D gm

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IS IIT
X
date Vf 25in0 Cos0 doff Vosign20 w

hadIsec
Lz 2 33 43
MR 233 6224
5
IT O
is thrownfromground initial speed lombec at an angle
eg A45Particle
with
with horizontal Find angular velocity of particle Wrt point of
H SF

Projection when
a Particle is at highest point
b u about to hit Ground
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A L
oil locos45 hmax 4074115232 2.5M
SEMISec
R
2 zg
tom
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ago
y5Y E et
Oe 1212
frontSec tons hey Iz
P

a semiSec
a At highest point
Vt 552Sind
r hmax
Sind
w Shima Sink 527g 252 hadBec

b Just before striking Ground

W hadHec
Lfa
A Particle is moving a circular path with const angular
eg w along
find angular velocity of Particle Wrt a point lying on
speed
 I ce of he circle
B
I
9W
w o
i
a7 iiiia n.A
ftp

Typesof circularMotions
u u u
Uniformcircular Non Uniform circular Non Uniform circular
Motion Motion with const AT Motion with variable at
speed v const AT d de

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w const ds
B distance a dw w dw
AT_const a const at TO

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af sV w Wot Xt W V DI
RI
a
Uta f

IS IIT
o
AR if ar
S ut Izatt wwootfztzf.tt
dt
g a w dI
f nar MR VE n zaps dt
V scalerform Variable
D I s distance
AR
YI
IT O
R
nu
w _d
H SF

atotal
rat
AT O v i Pa
e.az
ar
E
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JIaisgepays
atotal speed

D japitate
atotal Variable
diff dd ARI O
BYHY

tamp _AI AR
A Particle starts moving a circle of radius 2M with constant
eg along
P

acceleration 2misec2 Find

a time taken to cover first tworevolution
b Atotal at the end of 4 Secs
c time t distancetravelled till at _AR

at _2m15 const Using V Ut att

U O O 12 4
9 S 2 21112 ATM 8m15sec
AR v2
64m15
Using s ut z
Izatt at 2m15
41T 2 t t FT Sees atotal 4 25
 T 1
2 I misec

Using v ut att
52 0 Zt t LuzSec

S
I 2
E Im
eg when ceiling tan is switched off it makes
a 12 rotations till
its speed reduces to half of initial speed How many more
rotations will it make before coming to rest Assume uniform
retardation Using
wz Wor zag 2
WE 12 112 2T i

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2
further 02 122 n x21T ii
from i Cii
n 4

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Eg Particle is moving along circle such that at _4h5m1sec

IS IIT
find time t when total acceleration is inclined at 530with
radial direction R Im 2 Given a 1 41
ton 530 4th
MR
apetotal
i
na Eh g r
date
t
IT O
i
530K
here at 4th Sdv 14 Edt
k and AR o o
H SF

ar get 413
t 4 3
ROSIC

g
BYHY

S ITI Lt Oost i
2fotR
P

5 AT O 5m15 const
dat
hah
8 s mis2 O 1 210.531M

YI Is AR

2
a total attape jo s t
 to
eg

lait tart I
2 R
a AT deaf

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I t to I3

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AT
Again YI

IS IIT
VdIdg
s
J
Vo
d
Jdo
S
MR SIR
In v v e
sp
IT O
H SF

b Atotal jaffar BAR 52

25112
atotal rz.ve
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BYHY
P

cg
MABEE

K the const
D K
nv at Valdes CKrs C K 2fg KI
Aff
i
are k2S
R
 v
atotal AT ta µ Tz
g tant 292
x a D a tant

eg
Batio

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H J
0 0.21 2 i

IS IIT
att 2 5Secs w d 0.4T
W Oo4 2 5
I radlsec MR A ddhf 0.49rad sect
D hw r mise
r O 65m15sec 91 92 CO 4 r ii
IT O

O 4 0 65
H SF

AR 0 65 O 65
O 65
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Atotal jar 197 0.8 mised

BYHY
P
eg
a

Va VB

B A B

B
VA MAW
VB 913W

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velocity of B ITB Att VB VA
angular
wht A 1h73Al MB ha
when they are moving in opp W trad

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direction NB VA Sec

IS IIT
A T
B AthB A rest B
o
J f rest B ha
VB
MR
WBA VAL W
VB1VA GB19A
IT O
4radbee
H SF

b B B
VB VB
a o
s
E ga
v
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2h01
I NA ugoftp.A MB
r
r WBsA
h A A
BYHY

rest
WB a BOSO VASind
Gars
P

tanD I W
2
Sino
fg cos0
35
Dynamics of circular motion
i when Particle is moving along circular path Uniform1 Non uniform
a
there must be force acting always towards centre of circle which
provides radial accelerationcard to the Particle This force is also
known as centripetal force or Mosmalforce
Fcentsipetal MAR MI
R
Ii In Non Uniform circular Motion there must be a force
actingalways
tangential direction to the circular path which will provide tangential
acceleration to the particle This is also known as tangential force
FTangential MAT

iii while drawing FBD of a Particle in circular Motion forces should always
be resolved in following threedirections
a along radial direction Freet MAR
b along tangentialdirection Freet mat
c perpendicular to the Plane of circle FMet o

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w
eg i e e
2 I Gravity fseespace
are Rw2 s Ti m T2 2M

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Find ratio of Tension force in

IS IIT
AR 2lw2 the
two strings
aR lw
a Tz
a a 12 a
I 2M
MR m
T2 2M 2ewz
IT O
T Tz m ew2
72 4me w
H SF

we get
solving T SmeWZ
T Tz 5 4
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eg w Find Max angularspeed of

U
Disc such thatblockdoesnot
BYHY

u
m prough slip Wrt Disc
Disc nH M i
fs mg
P

AR MAR Mh 2
m fsmax Merwinax Umg
ts
Wmax
w
mg fge
r
EV
eg I u
enough
find comin such that block does not
m
ER
i slip Wrt cylinder
isfsmax UM
M MAR MRWZ
m UN
L e m radial Mg
a
AR
 mg mg
w comin

eg Mass is given Tangentialvelocity such that

p it revolves in a circular Path with speed
I l w find ang
I w a Angle 0
I E b Tension in
i m
string
r T r i
ai
conical Pendulum I radial

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h LSino ar
ar hw sin w v
mg

H J
IS IIT
Tcoso sis
1since_mg
maps mcesino w ii
MR solving we get s g mew
IT O
eg Cosa 0 cos169g
_e
H SF

A Particle describes a horizontal

a role on the smooth inner surface
ng of a conical funnel as shown at
ROSIC

a height h find speed of Particle

AR Ir r i NSind
r i Mcoso
mg MI
MAR
BYHY

radial r
ar tano 8 Ig
P

Fg v
fgh
eg

do12 REE
DOK
dm Rdf
p
Tet
2

Kif
a e

AR
R
I
Tsin Tsin dm AR
 centripetalforce
2T Sin doz
offdo
47 do MI
227
d 1
21172

Vehicle on a horizontal turn

A car with const speed on a horizontal turn
travelling
const
fs Mar MI
R

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ar Max safe speed during a turn
fsa V U
R fsMax mVmkx

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IS IIT
m ma Vmax
umg JugR
CIT on a
MR
horizontal road with variable speed
IT O
fs p ar fs matotal jar 1912
aAT
H SF

atotal
radial cahfswiffsyjnasxtftuahtmg.sn'pping
ROSIC

Tangential
Topview
BYHY

CIT Hoeidontal Turn on a banked road

µ speed u when friction
roi is not required
P

c i radial Mcoso
is Msince mg
Mar M 2
yo
tone
g KrgRiano
ring
Max speed limit if friction coeff is ee
µ Moose Mgtunesino
ro resinoteerecoso MvYzax_
c i radial
is
an 0 VMax grcsinotucoso
 Cosa usino
MM
ing
mat

eg A car starts moving along a rough circular path U with const

acceleration ao find the Total distance travelled till it starts
slipping on the track radius R Car will start slipping when
ay ra
fs fs Max
a fSnax Ma total
car 40
AT ao Umg Mj aft ape

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to af 1
g
v2 cis
RjµgJ af

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IS IIT
Using
v2 u2 12 as
2 O t 2905
MR s
zaojeegj a.ie
yato
IT O

concept of centrifugal force

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UW FBD of mass wht ground

O m nM
ROSIC

ar N
s fs
ng
maps Mhw2
Disc fs
ing
BYHY

FBD of mass coat Disc

nm M
fs Mg
P

rest 0
correct FBD Wrt Disc X
nm
umg
fcentrifugal
a mawz i It is applied Shown in FBD when
fs rest it is drawn wrt Non inertialrotating
my fs Maw frame
ii fcentrifugal Mhw2 m mass ofbody
9 distanceofbody
from axis ofrotation of
fsame
angularvelocity of fame
W
 s d radial ne goi ax of tot on
fromngaxis
along ng 0dg
away
smooth
eg
no f find w such that block
Efw l restrn 43
remains at rest wet elevator

I 10
FBD of block wht Elevator
H r g
the mewl
rest resino meat

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Moosa
mg
MG tone w

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egad

IS IIT
jgtqno­ww
loradls.ec

MR 10kg 5kg
Mt f 0.3M m2
IT O
e 0.12amto
se0.176Ms
Tho
H SF

µ 0.5
fsmax 0.5 10 10
ROSIC

50M
Wht Table
nM nM2
Min w Ml T T
BYHY

e s e o
126M a Mzhzw2
fs 88M
v
P

Mig Mag
2
Min w 10 0.124 102 fs 124 88
124M
mzhzw2 5 0.176 102 fs 36M
88M
b At the moment of just slipping
fs fsmax MiG WI Mz92W
Wo 50 11.67 had Sec
10 0.124 5 0.176

c If resultant of centrifugal forces is zero then fs O

 Mini w Mah's w Iori Sh's i
and h t h's O3 m ii

solving hi 0 im
Mj 0.2M

NOTE centrifugalforce is sufficient Pseudoforce if Body is at rest

Wrt rotating Platform But if body is not at rest then we
also apply one more pseudo force called Coriolis force along
with centrifugal
FToriolis 2M Tx F T velocity of body Wrt
Rotating frame

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I
Floriolis can not ang velocityof frame
changespeed of object
it can only direction of Motion

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change
because force is always 1 to velocity

IS IIT
a
Nz will balance
MR
Fcoriolis
IT O
N will balance
H SF

Mg
W oM M2
v
0
ROSIC

m Mhw2
ha
Mr
I r 0 a
Tzonidisa
BYHY

mg
A Udder Mh WZ hw 2
M U L
P

Judo
o a
hw dr D Wj v ar
 P
BYHY
ROSIC
H SF
IT O
MR
IS IIT
H J
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