1
Chapter 01: Circular Motion
Subtopics 1.0 Introduction
1.1 Angular displacement
1.2 Angular velocity and angular
acceleration 1.3 Relation between linear
velocity and angular velocity 1.4 Uniform Circular Motion
1.5 Acceleration in U.C.M (Radial acceleration)
1.6 Centripetal and centrifugal
forces 1.7 Banking of roads
1.8 Vertical circular motion due to earth’s gravitation
1.9 Equation for velocity and
energy at different positions in vertical circular motion
1.10 Kinematical equation for
circular motion in analogy with linear motion
1.0 Introduction Q.1. Define circular motion. Give its examples. Ans: Definition: Motion of a particle along the circumference
of a circle is called circular motion.
Examples:
i. The motion of a cyclist along a circular path.
ii. Motion of the moon around the earth.
iii. Motion of the earth around the sun.
iv. Motion of the tip of hands of a clock.
v. Motion of electrons around the nucleus
in an atom. 1.1 Angular displacement Q.2. What is radius vector?
Ans: i. A vector drawn from the centre of a
circle to position of a particle on
circumference of circle is called as
‘radius vector’.
ii. It is given by,
| r |→
= δθ
δs
where, δs = small linear distance
δθ = small angular displacement
iii. It is directed radially outwards.
iv. Unit: metre (m) in SI system and
centimetre (cm) in CGS system.
v. Dimensions: [M0L1T0] Q.3. *Define angular displacement. OR
What is angular displacement?
Ans: i. Angle traced by a radius vector in a
given time, at the centre of the circular
path is called as angular displacement. ii. Consider a particle
performing
circular motion in
anticlockwise
sense as shown in
the figure.
Let, A = initial
position of particle
at t = 0
Y′
s
A
B
r
θ O
Y
B
A O δs
δθ →
r
Circular Motion 01
2
Std. XII Sci.: Perfect Physics - I
B = final position of particle after time t
θ = angular displacement in time t
r = radius of the circle
s = length of arc AB
iii. Angular displacement is given by,
θ = Length of arc
radius of circle
∴ θ = s
r
iv. Unit: radian
v. Direction of angular displacement is
given by right hand thumb rule or right
handed screw rule.
Note:
1. If a particle performing circular motion
describes an arc of length δs, in short time
interval δt then angular displacement is given
by δθ = s
r
δ.
∴ δs = δθ r
In vector form, s
→
δ = →
δθ × r→
2. If a particle performing circular motion
completes one revolution then angular
displacement is given by θ = 360° = 2πc
where, πc represents angular displacement in
radians.
3. One radian is the angle subtended at the centre
of a circle by an arc of length equal to radius
of the circle. *Q.4. State right hand thumb rule to find the
direction of angular displacement.
Ans: Right hand thumb rule:
Imagine the axis of rotation to be held in
right hand with the fingers curled around it
and thumb out-stretched. If the curled
fingers give the direction of motion of a
particle performing circular motion then
the direction of out-stretched thumb gives
the direction of angular displacement
vector.
*Q.5. Explain right handed screw rule to find the
direction of angular displacement.
Ans: i. Imagine the right handed screw to be
held in the place in which particle is
performing circular motion. If the right
handed screw is rotated in the direction
of particle performing circular motion
then the direction in which screw tip
advances, gives the direction of angular
displacement.
ii. The tip of the screw advances in downward direction, if sense of rotation
of the object is clockwise whereas the
tip of the screw advances in upward
direction, if sense of rotation of the object is anticlockwise as shown in the
figure.
Direction of angular displacement
d
→
θ
A
B
θ
Tip of screw advancing
in upward
direction
Y′
Right handed screw rule
Y
A
BθO
d
→
θ
Additional Information
Characteristics of angular displacement: i. Instantaneous angular displacement is a vector
quantity (true vector), so it obeys commutative and associative laws of vector addition.
ii. Finite angular displacement is a pseudo
vector.
iii. Direction of infinitesimal angular displacement is given by right hand thumb
rule or right handed screw rule.
iv. For anticlockwise sense, angular displacement is taken as positive while in clockwise sense,
angular displacement is taken negative.
3
Chapter 01: Circular Motion
#Q.6. Are the following motions same or different?
i. Motion of tip of second hand of a clock.
ii. Motion of entire second hand of a clock.
Ans: Both the motions are different.
The tip of the second hand of a clock performs
uniform circular motion while the entire hand
performs rotational motion with the second
hand as a rigid body. 1.2 Angular velocity and angular acceleration Q.7. *Define angular velocity.
OR
What is angular velocity? State its unit and
dimension.
Ans: i. Angular velocity of a particle
performing circular motion is defined as
the time rate of change of limiting
angular displacement.
OR
The ratio of angular displacement to
time is called angular velocity.
ii. Instantaneous angular velocity is given
by,
→
ω = t 0
limt
→
δ →
δ θ
δ =
d
dt
→
θ
Finite angular velocity is given by,
ω = t
θ
iii. It is a vector quantity.
iv. Direction: The direction of angular
velocity is given by right hand rule and
is in the direction of angular
displacement.
v. Unit: rad s−1
vi. Dimensions: [M0L0T−1]
Note:Magnitude of angular velocity is called angular
speed. Q.8. *Define angular acceleration.
OR
What is angular acceleration? State its unit
and dimension.
Ans: i. The rate of change of angular velocity with respect to time is called angular
acceleration.
It is denoted by→
α .
ii. If 0
→
ω and →
ω are the angular velocities
of a particle performing circular motion at instant t0 and t, then angular
acceleration is given by,
→
α = 0
0t t
→ →
ω −ω
−
= t
→
δω
δ
iii. Direction: The direction of →
α is given
by right hand thumb rule or right handed
screw rule.
iv. Unit: rad /s2 in SI system.
v. Dimensions: [M0L0T−2]. Q.9. Define the following terms.
i. Average angular acceleration
ii. Instantaneous angular acceleration
Ans: i. Average angular acceleration:
Average angular acceleration is defined
as the time rate of change of angular
velocity.
It is given by avg
→
α = 2 1
2 1t t
→ →
ω −ω
−
= t
→
δω
δ
ii. Instantaneous angular acceleration:
Instantaneous angular acceleration is
defined as the limiting rate of change of
angular velocity.
It is given by →
α = t 0
limδ → t
→
δω
δ=
d
dt
→
ω
Concept Builder
Positive angular acceleration:
When an angular acceleration will have the same
direction as angular velocity, it is termed as positive
angular acceleration.
eg: When an electric fan is switched on, the angular
velocity of the blades of the fan increases with time. Negative angular acceleration:
When an angular acceleration will have a direction
opposite to that of angular velocity. It is termed as
negative angular acceleration.
eg: When an electric fan is switched off, the angular
velocity of the blades of fan decreases with time.
4
Std. XII Sci.: Perfect Physics - I
Note:
1. When a body rotates with constant angular velocity its instantaneous angular velocity is
equal to its average angular velocity, whatever
may be the duration of the time interval. If the
angular velocity is constant, we write
ω = | |→
ω = d
dt
θ
2. If a body completes one revolution in time
interval T, then angular speed, ω = 2
T
π
= 2πn,
where n = frequency of revolution.
3. d
→
θ , →
ω and →
α are called axial vectors, as they
are always taken to be along axis of rotation.
4. The direction of d→
θ and →
ω is always given by
right handed thumb rule. 1.3 Relation between linear velocity and
angular velocity Q.12. *Show that linear speed of a particle
performing circular motion is the product
of radius of circle and angular speed of
particle.
OR
Define linear velocity. Derive the relation between linear velocity and angular
velocity. [Mar 02, Mar 96, 08, 12, Oct 09]
Ans: Linear velocity:
Distance travelled by a body per unit time in a given direction is called linear velocity.
It is a vector quantity and is given by,
v
→
= ds
dt
→
Relation between linear velocity and angular velocity:
i. Consider a particle moving with uniform circular motion along the circumference of a circle in anticlockwise direction
with centre O and radius r as shown in
the figure.
ω
δθ
B
A O
r
→
r
→
v
→
v
→
δs
ω increases
→
ω
→
α
→
ωd
O
Figure (a)
Figure (b)
ω decreases
→
ω
→
α
→
ωd
Additional Information
Q.10. What happens to the direction of angular acceleration
i. if a particle is speeding up?
ii. if a particle is slowing down?
Ans: i. Direction of →
α when the particle is
speeding up: Consider a particle
moving along a
circular path in
anticlockwise direction
and is speeding up.
Magnitude of →
ω keeps
on increasing which
results in d
→
ω to be
directed up the plane.
Hence, direction of →
α is upward. As →
ω
and →
α are ⊥ar to the plane, they are
parallel to each other. [See figure (a)].
ii. Direction of →
α when the particle is
slowing down: Consider a particle is
moving along a circular path in
anticlockwise direction
and is slowing down.
Magnitude of →
ω keeps
on decreasing which
results in d
→
ω to be
directed down the plane.
Hence, direction of →
α is downward. [See
figure (b)]. Q.11. Write down the main characteristics of
angular acceleration.
Ans: Characteristics of angular acceleration:
i. Angular acceleration is positive if
angular velocity increases with time.
ii. Angular acceleration is negative if angular velocity decreases with time.
iii. Angular acceleration is an axial vector.
iv. In uniform circular motion, angular velocity is constant, so angular
acceleration is zero.
5
Chapter 01: Circular Motion
ii. Let the particle cover small distance δs
from A to B in small interval δt.
In such case, small angular displacement
is ∠AOB = δθ.
iii. Magnitude of instantaneous linear
velocity of particle is given by,
v = δt 0
lim→
s
t
δ
δ
But δs = r δθ
∴ v = r t 0
limtδ →
δθ
δ [ Q r = constant]
Also t 0
limδ →
t
δθ
δ = ω
∴ v = rω
In vector form, v→
= →
ω × r→
Q.13. *Prove the relation v→
= →
ω × r→
, where
symbols have their usual meaning.
OR
In U.C.M. (Uniform Circular Motion),
prove the relation v r
→ → →
= ω× , where symbols
have their usual meanings. [Mar 16]
Ans: Analytical method:
i. Consider a particle performing circular
motion in anticlockwise sense with centre O and radius r as shown in the figure.
ii. Let, →
ω = angular velocity of the particle
v→
= linear velocity of the particle
r→
= radius vector of the particle
[Diagram - ½ Mark]
iii. Linear displacement in vector form is given by,
s
→
δ = r
→ →
δθ × [½ Mark]
Dividing both side by δt,
s
t
→
δ
δ =
t
→
δθ
δ × r→
….(1)
iv. Taking limiting value in equation (1)
t 0
limδ →
s
t
→
δ
δ =
t 0
limδ → t
→
δθ
δ × r→
∴ ds
dt
→
= d
dt
→
θ × r→
[½ Mark]
But, ds
dt
→
= v→
= linear velocity,
d
dt
→
θ = →
ω = angular velocity
∴ v
→
= →
ω × r→
[**Explanation – ½ Mark] Calculus method:
i. A particle is moving in XY plane with position vector,
r
→
= r i cos ωt + r jsin ωt ….(1)
ii. Angular velocity is directed perpendicular to plane, i.e., along
Z-axis. It is given by →
ω = ω k ,
where, k = unit vector along Z-axis.
iii. →
ω × r
→
= ω k × (r i cos ωt + r j sin ωt)
[From equation (1)]
= ωr cos ωt ( k × i ) + ωr sin ωt ( k × j )
= ωr j cos ωt + ωr (– i ) sin ωt
k i j and k j i∧ ∧ ∧ ∧ ∧ ∧ × = × =− Q
∴ →
ω × r→
= – rω i sin ωt + ωr j cos ωt
∴ →
ω × r
→
= rω (− i sin ωt + j cos ωt) .…(2)
Also v→
= d r
dt
→
= r ( )ˆ ˆi sin t jcos t−ω ω + ω ω
v
→
= rω ( )ˆ ˆi sin t jcos t− ω + ω ….(3)
From equation (2) and (3),
∴ v
→
= →
ω × r→
1.4 Uniform Circular Motion Q.14. *Define uniform circular motion. OR
What is uniform circular motion?
Ans: i. The motion of a body along the circumference of a circle with constant
speed is called uniform circular motion.
v
→
r
→
→
ω
O
6
Std. XII Sci.: Perfect Physics - I
ii. In U.C.M, direction of velocity is along
the tangent drawn to the position of particle on circumference of the circle.
iii. Hence, direction of velocity goes on
changing continuously, however the magnitude of velocity is constant.
Therefore, magnitude of angular
velocity is constant.
iv. Examples of U.C.M:
a. Motion of the earth around the sun.
b. Motion of the moon around the earth.
c. Revolution of electron around the nucleus of atom.
Q.15. State the characteristics of uniform circular
motion.
Ans: Characteristics of U.C.M:
i. It is a periodic motion with definite
period and frequency.
ii. Speed of particle remains constant but
velocity changes continuously.
iii. It is an accelerated motion.
iv. Work done in one period of U.C.M is zero.
Q.16. Define periodic motion. Why U.C.M is
called periodic motion?
Ans: i. Definition:
A type of motion which is repeated after equal interval of time is called periodic
motion.
ii. The particle performing U.C.M repeats
its motion after equal intervals of time
on the same path. Hence, U.C.M is called periodic motion.
Q.17. Define period of revolution of U.C.M. State
its unit and dimension. Derive an expression
for the period of revolution of a particle
performing uniform circular motion.
Ans: Definition:
The time taken by a particle performing uniform circular motion to complete one
revolution is called as period of revolution.
It is denoted by T and is given by, T = 2π
ω.
Unit: second in SI system.
Dimensions: [M0L0T1] Expression for time period:
During period T, particle covers a distance
equal to circumference 2πr of circle with linear velocity v.
∴ Time period = Distance covered in one revolution
Linear velocity
∴ T = 2 r
v
π
But v = rω
∴ T = 2 r
r
π
ω
∴ T = 2π
ω
Q.18. What is frequency of revolution? Express
angular velocity in terms of frequency of
revolution. Ans: i. The number of revolutions performed by
a particle performing uniform circular
motion in unit time is called as frequency of revolution.
ii. Frequency of revolution (n) is the
reciprocal of period of revolution.
n = 1
T=
1
2π ω
= 2
ω
π =
v
2 rπ
iii. Unit: hertz (Hz), c.p.s, r.p.s etc.
iv. Dimensions: [M0L0T−1]
Angular velocity in terms of frequency of
revolution:
ω = 2
T
π = 2π ×
1
T
But 1
T = n
∴ ω = 2πn *Q.19.Define period and frequency of a particle
performing uniform circular motion. State
their SI units. Ans: Refer Q.17 and Q.18 1.5 Acceleration in U.C.M (Radial acceleration) Q.20. Define linear acceleration. Write down its
unit and dimensions. Ans: i. Definition:
The rate of change of linear velocity
with respect to time is called linear
acceleration.
It is denoted by a→
and is given by
a
→
= dv
dt
→
ii. Unit: m/s2 in SI system and cm/s2 in CGS system.
iii. Dimensions: [M0L1T−2]
7
Chapter 01: Circular Motion
Q.21. U.C.M is an accelerated motion. Justify this
statement. Ans: i. In U.C.M, the magnitude of linear
velocity (speed) remains constant but
the direction of linear velocity goes on changing i.e., linear velocity changes.
ii. The change in linear velocity is possible
only if the motion is accelerated. Hence, U.C.M is an accelerated motion.
Q.22. *Obtain an expression for acceleration of
a particle performing uniform circular
motion.
OR
Define centripetal acceleration. Obtain an
expression for acceleration of a particle
performing U.C.M by analytical method.
Ans: Definition:
The acceleration of a particle performing U.C.M which is directed towards the centre
and along the radius of circular path is called
as centripetal acceleration.
The centripetal acceleration is directed along
the radius and is also called radial
acceleration.
Expression for acceleration in U.C.M by
analytical method (Geometrical method):
i. Consider a particle performing uniform circular motion in a circle of centre O
and radius r with a uniform linear
velocity of magnitude v.
ii. Let a particle travel a very short distance
δs from A to B in a very short time
interval δt.
iii. Let δθ be the angle described by the
radius vector OA in the time interval δt as shown in the figure.
iv. The velocities at A and B are directed
along the tangent.
v. Velocity at B is represented by BC
while the velocity at A is represented by
AM . [Assuming AM = BD]
vi. Angle between BC and BD is equal to
δθ as they are perpendicular to OB and
OA respectively.
vii. Since ∆BDC ∼ ∆OAB
∴ DC
BD =
AO
AB ∴
v
v
δ =
AB
r
viii. For very small δt, arc length δs of
circular path between A and B can be taken as AB
∴ v
vδ=
r
sδ or δv =
v
r δs
where, δv = change in velocity
Now, a = t
limδ →0
v
t
δ
δ =
t
limδ →0
v
r s
t
δ
δ
∴ a = v
r
t
limδ →0
s
t
δ
δ =
r
v × v =
2v
r
As δt → 0, B approaches A and δv becomes perpendicular to the tangent
i.e., along the radius towards the centre.
ix. Also v = rω
a = 2 2r
r
ω
= ω2r
x. In vector form, →
a = − ω2→
r
Negative sign shows that direction of a→
is opposite to the direction of r→
.
Also a = 2
0
v
r
r
∧
, where 0r
∧
is the unit
vector along the radius vector. Q.23. Derive an expression for linear acceleration of
a particle performing U.C.M. [Mar 98, 08]
Ans: Refer Q.22 Q.24. Derive an expression for centripetal
acceleration of a particle performing uniform
circular motion by using calculus method.
Ans: Expression for centripetal acceleration by calculus method:
i. Suppose a particle is performing U.C.M
in anticlockwise direction.
The co-ordinate axes are chosen as
shown in the figure.
Let,
A = initial position of the particle
which lies on positive X-axis
P = instantaneous position after time t
θ = corresponding angular displacement
C D
vM
δθ
δθ
B
A O r
δθ
v
8
Std. XII Sci.: Perfect Physics - I
ω = constant angular velocity
r
→
= instantaneous position vector at time t
ii. From the figure,
r
→
= i x + jy
where, i and j are unit vectors along
X-axis and Y-axis respectively.
iii. Also, x = r cos θ and y = r sin θ
∴ →
r = [r i cos θ + r j sin θ]
But θ = ωt
∴ →
r = [r i cos ωt + r j sin ωt] ....(1)
iv. Velocity of the particle is given as rate of change of position vector.
∴
→
v = dt
dr
→
=
dt
d[r i cos ωt + r jsin ωt]
= r dcos t
dt
ω
i + r
dsin t
dt
ω
j
∴ v
→
= − rω i sin ωt + rω j cos ωt
∴ v
→
= rω (− i sin ωt + j cos ωt) ....(2)
v. Further, instantaneous linear
acceleration of the particle at instant t is
given by,
→
a = dt
dv
→
=
dt
d [rω (− i sin ωt + jcos ωt)]
= rω d ˆ ˆ( i sin t jcos t )dt
− ω + ω
= rω d dˆ ˆ( sin t )i (cos t ) jdt dt
− ω + ω
= rω (− ω i cos ωt − ω j sin ωt)
= − rω2 ( i cos ωt + j sin ωt)
∴ →
a = − ω2 (r i cos ωt + r jsin ωt) ....(3)
vi. From equation (1) and (3),
→
a = − ω2→
r
Negative sign shows that direction of
acceleration is opposite to the direction
of position vector.
vii. Magnitude of centripetal acceleration is
given by,
a = ω2r
As ω = v
r
∴ a = 2
v
r Note:
To show a→
= →
ω × v→
, →
ω × v→
= ω k × ( – rω i sin ωt + rω j cos ωt)
= – rω2 sin ωt ( k × i ) + rω2 cos ωt ( k × j )
= – rω2 sin ωt j + rω2 cos ωt (– i )
k i j and k j i∧ ∧ ∧ ∧ ∧ ∧ × = × = − Q
= – rω2 i cos ωt – rω2 j sin ωt
= a→
[From equation (3)] Q.25. Derive an expression for centripetal
acceleration of a particle performing
uniform circular motion. [Mar 02, Mar 06]
Ans: Refer Q. 24 *Q.26.Derive the relation between linear
acceleration and angular acceleration if a
particle performs U.C.M.
Ans: Relation between linear acceleration and
angular acceleration in U.C.M:
i. Consider a particle performing U.C.M.
with constant angular velocity ω with path radius r.
ii. Magnitude of linear acceleration is
given by,
a = t 0
limδ →
v
t
δ
δ
∴ a = dv
dt
iii. But, v = rω
∴ a = d
dt(rω) = r
d
dt
ω
+ ωdr
dt
iv. Since, r = constant
∴ dr
dt = 0
∴ a = rd
dt
ω
Y
N →r
X O
θ A
P(x, y) →
r
x
y
ω
v
9
Chapter 01: Circular Motion
But, d
dt
ω
= α
∴ a = rα In vector form,
a
→
= r→ →
α ×
This is required relation. Q.27. Define non-uniform circular motion. Derive
an expression for resultant acceleration in non-uniform circular motion.
Ans: Non-uniform circular motion: Circular motion with variable angular speed
is called as non-uniform circular motion. Example: Motion of a body on vertical circle.
Expression for resultant acceleration in
non-U.C.M:
i. Since, v→
= →
ω × r
→
....(1)
Differentiating equation (1) with respect
to t, we get d v
dt
→
= d
dt (→
ω × r
→
)
d v
dt
→
= →
ω × d r
dt
→
+ d
dt
→
ω
× r→
....(2)
ii. But d v
dt
→
= →
a , d r
dt
→
= v→
and d
dt
→
ω
= →
α
∴ Equation (2) becomes,
a
→
= →
ω × v→
+ →
α × r→
....(3)
iii. ω
→
× v
→
is along the radius of the circle,
pointing towards the centre, hence it is
called radial acceleration R
a
→
.
∴ R
a
→
= →
ω × v
→
….(4)
iv. →
α × r
→
is along the tangent of the
circumference of the circular path, hence
it is called tangential acceleration T
a
→
.
∴ T
a
→
= →
α × r→
….(5)
v. From equation (3), (4) and (5),
a→
= R
a
→
+ T
a
→
From figure,
Magnitude of resultant linear
acceleration is given by | a |→
= 2 2
R Ta a+
Q.28. For a particle performing uniform circular
motion → → →
v = ω × r . Obtain an expression for
linear acceleration of the particle performing non-uniform circular motion. [Mar 14]
OR
In circular motion, assuming v r
→ → →
= ω× ,
obtain an expression for the resultant acceleration of a particle in terms of
tangential and radial component. [Mar 15] Ans: Refer Q.27
[Expression for relation between v→
and →
ω – ½
Mark, Derivative of the relation between v→
and →
ω – ½ Mark, Expression for relation between
a
→
and →
α – ½ Mark, Final expression for linear
acceleration – ½ Mark]
Note: 1. Resultant linear acceleration in different
cases
Situation Resultant motion
Resultant linear acceleration
aR = 0, aT = 0 Uniform linear
motion
a = 0
aR = 0, aT ≠ 0 Accelerated linear motion
a = aT
aR ≠ 0, aT = 0 Uniform
circular motion
a = aR
aR ≠ 0, aT ≠ 0 Non-uniform circular motion
a = 2 2
R Ta a+
2. In non-uniform circular motion, aR is due to
change in direction of linear velocity, whereas aT is due to change in magnitude of linear
velocity.
3. In uniform circular motion, particle has only
radial component aR due to change in the direction of linear velocity. It is so because
ω = constant
∴ α = d
dt
ω
= 0 so, aT = α × r = 0
4. Since the magnitude of tangential velocity does not change, there is no component of
acceleration along the tangent. This means the
acceleration must be perpendicular to the
tangent, i.e., along the radius of the circle.
Oθ
P R
a
→
Ta
→
a
→
→
α
ω
