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UNIT 03 ))
Circular Motion:
Examples:
"Motion of a body moving in circular path or motion of abody whose distance from axis of
rotation remains constant is called circular motion ",
Motion of satellites around the earth.
Motionof car moving on a circular track.
Motion of stone tied with a string, rotating in acircular path.
Examples:
ROTATIONAL AND
"Angle subtended at the center in small internal of time or angle Ae which gives the change in
angular position of a body is called angular displacement".
CIRCULAR MOTION
)
If body moves from point Ato Bon circular path then its angular displacement is A
P:
P
ANGULAR DISPLACEMENT
(a)90°
Note One radian is the angle betwe n two radii which cut off on the circumference an arc equal to radius
SI unit of angular displacement is radian and other units are degree, revolution etc. For small value, angular displacement is vector quantity.
Right hand rule :
For large value of angle angular displacement is not vector because it does not obey the
vectors lawssuch ascommutative law (8, + , = 0, +8, ).
Example: If a body moves from one end of the diameter to other, then angular
displacement of the body will be (b) 1800
Note Rotational motion is either two or three dimensional motion and cannot be one dimensional.
(c) 270°
Direction of angular displacement is along axis of
rotation and it is determined by right hand rule. (Rotate fingers in direction of rotation while keeping the thumb erect then thumb indicates the direction of angular displacement).
(d) 360°
angular displacement
The angular displacement is assigned a +ve sign when sense of rotation is counter
clock wise (Anti-clockwise). Angular displacement is assigned -ve sign when sense of rotation is clockwise.
Relation between Linear and Angular displacement:
If a particle moving in circular path of radius r. covers
an arc length S and angular displacement then
If abody is rolling without slipping then its linear
distance is equal to arc length.
Examples: If a (^) wheel of radius (^) 0.Sm is (^) rolling (^) without slipping then its linear distance covered in 3-revolutions will be (a)l.5m (c) 5.4m
Conversion of degree into radian:
30° =30 ×
usuoSdegree
45° = 45 X
60° = 60 x
S= r
zrad =
1800 2
(a) 2 rad
180
180
180
180
(b)2m (d) 9.4m
Conversion of Radian into Degree:
Trad = 1800
T rad
L rad
3
= 900
rad
180°4 Snrad
3 2 rad =
Sehold j"S = r8"
-SrtRadians
45º
Some important conversions
60°
solution:
1rev = 360° = 2I rad
1rad = 57.30 =
S = r0= 0.5 x3 x 2T =3m = 9.4 m
(c) 6TI rad
(1rev = 27trad)
(^) 30°= (^) rad= 1 6
3x 2
1° = 0.0174 rad =
4
3
12 1
90° = rad =
=*rad= 7rev 1 6
2700
;rad rev
4 180° = nrad =
1800
(d) 9 rad
1
180
2
rev
rev
Examplel: if^ a^ body (^) covers (^) three (^) revolutions in (^) 5seconds (^) then its
Solution:
angular displacement in SI units will be (b) 3n rad
rev
3 rev 3x 2 rad =67 rad
Uniform Angular Velocity:
If body covers equal angular displacement in equalintervals of time then bcdy is rotating with uniformangular ve.vcii. a= 0
a¡y
I= 0
aeiAngular acceleration"»neeAngular velocity"'bUn Constant&4eSYS
Rate of change of angular velocity of a body is
called angular acceleration.
etc.
Instantaneous Angular velocity:ib
Angular velocity of a body at any particular
instant of time is called instantaneous angular
velocity.
ANGULAR ACCELERATION
Total change in angular velocity Total time
At Angular acceleration is a vector quantity and its direction is always along the
direction of torque.
If w is increasing
SI unit of angular acceleration is rads-
and other units are degs- and revs-?
Uniformn Angular Acceleration; If angular velocity of a body changes equally in equal intervals of time then body is moving with uniform angular acceleration.
|fangular velocity is
increasing then angular acceleration is positive and parallel to angular velocity.
w = lim At’0 At
If angular velocity is decreasing then angular acceleration is negative and anti-parallel toangular
velocity.
If w is decreasing
ens,SeybtAngular acccleration
-esuntleGAngular velocity
Linear acceleration is caused by
force, similarly angular
acceleration is caused by torque
Instantaneous Angular Acceleration: Angular acceleration of a body at any particular instant of time is called instantaneous angular acceleration.
T= la
Cins
Aw =lim At’0 t
If angular velocity is constant then angular acceleration is zero and net torque acting on the body is also zero
a= 0 If wis constant
Note:4body moving in a circular path may have:
i.
ii.
iii.
|0v.
ii.
i.
I.
I.
Tangential acceleration (due to changing speed of the body). Angular acceleration
(due to changing angular velocity of the body).
Centrinetal acceleration
(due to changing direction of linear velocity of the body).
If body is moving in circular path with uniform speed or uniform angular velocity then body has only centripetal acceleration due to changing direction of velocity and a =0 and a, 0
i.
RELATION BETWEEN LINEAR AND ANGULAR VARIABLES
S= r V,=w or a, =ra^ or ,^ =^ ßx
Note: For a rotating^ rigid^ body,^ all^ particles^ of^ rigid^ body^ will^ have^ same^ angular displacenment , angular velocity^ w^ and^ angular^ acceleration^ a^ but^ values^ of^ S,^ v^ and a,may be^ different^ depending^ upon^ the^ distance^ r.
84 = g= c WA = Wp = Wc
Equation:
Equations of motion for angular motion:
Equation:
II. Equation:
Limitations:
Wf = W; + at
= W;t+
1 2
2a8 = w t w
These equations^ areapplicable^ only^ if Angular acceleration^ a^ is^ uniform.
tAngular displacement"s 0= w,t+« 2 t
(V,,Y and Ware^ always^ perpendicular^ to^ eachother^ ) (a,, Yand ä are^ always^ perpendicular^ to^ eachother^ )
But Sç > Sp SA Vc >VB> VA aç > aB > aA
a , aç and a are always mutually perpendicular.
ii.
-uuse "cquation rtd without 8
-u/use 2nd equation t/p without W;
-yuse d cquation rtsd without time
Axis of rotation does not change.
7xtAngular displacement"nh
8, = Wt (2n - 1)
R
Centripetal force depends upon Mass of the body
Fc
Speed of the body
Radius of circular path F x
Various types of graph for centripetal force :
Examplel:
Example2:
Example3:
F, « mn
Satellites revolving around the earth. Force of gravity
provide the required centripetal force.
F =g
my
=
ExYmple4:
F. = F my?
GMm
Electrons revolving around the nucleus. Electric
force provides the required centripetal force.
Kqi r
F = T
qvB =
1
|GM
r
A stone tied to a string moving in circular path. Tension in string provide required centripetal force.
Fm= Fe my
qBr
m
For greater^ masSS,^ greater amount^ of^ force^ is required to^ bend^ the^ body^ in^ a^ circular^ path Withgreater^ speed, greater^ amount^ of^ force^ is required to bend the body in a circular path greater amount^ of^ force^ is^ required^ to^ bend^ thbe body in^ a^ circular^ path^ of^ shorter^ radius
A charge moving in circular path in a magnetic field.
Magnetic field force provides the required centripetal
force.
B in
X
Fo
ELECTRON
X
Fe
(invilatonal forne
NUCLEUS
X
X
1
X
(enintug:force
Example5: A car moving in circular road. Force of friction provides the required centripetal force. Banked (^) tracks are (^) needed for turns (^) that (^) are taken so (^) quickly that (^) friction alone (^) cannot provide required centripetal force
Case i.^ If force acting on a body is equal to required
centripetal force then body will move in circular path.
Case i.fforce acting on a body is greater than
required centripetal force then body will fall towards the center of the circle. Case ii. If (^) force (^) acting on (^) a body is (^) less than (^) the required centripetalforce then body will move out of circular.
Centripetal acceleration:
Centripetal acceleration is always perpendicular to velocity.
Instantaneous acceleration of thebody moving in a circular path with uniform speed is
always directed towards the center of the circle. It is known as centripetal acceleration.
Direction of centripetalacceleration is always directed towards the center but its direction
iscontinuously changing with time
º Centripetal acceleration is due to changing the direction of velocity.
F
In terms of speed my
Important Expressions for CentripetalForce and Acceleration
i
Ify = Constant 1 F, a
a, =
In terns if: angular speed
F= mru? If w = Constant F,ar
a, = rw
In terms of In terms of time period Imomentum
F, =
41'mr
T
4n'r T?
Various types of graph for centripetal Acceleration :
F, =
a, =
(Earth
p m'r
In terms of K.E
F, =
a, =
2K.E
2K.E mr
tuctiottalforce
1
In terms of p & v
F=
pv
pv mr
Centripetal force
Constantquantities
IMomentum
-my =-t=
Under the action of only
centripetal force following
quantities remains constant Speed, kinetic energy,
angular speed ,time period ,
angular momentum, magnitude of velocity and magnitude of linear
D
rcos
-my
C
El
r
A
If a body of mass 'nm' tied with a string of length 'r ismoving in a vertical circle in the gravitational field as shown in the figure below:
Under the action of only
centripetal force following
quantities remains zero.
Work done , change in kinctic
energy ,^ tangential^ acceleration,
angular acceleration , tangential
force, torque produced by
centripetal force, change in
angular velocity and change in angular momentum.
MOTION IN AVERTICALCIRCLE
T
Quantities which are zero
mgsinb
In Vector Form
mg
Centripetal acceleration
B
mgcose
º Velocityof the body at any point is given as
v=2g(hy - h) + v?
v=/2g(r + rcos0) +gr
v=/3gr +2grcos
Since vertical distance = h - h, =r+ rcose
and at point C° vf = gr. Hence
Quantities which are changing direction Under the action of only
centripetal force
magnitude of following
remain constant but their
direction changes
velocity acceleration, momentum and force.
For minimum velocity at point C° just source of gravity will provide the required centripetal force. myz =mg
V=/gr
At point 'P the resultant force along the radius will provide require centripetal force.
F =T- mgcose
Tension in a string at any point is given as
T=F + mgcose
my T= + mgcose
OR
Position
Velocity
Tension
v=/3gr +2grcos v=/5gr(maximum)
T=
A 0=
my
my
my
V=
T= + mg (max.)
T= 6mg(v=/5gr)
-+ mgcos
Minimumn velocity required to put a satellite in a circular orbit is called orbital velocity.
F = Ig GMm r
r
GM
v=/3gr+ 2grcos
T=
B 0= 90
ORBITAL VELOCITY
If velocity is equal or greater than escape velocity it will escape from earth's gravity
3gr my
T=3mg(v=3gr)
º If velocity of satellite is less than critical velocity (v< 27000km/h) it will fall towards earth.
If velócity is equalto critical velocity (v = 27000km/h)then it will move in circular path.
T=
If velocity is greater than critical velocity but less than escape velocity it willmove in an elliptical path.
my
1
satellite.
GPS system.
GM R
V=/3gr + 2grcos
Note: V « and independent of mass of
C 0 180
v=/gr minimum)
T=
V<27000 kmh
(Earth)
T= + mgcos
my
If satellite is revolving around the earth near its surface then r = R
myz
and T = 5060 sec s 84 min Minimum height of satellite' revolving around iisorbital abnjheight ýsatellite so*the earth is 400 kmand 24 such satellitesform
T= 0(v= gr)
V= =/gR =7.9 km/s
V=27000km/h
Sy>40000 km/h
V>27000 km/h
Angular displacement
Angular displacement nth second
Angular velocity
Angular acceleration
1 Equation
2md Equation
IMPORTANT FORMULAS SUMMARY
3 Equation
Newton 2nd Law for angular motion
Angular Momentun
Work done
Rota ional Kinetic Energy
Centripetal Force
Centripetal force for vertical circle
Centripetal acceleration
Tension in^ a^ string at any
point of yertical circle
a= At
At
my F =:
a, =
t
T=la
a, =
p?
8,, = W; t(2n - 1) 2
m'r
T=F +mgcos
2rad
a=
Wf = W; + at
F = mrw?
8= w;t +at' 2
2a0 = w + w
1
W= T
L=lw
K. Erot 5lu
1
a, =rw?
F, = (^) mr
F,=T- mgcos®
T=
9= wt
T=
aç =
a, =
AL At
my?
F =
4nr T?
2K. E
at
mr
2K. E
