rotational motion in physics for class 11 ppt by arjith jp

2. NAME :
CLASS :
TOPIC :
AD.NO :
S.NO :

5. 2. Angular displacement (θ) - The radian value of the
angle displaced by an object on the center of its path in circular
motion from the initial position to the final position is called the
angular displacement.
1. Angular position (Φ) - The angular position of a particle is
the angle ɸ made between the line connecting the particle to
the original and the positive direction of the x-axis, measured
in a counterclockwise direction
ɸ = l / r

6. 3. Angular Velocity (ω)- Angular velocity of an object in
circular motion is the rate of change of angular
displacement
Θ = ΘF- ΘI
ω = θ / t
Unit - rads-1
 Vector direction by
Right hand rule

7. 4. Angular acceleration- Angular acceleration of an object
in circular motion is the rate of change of angular
velocity
θ
ω0
r
t = 0
ω
t = t
α = (ω – ω0) / t
Unit- rads-2
Direction- By right
hand rule

8. α = (ω – ω0) / t ω = ω0 + αt
(ω + ω0 )/2 = θ / t θ = (ω + ω0 )t / 2
θ = ω0t + ½ αt2
(ω + ω0 )/2 = θ / t , ωt = 2θ + ω0t
, (ω0 + αt)t = 2θ + ω0t
θ = (ω + ω0 )t/2, θ = (ω + ω0 )(ω - ω0 )
2 α
ω2 = ω0
2 + 2αθ

9. Therefore the four equations of angular movement are-
1. Ω = Ω0 + ΑT
2. Θ = (Ω + Ω0 )T/2
3. θ = ω0t + ½ αt2
4. ω2 = ω0
2 + 2αθ
It should be noted that these four equations
are analogous to the four linear equations of
motion:
1. V = U + at
2. S = (V + U)t/2
3. S = Ut + ½ at2
4. V2 = U2 + 2as

10. RIGHT HAND
RULE
Take your right hand and curl your fingers along the direction of the
rotation. Your thumb directs along the specific vector you need.
(angular velocity, angular acceleration, angular momentum etc.)
Direction of
rotation
Axis of rotation
Thus, right hand rule
is used whenever, in
rotational motion, to
measure the
direction of a
particular vector

11. 1. Linear displacement- Angular displacement
2. Linear velocity- Angular velocity
S / t = r θ / t V = rω
θ
r
S S= (2πr / 2π ) × θ = rθ
S = rθ

12. 3. Linear acceleration- Angular acceleration
αr = a a = rα
Displacement Velocity Acceleratiom
Translational motion S V a
Rotational motion θ ω α
Relationship S = rθ V = rω a = rα
Α = (Ω – Ω0) / T
ΑR = (Ω – Ω0)R / T
ΑR = (ΩR – Ω0R) / T
ΑR = (V - V0) / T

13. Period (T) is the time taken by an object
in rotational motion to complete one
complete circle.
Frequency (f) is the no. of cycles an object
rotates around its axis of rotation
Thus, f = 1/ T . However, ω = θ / t
Therefore, ω = 2π / T
Therefore, ω = 2π / (1/f)
Thus, ω = 2πf

14. Unlike in the case of linear movement’s inertia (reluctance
to move or stop) , inertia of circular/rotational motion
depends both upon the mass of the object and the
distribution of mass (how the mass is spread across the
object)
Moment of Inertia of a single object-
i
s
I = mr2
Moment of Inertia
a scalar quantity
r
Axis of
rotation
Path of the
object

15. Suppose a body of mass M has moment of Inertia I
about an axis. The radius of gyration, k, of the body
about the axis is defined as
I= Mk2
That is k is the distance of a point mass
M from the axis of rotation such that this
point mass has the same moment of inertia
about the axis as the given body.

16. Moment of Inertia of some common shapes
Body Axis Figure I k
Ring
(RadiusR)
Perpendicular
to the plane
at the center
MR2 R
Disc (Radius R) Perpendicular
to the plane
at the center
½ MR2 R / √2
Solid Cylinder
(Radius R)
Axis of
cylinder ½ MR2 R / √2
Solid Sphere
(Radius R)
Diameter
⅖MR2 R√(⅖)

17. Angular momentum is the product of the moment of inertia and the
angular velocity of the object.
θ
r P
y
x
z
L = r sin θ × P
L = r P Sinθ
However, P= mv ,
(Linear moment= mass × linear velocity )
(as V= rω)
Therefore, L= m r v Sinθ
= m r (ω r ) Sinθ
= mr2 ω Sinθ
Therefore, L = Iω Sin θ (as I =mr2)L= Iω Sinθ

18. But in most cases the radius of rotation is perpendicular
to the momentum.
THUS, Θ = 90° , L = IΩ
18
P
r
Axis of
rotation

19. •THE RATE OF CHANGE OF ANGULAR MOMENTUM
OF AN OBJECT IN ROTATIONAL MOTION IS
PROPORTIONAL TO THE EXTERNAL UNBALANCED
TORQUE. THE DIRECTION OF THE TORQUE ALSO
LIES IN THE DIRECTION OF THE ANGULAR
MOMENTUM.
•TORQUE IS CALLED THE MOMENT OF FORCE AND IS
A MEASURE OF THE TURNING EFFECT OF THE FORCE
ABOUT A GIVEN AXIS.
•A TORQUE IS NEEDED TO ROTATE AN OBJECT AT
REST OR TO CHANGE THE ROTATIONAL MODE OF
AN OBJECT.

20. Τ = (IΩ – IΩ0 )/T
Τ = I [(Ω - Ω0 )/T]
20
By Newton’s second
law of motion
F = ma
Fr = mra = mr (rα)
( as a=rα)
Fr= mr2 α
Fr = Iα (as I= mr2) However,
from the above derivation,
Iα =τ .
Therefore, Fr = τ
Thus, τ = Fr
THEREFORE, Τ = IΑ

21. θ
d
r
F
y
This theory can be
expressed as:
z
x
τ = Fd
τ = Fr Sinθ
O
τ = F × r

22. Relationship between Torque and Angular
Momentum-
τ = dL / dt
Τ – NET TORQUE
L- ANGULAR
MOMENTUM
This result is the rotational analogue of
Newton’s second law:
F = dP / dt

23. 1. Newton’s First law and equilibrium-
If the net torque acting on a rigid object is zero, it will
rotate with a constant angular velocity.
Concept of equilibrium-
M m
d 3d
If the system is
at
equilibrium, total
torque around O
should be zero
O
Therefore, τM + τm = 0
Mgd + [ -(mg(3d))] = 0
Therefore, m = M/3

24. 2. Newton’s second law of motion
The rate of change of angular momentum of an object in
rotational motion is proportional to the external unbalanced torque.
The direction of the torque also lies in the direction of the angular
momentum.
τ α α
τ = Iα
This is analogous to the linear equivalent, which states,
The rate of change of momentum is directly proportional to the
external unbalanced force applied on an object and that force lies
in the direction of the net momentum.
Thus, torque could be treated as the rotational
analogue of the force applied on an object.

25. THEOREM OF PARALLEL AXES :
Let ICM be the moment of inertia of a body of mass M
about an axis passing through the center of mass and let I
be the moment of inertia about a parallel axis at a
distance d from the first axis.
Then, I = ICM + Md2
Thus
the minimum moment of
inertia for any object
is at the center of mass,
as x in the above
expression is zero.

26. The moment of inertia of a body about an
axis perpendicular to its plane is equal to
the sum of moments of inertia about two
mutually perpendicular axis in its own plane
and crossing through the point through
which the perpendicular axis passes
• Iz = Ix + Iy

27. Ek = Σ ½ mi vi
2
But, vi = riω
So, Ek = Σ ½ mi ri
2ω2
= ½ (Σ m r 2 )ω2
i i
Therefore, Ek = ½ I ω2
1r m1
r2 m2
rn
mn
ω
Ek = ½ I ω2
Rotational kinetic energy is the rotational
analogue of the translational
kinetic energy, which is EK = ½ mv2. In
fact, rotational kinetic energy equation
can be deduced by substituting v =rω
and viceversa.

28. Law- If the resultant external torque on a system
is zero, its total angular momentum remains
constant.
That is, if τ = 0, dL / dt = 0 , which means that L is a constant
This is the rotational analogue of the law of conservation of linear
momentum.
25

29. Therefore, W= τ [θ2 - θ1 ]
Therefore, the total work done in
rotating the body from an
angular displacement of θ1 to an
angle displacement θ2 is
θ2​
W= ∫ τ dθ
θ1​
θ2​
W= τ ∫ 1. dθ
θ1​
DW = Τ DΘ

30. The rate at which work is done by a torque is
called Power
P= dw/dt = τ dθ/dt = τ ω
Therefore, P = τ ω

31. From ω2 = ω 2 + 2αθ and W= τθ , it is clear that the work0
done by the net torque is equal to the change in
rotational kinetic energy.
τ = Iα and ω2 = ω0
2 + 2αθ .
Therefore, ω2 = ω0
2 + 2(τ/I)θ .
+ 2[(τθ)/I] .
+ 2W/I
Thus, ω2=ω0
2
Thus, ω2 = ω0
2
OR
W= ½ I (ω2 - ω0
2 ). This is called
the work-energy principle.

32. •τ = Iα = I dω/dt = d(Iω)/dt But,
τ = dL/dt​
•Thus, dL/dt = d(Iω)/dt​
•By integrating both sides,​
•It can be shown that, L = Iω
•This is the rotational analogue
of P = mv​

33. 33
Rolling is a combination of rotational and transitional(linear)
motions.
Suppose a sphere is rolling on a plane surface, the velocity
distribution can be expressed as,

34. V V = 0
This two systems (rotational and transitional) can be
combined together to understand how actually the
sphere above moves in the plane. The
final distribution shows clearly that in reality the
ball always instantly experiences a zero velocity at
the point of contact with the surface and the
maximum velocity is at the top
31

35. Thus, Etotal = Etransitional + Erotational
E = ½ MV2 + ½ IΩ2
E = ½ mR2 ω2 + ½ mk2 ω2
(Where k- radius of gyration)
E= ½mω2 (R2 + k2)

36. From the conservation of energy,
θ
h
Mgh = ½mv2 + ½Iω2
s
mg

37. •
• 2.ROTATIONAL EQUILIBRIUM
•
1. Transitional Equilibrium
For a body to be in transitional equilibrium, the vector sum
of all the external forces on the body must be zero.
Fext = M acm
τext = Iα and α must be zero for rotational equilibrium.

38. 38
Term Definition
Angular position The angular position of a particle is the angle ɸ made between
the line connecting the particle to the original and the positive
direction of the x-axis, measured in a counterclockwise direction
Angular
displacement
The radian value of the angle displaced by an object on the
center of its path in circular motion from the initial position to the
final position is called the angular displacement.
Angular velocity Angular velocity of an object in circular motion is the rate of
change of angular displacement
Angular acceleration Angular acceleration of an object in circular motion is the rate of
change of angular velocity
Angular equations of
motion ω = ω0 + αt , θ = (ω + ω0 )t/2,
θ = ω0t + ½ αt2 , ω2 = ω0
2 +2αθ

39. 36
RIGHT HAND
RULE
Take your right hand and curl your fingers along the direction of the
rotation.Your thumb directs along the specific vector you need
Relationship
between linear
and angular
qualities
S = rθ , V = rω , a = rα
(s- translational displacement , v- translational velocity,
a-translational acceleration )
Period and
frequency
Period (T) is the time taken by an object in circular motion to complete
one complete circle.
Frequency (f) is the no. of cycles an object rotates around its axis of
rotation
Thus, f = 1/ T
Moment of
inertia
Inertia of circular/rotational motion depends both upon the mass of the
object and the distribution of mass.
Radius of
gyration
The radius of gyration, k, of the body about the axis is defined as
I= Mk2
Angular
momentum
Angular momentum is the product of the moment of inertia and the
angular velocity of the object.

40. Torque Torque is called the moment of force and is a measure of the
turning effect of the force about a given axis.
Theorem of parallel
axes
Let ICM be the moment of inertia of a body of mass M about an
axis passing through the center of mass and let I be the moment
of inertia about a parallel axis at a distance d from the first axis.
Then, I = ICM + Md2
Theorem of
perpendicular axes
The moment of inertia of a body about an axis perpendicular to its
plane is equal to the sum of moments of inertia about two
mutually perpendicular axis in its own plane and crossing through
the point through which the perpendicular axis passes
Iz = Ix + Iy
Rotational kinetic
energy
Ek = ½ Iω2
Law of conservation
of angular
momentum
If the resultant external torque on a system is zero, its total
angular momentum remains constant.
Work done W = τ [θ2 - θ1 ]

41. Power The rate at which work is done by a torque is called
Power
P = τ ω
Work-Energy principle W= ½ I (ω2 - ω0
2 ). This is called the work-energy
principle.
Rolling body Rolling is a combination of rotational and
transitional(linear) motions.
Equilibrium of a rigid body 1. Transitional Equilibrium
For a body to be in transitional equilibrium, the vector
sum of all the external forces on the body must be zero.
Fext = Macm
and acm must be zero for transitional equilibrium.
2. Rotational Equilibrium
For a body to be in rotational equilibrium, the vector
sum of all the external torques on the body about any axis
must be zero.
τ = Iα and α must be zero for rotational equilibrium.

42. Thank you...