Important Formulas: Circular Motion | Physics for JEE Main & Advanced PDF Download

 Page 1


FORMULAE SHEET
?(a)	
1
F Fcos component of F along AC = ?=
?
    
2
F Fsin component of F perpendicular to AC = ?=
?
 
F1
F2
F B
AC

Figure: 4.73
?(b)	
dp d(mv)
F
dt dt
= =
?(c)	
net
net
F
F F ma or a
m
= = =
?
?
?? ? ?
     
T
F=ma
P
mg
Figure: 4.74
        
x
y
1
F 0 Tsin ma
F 0 Tcos mg
a
tan
g
-
= ? ?=
= ? ?=
? ?
?=
? ?
? ?
?
?
? ( d ) 	 Impulse	=	change	in	momentum        
fo
F t mv mV ?= -
?? ? ?? ?
?(e)	
max
f
R
µ=
 
    
A
Force of
Friction
Force Applied, F
O
f
max
f
K
f
S
?(f )	
max limiting s
ff R = = µ
A
R
O
f
m

Figure: 4.76
? ( g ) 	 Angle	o f	Fr iction:         
max
f
tan
R
?= = µ
     
1
or tan ( )
-
?= µ
? ( h ) 	 Pseudo	fo r ce:	  F = –ma;   where m = mass of the object, a = acceleration of the reference frame
?(i)	 A particle in circular motion may have two types of velocities as listed hereunder.
        (i)  Linear velocity v and
        (ii) Angular velocity ? .These two are related by the equation     v = R ? (R = radius of circular path)
?( j)	 Acceleration of a particle in a circular motion may have two components as listed hereunder.
        (i) Tangential component (
t
a ) and
        (ii) Normal or radial component (
n
a ).
As the name suggests, the tangential component is tangential to the circular path, given by 
t
a
= rate of 
change of speed
Figure: 4.75
Page 2


FORMULAE SHEET
?(a)	
1
F Fcos component of F along AC = ?=
?
    
2
F Fsin component of F perpendicular to AC = ?=
?
 
F1
F2
F B
AC

Figure: 4.73
?(b)	
dp d(mv)
F
dt dt
= =
?(c)	
net
net
F
F F ma or a
m
= = =
?
?
?? ? ?
     
T
F=ma
P
mg
Figure: 4.74
        
x
y
1
F 0 Tsin ma
F 0 Tcos mg
a
tan
g
-
= ? ?=
= ? ?=
? ?
?=
? ?
? ?
?
?
? ( d ) 	 Impulse	=	change	in	momentum        
fo
F t mv mV ?= -
?? ? ?? ?
?(e)	
max
f
R
µ=
 
    
A
Force of
Friction
Force Applied, F
O
f
max
f
K
f
S
?(f )	
max limiting s
ff R = = µ
A
R
O
f
m

Figure: 4.76
? ( g ) 	 Angle	o f	Fr iction:         
max
f
tan
R
?= = µ
     
1
or tan ( )
-
?= µ
? ( h ) 	 Pseudo	fo r ce:	  F = –ma;   where m = mass of the object, a = acceleration of the reference frame
?(i)	 A particle in circular motion may have two types of velocities as listed hereunder.
        (i)  Linear velocity v and
        (ii) Angular velocity ? .These two are related by the equation     v = R ? (R = radius of circular path)
?( j)	 Acceleration of a particle in a circular motion may have two components as listed hereunder.
        (i) Tangential component (
t
a ) and
        (ii) Normal or radial component (
n
a ).
As the name suggests, the tangential component is tangential to the circular path, given by 
t
a
= rate of 
change of speed
Figure: 4.75
?(k)	
dv
dv
R
dt dt
= = a
?
   where a = angular acceleration = rate of change of angular velocity =
d
dt
?
The normal or radial component, also known as centripetal acceleration is toward the center and is given by 
2
2
n
v
aR
R
= ?=
?(l)	 Net acceleration of a particle is the resultant of two perpendicular components, 
n
a
 and
t
a
. Hence, 
22
nt
aa a = +
?(m)	Tangential component 
t
a
is responsible for change of speed of a particle. This can be positive, negative or 
zero, depending upon the situation whether the speed of the particle is increasing, decreasing or remains 
constant.
?(n)	 In general, in any curvilinear motion, direction of instantaneous velocity is 
tangential to the path, while acceleration may assume any direction. If we 
resolve the acceleration in two normal directions, one parallel to velocity and 
another perpendicular to velocity, then the first component is 
t
a while the 
other is 
n
a .
Thus, 
t
a
= component of a
?
along v
?
 = acos ? =
a.v
v
??
 
dv
dv
dt dt
= =
? ?
= rate of change of speed. 
Further, 
n
a
= component of 
1x 2x 3x
aa a ?= =
perpendicular to v
?
 
2
22
t
v
aa
R
= -=
Here, v is the speed of the particle at that instant and R is called the radius of curvature to the curvilinear path 
at that point.
?(o)	 In 
t
a = a cos ? , if ? is acute, 
t
a will be positive and speed increases. However, if ? is obtuse 
t
a will be negative 
and speed will decrease. If ? is 90° , 
t
a
 is zero and speed will remain constant.
?(p)	 Now, depending upon the value of 
t
a , circular motion may be of three types as listed hereunder.
(i)   Uniform circular motion in which speed remains constant or 
t
a
= 0.
(ii)  Circular motion of increasing speed, in which 
t
a is positive.
a
a
n
a
n
v
a
t
a
a
t
a
n
v
a
v
Figure: 4.78
(iii)  Circular motion of decreasing speed, in which
t
a
is negative.
a

v


Figure: 4.77