Chapter 1: Rotational Dynamics
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Chapter 1
Rotational Dynamics and
Oscillatory Motion
Isaac Newton (1643–1727) AD
Isaac Newton, one of the most influential scientists in history, laid the
foundation of classical physics with his formulation of the laws of
motion and universal gravitation. His work unified the motion of
objects on Earth and in the heavens, marking a turning point in
scientific thought.
Newton made important contributions to rotational dynamics,
introducing fundamental concepts such as torque, moment of inertia,
and angular momentum. These ideas, rooted in his laws of motion,
form the basis for understanding the behavior of rotating bodies.
Beyond mechanics, Newton also advanced the fields of optics and
mathematics, co-developing calculus and exploring the nature of light
and color. His scientific legacy continues to shape our understanding
of the physical world.
Chapter 1: Rotational Dynamics
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1.1 Introduction
From a spinning top to the rotating Earth, rotation is part of our daily experience. In physics,
we study such motion using the concept of a rigid body—an ideal object that doesn’t deform
as it spins.
Rotational motion follows its own set of rules. Just like force causes linear motion, torque
causes rotation. And just like mass resists motion, a rotating object resists change through its
moment of inertia.
Rotation may seem simple but it reveals deep laws that govern how objects turn, balance, and
store energy.
Do You Know?
i) Why do skaters spin faster when they pull in their arms?
ii) How do wheels help vehicles move more easily?
iii) What keeps the Earth spinning day after day?
These questions take us into the world of rigid body rotation, where physics meets the
graceful, powerful motion of spinning objects.
Rigid Body
A rigid body is an object that does not deform or change its shape when external forces or
torques are applied. This means the distance between any two points within the body remains
constant throughout its motion. In simpler terms, the positions of its constituent particles
PHYSICS, Grade 12 1
relative to each other do not change. The study of how such bodies move under the influence
of external forces and torques is known as rigid body dynamics. It plays a vital role in various
fields, including the design of vehicles and mechanical systems, as well as in creating
realistic animations in computer graphics.
A rigid body is one in which the position of the constituent particles remain unchanged
throughout the motion.
[Illustration: Fig 1.1: Rotation of a rigid body with unchanged position of constituent particles.]
The motion of a rigid body can be classified into translatory and rotatory. Translatory
motion occurs when all points on the body move in parallel paths, resulting in straight-line
movement, such as a car driving straight. Rotatory motion involves the body rotating around
a fixed axis, with all points following circular paths around the axis, like a spinning wheel.
Chapter 1: Rotational Dynamics
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1.2 Moment of Inertia
In Newton’s second law, mass links force and acceleration. This mass, called inertial mass,
shows how much an object resists changes in its motion.
Similarly, in rotational motion, the moment of inertia, I, plays this role. It measures how hard
it is to start or stop an object spinning around an axis. Unlike mass, moment of inertia
depends on both the amount of mass and how far that mass is from the axis of rotation. The
farther the mass is spread out, the larger the moment of inertia, and the harder it is to spin.
For example, a solid wheel spins more easily than a ring of the
same size and mass because the ring’s mass is farther from the
center. Likewise, it’s easier to spin a bicycle wheel than a car wheel
because the car wheel has more mass distributed farther from its
axis, increasing its moment of inertia and resistance to spinning.
In short, moment of inertia is the rotational equivalent of mass,
representing an object’s resistance to changes in its spinning
motion.
[Illustration: Fig 1.2: Calculation of Moment of Inertia]
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Moment of inertia, I is mathematically defined as the sum (or integral) of each mass element
multiplied by the square of its distance from the axis of
rotation.
In other words:
I=∑miri 2
where mi is a small mass element ri is theType equation here. distance from the axis.
For a point mass at a distance R from the axis has a moment of inertia I=MR2, but for
extended objects like discs, rods, or rings, the moment of inertia varies depending on their
shape and mass distribution. Therefore, the farther the mass is from the axis, the greater the
moment of inertia and the harder it is to rotate the object. So, it's not simply the product of
total mass and the square of the distance — the distribution of mass matters, and you sum or
integrate over all mass elements.
Let’s demonstrate how the distribution of mass affects the moment of inertia and rotational
motion:
Activity 1.1
Materials Required: A ramp, two discs (A convex lens and a concave lens) of equal mass and
radius, a stop watch.
Method:
(i) Take a ramp (an inclined plane) and place it on a flat surface so that a disk can roll
down it easily. Make sure the ramp has a gentle incline.
(ii) Take two discs of equal mass and radius but different mass distributions (e.g., one
with mass concentrated near the center, a convex lens and the other with mass
distributed towards the edge, a concave lens). Predict which disk will reach the
bottom of the ramp first, based on their understanding of mass distribution and
moment of inertia.
(iii) Place both disks at the top of the ramp.
(iv) Release them simultaneously and observe which disk reaches the bottom first.
(v) Explain that the disk with mass concentrated near the center (the convex lens) has a
smaller moment of inertia and thus accelerates faster down the ramp.
(vi) The disk with mass distributed towards the edge (the concave lens) has a larger
moment of inertia and takes longer to reach the bottom.
(vii) Hence summarize that the distribution of mass affects the moment of inertia and the
rotational motion of objects.
(viii) Answer the following question:
Guess why the disk with mass concentrated near the center has a smaller moment of
inertia.
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[Illustration: Fig 1.3: Rotation of a rigid body]
The moment of inertia of a sphere depends on the axis of rotation. For a solid sphere, the
moment of inertia about the diameter is different from the moment of inertia about the
tangent to the sphere due to the different positions of the masses. The moment of inertia
about the tangent is greater because the constituent masses are at a greater distance from the
axis of rotation.
Our inner ear helps us keep our balance and sense of direction. If it's not working perfectly,
or if we try to walk with our eyes closed, small differences in how our body is balanced can
make us go off the track.
In short, the moment of inertia influences how well we control our movements and stay
balanced. Even tiny differences can make it hard to walk in a straight line.
1.2.1 Radius of Gyration
Radius of gyration is a way to describe how the mass of an object is spread out around its
axis of rotation.
Conceptually, it is the distance from the axis at which we can imagine the entire mass of the
object to be concentrated, so that it would have the same moment of inertia as it does in
reality.
In simple terms, it helps us understand how far the mass is, on average, from the axis when
the object is rotating.
Mathematically the radius of gyration K is written as:
I
K = √M
where I is the moment of inertia and M is the total mass. This concept simplifies the
analysis of rotational motion and helps understand the stability and strength of structures.
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[Illustration: Fig 1.4: Radius of gyration]
For a rigid body consisting of n particles is undergoing rotational motion about an axis AB.
The particles are m1, m2,….… , mn at the respective position r1, r2,……, rn from the axis of
the rotation.
( 𝑟12 + 𝑟22 +⋯…+ 𝑟𝑛2)
I = MK2 and K = 𝑛
( 𝑟12 + 𝑟22 +⋯…+ 𝑟𝑛2)
Hence K = √ is the radius of gyration and is the root mean square distance of
𝑛
all the particles of a body.
The radius of gyration is a scalar quantity that helps us visualize how mass is spread out in a
rotating object and depends on the position and axis of rotation.
Let’s do the activity 1.2 that follows to demonstrate how mass distribution affects the radius
of gyration and its impact on rotational motion:
Activity 1.2
Materials Required: Two small masses, say coins, a long ruler and pivoting arrangement, a
tape bundle, a stop watch.
Method:
(i) Attach the two small masses, say coins near the ends of a ruler and secure them
with tape as in fig below:
[Illustration: Fig 1.5]
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(ii) Attach the two coins closer to the center to the other ruler and secure them with
tape
(iii) Place the rulers on the smooth surface with the pivot in the center, allowing them
to rotate freely.
(iv) Predict which ruler will spin more easily and why.
(v) Spin each ruler by applying a small force at the end.
(vi) Observe the ease of spinning and the time it takes for each ruler to come to rest.
(vii) Record the times using the stopwatch it takes for each ruler to stop spinning and
compare the results.
(viii) The ruler with masses near the center has a smaller radius of gyration and spins
more easily, while the ruler with masses near the ends has a larger radius of
gyration and takes longer to spin and stop.
(ix) Summarize that the radius of gyration affects how mass distribution influences
rotational motion. The closer the mass is to the axis, the smaller the radius of
gyration and the easier it is to rotate.
Predict the following:
(a) Why the ruler with masses near the center spin more easily and
(b) How the results would change if the masses were heavier or lighter.
This activity provides a hands-on way for students to visualize and understand the concept of
the radius of gyration and its impact on rotational motion.
Chapter 1: Rotational Dynamics
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1.3 Moment of Inertia of a Uniform Rod
To calculate the moment of inertia of a uniform body, we consider a small mass dm of an
infinitesimally small particle and the perpendicular distance r of the particle from the axis
of rotation. We then integrate this over the entire body. i.e, I = ∫ dm r2
1.3.1 Moment of Inertia of a Uniform rod about an axis passing through the Centre of
Gravity:
[Illustration: Fig 1.6: M.I of a thin uniform rod about an axis passing through CG]
Consider a uniform rod with mass M and length l. To find its moment of inertia about an axis
through its center and perpendicular to its length, we start by setting the origin of our
coordinate system at the rod's center. The rod lies along the x-axis. We consider an
infinitesimally small mass element (dm) at a distance x from the origin. The moment of
inertia (dI) of this tiny mass element about the axis is
dI = dm x2. (1.1)
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By integrating this expression over the entire length of the rod, we can calculate the total
moment of inertia.
𝑀
For the uniform rod, the mass per unit length of the rod is = 𝑙 .
𝑀
Therefore, the mass of the infinitesimally small length dx = 𝑙 dx = dm.
As the mass is distributed on either side of the origin, the limits for integration are taken from
- l/2 to +l/2.
Therefore, the moment of inertia (I) of the entire rod can be found as
𝑙/2 𝑀 𝑙/2 𝑀 𝑥3
I = ∫−𝑙/2 𝑑𝑚𝑥 2 = ∫ 𝑥 2dx = [ ]l/2-l/2
𝑙 −𝑙/2 𝑙 3
(1.2)
Ml2 𝑙
If K be the radius of gyration, MK2 = 12 ⇒ K =2 3 (1.3)
√
1.4.2 Moment of Inertia of a uniform rod about an axis passing through its one end and
perpendicular to its length:
[Illustration: Fig 1.7: M.I of a thin uniform rod about an axis passing through one end.]
Let us consider a uniform rod with mass M and length l. To find its moment of inertia about
an axis through its one end and perpendicular to its length, we start by setting the origin of
our coordinate system at the rod's end. The rod lies along the x-axis. We consider an
infinitesimally small mass element (dm) at a distance x from the origin. The moment of
inertia (dI) of this tiny mass element about the axis is
dI = dm x2. (1.4)
By integrating this expression over the entire length of the rod, we can calculate the total
moment of inertia.
𝑀
For the uniform rod, the mass per unit length of the rod is = 𝑙 .
𝑀
Therefore, the mass of the infinitesimally small length dx = 𝑙 dx = dm.
As the mass is distributed on one side of the origin, the limits for integration are taken from 0
to l.
Therefore, the moment of inertia (I) of the entire rod can be found as
𝑙 𝑀 𝑙 𝑀 𝑥3 𝑀𝑙2
I = ∫0 𝑑𝑚𝑥 2 = ∫ 𝑥 2 dx = 𝑙 [ 3 ]l0 = 3
𝑙 0
𝑀𝑙2
∴I= (1.5)
3
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Ml2 𝑙
If K be the radius of gyration, MK2 = 3 ⇒ K = 3 (1.6)
√
Chapter 1: Rotational Dynamics
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1.5 Rotational Kinetic Energy of a Rigid Body
Consider a rigid body consisting of n particles is undergoing rotational motion about an axis
XY. The particles are of masses m1, m2,….…,mn at perpendicular distances r1, r2,……, rn
from the axis of the rotation and moving with velocities v1, v2,……,vn respectively about the
axis. The rotational kinetic energy KErot of a rigid body undergoing rotational motion is
indeed the sum of the kinetic energies of all the particles that make up the body. Each particle
contributes to the total kinetic energy due to its mass, velocity, and position relative to the
axis of rotation.
[Illustration: Fig 1.8:Rotational KE of a rigid body]
For a single particle i with mass mi_ at a distance ri from the axis of rotation and moving with
a tangential velocity vi, the kinetic energy KE is given by:
1 1
KEi = 2mi (ωri)2 = 2mi ri2 ω 2
The total rotational kinetic energy KErot is then the sum of the kinetic energies of all n
particles:
1 1
KErot =∑𝑛𝑖=1 2 mi ri2 ω 2 = 2 ω2 ∑𝑛𝑖=1 𝑚𝑖 ri2
This summation ∑𝑛𝑖=1 𝑚𝑖 ri2 is known as the moment of inertia I of the body about the axis
XY. Therefore, we can write:
1
KErot = 2 𝐼 ω2 (1.7)
This confirms that the rotational kinetic energy of the body is indeed the sum of the kinetic
energies of all the particles, taking into account their masses, positions, and velocities relative
to the axis of rotation.
The greater the moment of inertia, the higher the kinetic energy (KE). As a result, more work
is required to accelerate the body from rest.
If ω = 1, I = 2KE i.e, M.I is twice the K.E
Example
Chapter 1: Rotational Dynamics
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1.1 A uniform rod of length 1.2 meters and mass 3 kg is rotating about an axis perpendicular
to its length and passing through one of its ends.
(i) Calculate the moment of inertia of the rod about this axis.
(ii) If the rod is rotating with an angular velocity of 5 rad/s, determine the kinetic
energy of the rod.
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Solution:
Here, length, l = 1.3 m
Mass, m = 3 kg
(i) Moment of inertia, I = ?
We have for the axis at the end of the rod and normal to the rod,
𝑀𝑙2 3×1.32
I= 3 = 3
= 1.44 kg-m2
(ii) Angular velocity, ω = 5 rads-1
1 1
Therefore, KE = 𝐼 ω2 = × 1.69 ×52
2 2
= 18 J
Hence the moment of inertia of the rod is 1.44 kg⋅m2, and the kinetic energy of the rod is 18 J.
Chapter 1: Rotational Dynamics
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1.2 Torque and Angular Acceleration of a Rigid Body
Consider a rigid body rotating about an axis XY with a uniform angular acceleration α, under
the action of a torque.
Let the body consists on n particles of masses m1, m2,….…,mn at perpendicular distance r1,
r2,……, rn from the axis as show in the fig 1.9 below:
[Illustration: Fig 1.9: Rotation of a rigid body]
As the body is rigid, angular acceleration α of all the particles about the axis is the same.
However, the linear acceleration of the particles varies due to their position relative to the
axis of rotation. If a1, a2,…., an be the respective linear acceleration of the particles, then
a1 = r1α, a2 = r2α,….and an = rnα
Hence in general, force on an ith particle of mass mi is Fi = miai = miriα.
The moment of this force i.e, product of a force times the perpendicular distance from the
point of rotation to the line of the force, is called the torque produced by the force, which is
usually represented by 𝜏 and 𝜏 = F × r
Therefore, torque on the ith particle, 𝜏𝑖 = Fi × ri = miri2α. (1.8)
𝑛 2
∴ Torque acting on the body, 𝜏 = ∑𝑖=1 𝑚𝑖 𝑟𝑖 ω = Iα (1.9)
Hence 𝜏 = Iα (1.10)
Equation 1.18 gives the relation between torque and the angular acceleration of a rigid body.
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The equation relating torque (τ) to angular acceleration (α) for a rigid body is indeed a
fundamental equation in rotational dynamics and is analogous to Newton's second law of
motion.
Note: The gravitational torque on a body can be found by treating the body as a particle with
all the mass M concentrated at the center of mass.
Example
Chapter 1: Rotational Dynamics
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1.2 A uniform rod of mass 2 kg and length 1.2 m is hinged at one end and is free to rotate in
a vertical plane without friction. It is held horizontally and then released from rest.
a) What is the torque acting on the rod about the hinge just after it is released?
b) Calculate the angular acceleration of the rod at that instant.
c) What is the linear acceleration of the free end of the rod immediately after release?
Solution:
Here, Mass of rod, M=2 kg
Length of rod, L=1.2 m
Acceleration due to gravity, g=9.8 m/s2
Moment of inertia of rod about one end:
1
I= 3ML2
(a) Torque about the hinge just after release
𝐿
Torque due to weight acts at the center of mass, which is at 2 from the hinge.
𝐿
τ=Mg⋅2
Substitute values:
1.2
τ= 2 × 9.8 × 2 = 11.76 Nm
b) Angular acceleration of the rod
Using the relation:
𝜏
τ=Iα ⇒α= 𝐼
1 1
To calculate moment of inertia, we use I= 3ML2 = 3 ×2×1.22 = 0.96 Kgm2
𝜏 11.76
Now α= = = 12.25 rads-2
𝐼 0.96
(c) Linear acceleration of the free end of the rod
The linear acceleration a at the end of the rod is related to angular acceleration α by:
a=α⋅L ⇒ a=12.25×1.2=14.7 ms -2.
Chapter 1: Rotational Dynamics
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1.3 The Earth has a radius of approximately 6371 km and completes one rotation every 24
hours. Calculate the tangential force needed to stop the Earth's rotation within 1 hour.( Mass
of the Earth = 5.97 ×1024 kg)
Solution: Here,
Radius of Earth R=6371 km = 6371×103 m
Rotational period T=24 hours=24×3600 s
Stopping time tstop=1 hour=3600 s
Mass of the Earth, M = 5.97 ×1024 kg
2π 2π
ω= 𝑇 = 24×3600 ≈ 7.27×10−5 rad/s
PHYSICS, Grade 12 10
𝜔 −𝜔 7.27×10−5
Angular acceleration, α = 𝑡 0 = s = 2.02 × 10−8 rads-2
𝑠𝑡𝑜𝑝 3600
The tangential acceleration, a = R. α = 6371×103× 2.02 × 10−8 = 0.1286 ms-2
Now tangential force, F = Ma = 5.97 ×1024 × 0.1286 = 7.68 × 1023 N
Chapter 1: Rotational Dynamics
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1.3 A string is wrapped around the rim of a wheel of moment of inertia 0.1kgm and of 2
radius 20cm. The wheel is free to and free to rotate about its axis and initially it is at rest.
The string is now pulled by a force of 20N. What is the angular velocity of the string
after 2sec.?
[Illustration: Fig 1.10]
Solution: We know that, τ =F х r…. (1.11)
Also, τ =I . α (1.2)
From equation (1) and (2)
Fхr=Iхα
20 х 0.2 = 0.1 х α
α = 40rads -2
Again, ω = ω + αt
o
or, ω = 0 + 40 х 2
ω = 80rads-1.
Test yourself
Chapter 1: Rotational Dynamics
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1.1 A uniform rod of mass 2.5 kg and length 1.2 m is hinged at one end and placed at an
angle of 60° with the vertical in a vertical plane. It is then released from rest.
a) What is the initial torque acting on the rod about the hinge due to gravity?
b) Calculate the angular acceleration of the rod just after release.
c) What is the initial linear acceleration of the free end of the rod?( Ans: a. 12.73 Nm, b.
10.61 rads-2, c. 12.73 ms-2)
Chapter 1: Rotational Dynamics
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1.3 Work and Power in Rotational Motion
In rotational motion, work is done when a torque acts on a body and causes it to rotate
through an angular displacement. This is similar to linear motion, where work is done when a
force causes displacement.
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Work Done in Rotational Motion
[Illustration: Fig 1.11: Showing work done in rotational motion]
In rotational motion, work is done when a torque causes a body to rotate through an angular
displacement. This concept is very similar to linear motion, where a force does work when it
causes a displacement.
Just as force causes a linear displacement in translational motion, torque causes angular
displacement in rotational motion.
When a torque is applied to a body and the body turns through an angle, energy is transferred
— this is the work done in rotation.
W=τθ
Where:
W = work done,
τ = torque applied,
θ = angular displacement
This formula is valid only when torque and angular displacement act about the same axis.
Energy Perspective:
In linear motion, work done by a force increases the kinetic energy of a moving body.
Similarly, in rotational motion, work done by torque increases the rotational kinetic energy of
the object.
If a rotating body starts from rest and spins up to a certain speed, the work done by the torque
is stored as rotational kinetic energy:
1
W= 2 𝐼 ω2
This is a special case, showing how the energy input through torque results in rotational
motion.
Power in Rotational Motion
In linear motion, power is the rate at which work is done by a force moving an object at velocity
v
i.e. Power =force ×velocity
In rotational motion, power is the rate at which work is done by a torque rotating an object
at angular velocity ω. So, in rotational motion,
Power, P = τω
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The two formulas are analogous and are connected through the relationship between force &
torque, and velocity & angular velocity.
Example
Chapter 1: Rotational Dynamics
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1.3 An aircraft engine delivers a power of 200 kW to the propeller. The propeller rotates at a
constant frequency of 1500 rpm. (a) Calculate the torque exerted by the engine on the
propeller. (b) Determine the work done in one revolution of the propeller.
Solution: Here, power, P = 200 Kw = 2 × 105 W
1500
Frequency, f = 1500 rpm = 60 = 25 revs-1
⇒ω = 2𝜋f = 2𝜋 ×25 = 50𝜋 rad s-1
(a) Torque, 𝜏 = ?
We have, P = 𝜏ω
𝑃 2 ×105
⇒𝜏=𝜔 = = 1272.72 Nm
50𝜋
(b) Work, W = ?
2 ×105
We have, W = 𝜏θ = 2𝜋 = 8000 J
50𝜋
Test yourself
1.2 An aircraft engine delivers a power of 300 hp to the propeller. The propeller rotates
at a constant frequency of 1800 rpm.(a) Calculate the torque exerted by the engine on
the propeller. (b) Determine the work done in one revolution of the
propeller.(Ans:a.1186.8 Nm, b.7460 J)
Chapter 1: Rotational Dynamics
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1.1 Angular Momentum, Conservation of Angular Momentum
Angular momentum is the moment of linear momentum of a body about a point or axis,
defined as the cross product L⃗= r⃗×p⃗, where r is the position vector and p⃗=mv⃗ is the linear
momentum. It represents how far and how fast an object moves in a circular path. For a
rotating rigid body, it is also given by L=Iω, where I is the moment of inertia and ω is the
angular velocity. Angular momentum is a vector quantity with both magnitude and direction,
the latter determined by the right-hand rule. It is the rotational equivalent of linear
momentum, and in a closed system, it remains conserved unless acted upon by an external
torque. In simple terms, the faster an object spins, the more massive it is, and the farther its
mass is from the axis, the greater its angular momentum.
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[Illustration: Fig 1.12]
Let’s do the activity 1.3 and feel the effects of angular momentum.
Activity 1.3
Materials Required:
A Bicycle tire with a handle attached to the axle.
A stable stand to hold the tire in place for spinning.
Method:
(i) Secure the bicycle tire on the stand so that it can spin freely.
(ii) Spin the bicycle tire as fast as possible.
[Illustration: Fig 1.13]
Observe the spinning tire and predict what will happen when its angle is changed.
(iii) Take turns spinning the tire and carefully tilting it to different angles.
(iv) As you tilt the tire, you will feel a force pushing in the opposite direction, trying to
maintain angular momentum. During the time you made plane of the tire horizontal, the
spinning of the tire makes you spin in the opposite direction to conserve the principle of
angular momentum.
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Discussion:
(i) Explain the principle of angular momentum and how the force you felt is the tire's attempt
to maintain its angular momentum.
(ii) Relate this to other examples, such as a figure skater pulling in their arms to spin faster or
a fidget spinner resisting changes in angle when spinning.
(iii) Ask all your friends to share their experiences and observations.
(iv) Discuss how this principle can be observed in everyday life and other scientific
applications.
Conclusion: In this activity, we saw that a spinning tire resists changes in its direction. When
we tried to tilt it, we felt a force in the opposite direction. This happened because the tire was
trying to keep its spinning motion. This helped us understand the idea of angular momentum
and how it is conserved.
Conservation of Angular Momentum
The conservation of angular momentum means that if no outside force (torque) is acting, the
total angular momentum of a system stays the same.
For example, when a figure skater pulls in their arms, they spin faster. That’s because their
mass is closer to the axis, and to keep angular momentum constant, their speed increases.
Torque is a measure of how much a force causes an object to rotate. It is also defined as the
rate at which the angular momentum of the object changes with time. This means if a force
acting at some distance from the axis of rotation makes the spinning speed or direction of an
object change, that force creates a torque.
Mathematically, this is written as:
Change in angular momentum
Torque = Time taken
dL
Or, 𝜏 = dt
Just like force changes how an object moves in a straight line, torque changes how an object
spins or rotates.
d(Iω)
And 𝜏 = dt
d(Iω)
If no torque is applied on a rotating body, 𝜏 = 0 ⇒ =0
dt
Or, Iω = constant
Or, (Iω) o = (Iω) f
Or, I1ω1 = I2ω2 ⇒ Iω = constant.
Hence, in the absence of the torque applied, total angular momentum remains conserved. It is
the principle of conservation of angular momentum.
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Illustration:
[Illustration: Fig 1.14]
1. Sit in a chair with weights in your hands, extended away from your body.
2. Have someone give you an initial spin.
3. Pull the weights towards your body to spin faster.
4. Extend the weights away from your body to slow down.
By bringing your arms in, you reduce your moment of inertia and speed up. In an ideal
system without external forces, you could continue adjusting your speed by moving your arms
in and out resulting to a conservation of angular momentum.
Examples:
The principle of conservation of angular momentum implies Iω = constant i.e, an increase in
moment of inertia leads to decrease in rotation speed, ω and vice – versa.
(i) Imagine an ice skater spinning. When she pulls her arms in, she spins faster; when
she extends them, she slows down. This happens because pulling her arms in reduces
her moment of inertia ( I = mr2), increasing her spin rate.
[Illustration: Fig 1.15: Ice - Skater]
PHYSICS, Grade 12 16
Conversely, extending her arms increases her moment of inertia, slowing her
down. This illustrates the conservation of angular momentum, I 1ω1 = I2ω2
(ii) When a planet like Earth moves around the Sun, its speed changes because of the
conservation of angular momentum. When the planet comes closer to the Sun, the
distance between them becomes smaller, which lowers its moment of inertia. To
keep the angular momentum, the same, the planet must spin faster, so it moves
quicker. When the planet moves farther away, its moment of inertia becomes larger,
so its speed slows down. This idea is shown by the equation
I1ω1=I2ω2
which means the planet’s angular momentum stays constant as it moves.
(iii) In a tornado, air in the inner layer spins faster because it moves closer to the center.
This is due to the conservation of angular momentum, similar to a skater spinning
faster when pulling in their arms.
(iv) Spokes in a bicycle wheel help reduce the moment of inertia by keeping most of the
wheel’s mass closer to the center. This makes it easier for the wheel to spin faster
and with less effort. Without spokes, the mass would be spread out toward the rim,
increasing the moment of inertia and making the wheel harder to spin. Spokes not
only make the wheel lighter and easier to rotate, but also provide strength and
stability for smooth and efficient movement.
(v) When polar ice melts, water moves from the poles to the equator, increasing Earth's
moment of inertia. To conserve angular momentum, Earth slows down its rotation,
making the days slightly longer. This shows how the conservation of angular
momentum affects Earth's rotation due to changes in mass distribution.
(vi) When a diver jumps into the water, they pull their arms and legs in to spin faster
(reducing moment of inertia) or extend them to slow down (increasing moment of
inertia).
[Illustration: Fig 1.16]
This helps them control their rotation, demonstrating the conservation of angular
momentum.
PHYSICS, Grade 12 17
Analogical relation between linear and angular kinematics:
S.N Linear Motion Angular Motion
1 Linear displacement, s Angular displacement, θ
2 Linear velocity, v Angular velocity, ω
3 Linear acceleration, a Angular acceleration, α
4 Inertial mass, m Moment of inertia, I
5 Linear momentum, P= mv Angular momentum, L= Iω
6 Force, F=ma Torque, 𝜏 = Iα
7 Work done, W= F.s Work done, W = 𝜏θ
8 Translational KE= 12mv2 1
Rotational KE = 2Iω2
9 Power = Fv Power = , 𝜏ω
10 Equations of linear motion Equations of angular
v = u + at motion
v2 = u2 +2as ω = ω0 + αt
1 ω2 = ω02 + 2αθ
s = ut +2at2
1
θ = ω0t + 2αt2
Project Work
Project Title: Understanding Gyroscopic Motion and Stability
Objective: To experimentally measure gyroscopic effects, angular momentum, and stability
using a spinning wheel or gyroscope.
Step 1: Measuring Angular Momentum (L = Iω)
Materials Needed:
i) A bicycle wheel or toy gyroscope
ii) A rod or axle for suspension
iii) Measuring tape
iv) Weighing scale
v) Stopwatch
Procedure:
1. Find the Moment of Inertia (I):
i) Use a weighing scale to measure the mass m of the wheel.
ii) Use a measuring tape to find the radius r of the wheel.
PHYSICS, Grade 12 18
iii) For a solid disc (like a bicycle wheel), moment of inertia is:
1
I = 2 mr2
2. Measure Angular Velocity (ω):
i) Spin the wheel manually and count the number of rotations in 10 seconds.
ii) Use a stopwatch to measure time precisely.
iii) Calculate ω (angular velocity) using the formula:
ω=2π×rotations/time
3. Calculate Angular Momentum (L):
L=I×ω
Step 2: Measuring Gyroscopic Stability
Procedure:
i) Spin the wheel rapidly and suspend it using its axle.
ii) Apply a sideways force and observe how the wheel resists tilting and instead moves in a
circular motion (precession).
iii) Vary the spin speed and note how increasing speed makes precession stronger and
maintains stability.
iv) Time how long the wheel remains upright before falling when spun at different speeds.
v) Compare results: A fast-spinning wheel is significantly more stable than a slow-spinning
one.
Results
i) Record all measurements in a data table (mass, radius, rotations, time).
ii) Compare the calculated angular momentum for different speeds.
iii) Explain how gyroscopic effects apply to bicycles, drones, aircraft, and spacecraft
using real-world examples.
Execute yourself
Project 1 Investigating the Moment of Inertia of Different Shapes
Objective:
To study how the moment of inertia varies for objects of different shapes and mass
distributions and verify theoretical predictions through experiments.
PHYSICS, Grade 12 19
Exercise
1. A disk and a ring, both with the same mass and radius, are released from the top of an
inclined plane at the same time. Which one will reach the bottom first?
a) Disk , b) Ring, c) Both will reach at the same time, d) It depends on the inclination
angle.
2. Two bodies of masses m1 amd m2 move in a circles of radii r1 and r2, respectively.
What will be the ratio of their linear velocities v1 and v2 if they complete the circles in
equal time?
m r
a) m1 b) r1
2 2
m r
c) 1 d) m1 r1
2 2
3. A figure skater pulls in her arms during a spin. What happens to her moment of
inertia and angular velocity?
a) Moment of inertia decreases, angular velocity increases
b) Moment of inertia decreases, angular velocity decreases
c) Moment of inertia increases, angular velocity increases
d) Moment of inertia increases, angular velocity decreases
4. Which one has the highest moment of inertia a ring, a solid sphere and a disc all of
same mass and radius?
a) Ring, b) Solid sphere,
c) Disc, d) Solid cylinder.
5. Which factor does not affect the moment of inertia of a rotating body?
a) Mass of the body
b) Distribution of mass around the axis
c) Shape of the body
d) Temperature of the body
6. Why does a diver tuck in their body during a dive to increase their rotational speed?
a) To increase moment of inertia, b) To decrease moment of inertia
c) To increase torque d) To decrease torque
7. Two wheels of different radii and same masses are rolling without slipping on a
horizontal surface. What happens to the rotational kinetic energy if their linear
speeds are equal?
a) Larger radius has more energy b) Smaller radius has more energy, c) Remains
same d) The one with greater angular speed
8. What are the components of the total kinetic energy of a solid cylinder rolling down
an incline without slipping?
a) Only translational b) Only rotational
c) Half translational, half rotational d) Two-thirds translational, one-third rotational
PHYSICS, Grade 12 20
9. What happens to the torque if the angle between r and F is 900?
a) Torque is zero, b) Torque is maximum, c) Torque is minimum, d) Torque is
constant
10. Which one has the highest moment of inertia a ring, a solid sphere and a disc all of
same mass and radius?
a) Ring, b) Solid sphere, c) Disc, d) hollow sphere.
11. Why does a gyroscope remain stable and resist changes in its orientation?
a) Due to its moment of inertia, b) Due to its angular velocity
c) Due to conservation of angular momentum, d) Due to its mass distribution
12. What happens to the angular momentum of a rotating object if the radius is halved
and the mass is kept constant?
a) It remains the same b) It is doubled c) It is halved d) It is quadrupled
13. What is the angular velocity of a minute hand in rads-1?
𝜋 𝜋 𝜋 𝜋
a) 60 b) 180 c) 1800 d) 360
14. While observing a gymnast performing a spin, one might wonder how she manages to
control her rotational speed.
a. Define moment of inertia and explain how it depends on the distribution of mass.
b. State the principle of conservation of angular momentum. Explain why a gymnast
spins faster when pulling in her arms.
c. During a dramatic stage performance, a dancer (mass 60 kg) begins spinning on a
frictionless turntable with her arms outstretched. She is approximated as a rotating
body with a moment of inertia of 5 kg·m² and an angular velocity of 2 rads-1.
Suddenly, she pulls her arms in, decreasing her moment of inertia to 2 kg·m² in 0.5
seconds.
i) What is her new angular velocity after pulling in her arms?
ii) What is the change in her rotational kinetic energy?
iii) What average torque would be required to achieve this change if it were done by
an external force (hypothetical) (Ans: a. 5 rads-1, b. 15 J, c. 30 Nm)
15. The motion of a wheel rolling down an incline without slipping combines rotation and
translation.
a. Define rolling motion and explain the condition for pure rolling.
b. Derive the relation between linear velocity and angular velocity for a body
undergoing pure rolling.
c. A uniform rod of length 1.2 m and mass 2.4 kg is placed on a smooth horizontal
surface. A force of 12 N is applied perpendicularly at one end of the rod.
i) What is the torque produced about the center of mass of the rod?
ii) What is the angular acceleration of the rod about its center of mass?
iii) What is the linear acceleration of the center of mass?
( Ans: a. 7.2 Nm, b. 25 rads-2, c. 5 ms-2)
PHYSICS, Grade 12 21
16. A rigid body rotates about a fixed axis, and its distribution of mass affects its
rotational inertia.
a. Define radius of gyration. How is it related to the moment of inertia?
b. Explain how moment of inertia varies for different geometrical bodies like a ring,
disc, and sphere (mention any two with reasoning).
c. A uniform rod of mass 2 kg and length 1 m lies on a smooth inclined plane making
30° with the horizontal. The rod is hinged at its upper end to the incline and initially
held parallel to the incline. It is then released and begins to rotate downwards.
i) Calculate the torque acting on the rod about the hinge just after it is released.
ii) Determine the angular acceleration of the rod at that instant. ( Ans: a. 4.9 Nm, b.
3.75 rads-2)
17. The torque required to rotate a rigid body plays a central role in its dynamics.
i. Define torque and give its SI unit. Write the vector formula and explain each term.
ii. State the relation between torque and angular acceleration. What is the rotational
analogue of Newton’s second law?
iii. A constant torque of 1000 N·m turns a wheel of moment of inertia 200 kgm2about
an axis through its centre. Calculate its angular velocity after 3 sec.( Ans: 15 rads-1)
18. A body is rotating with angular acceleration and constant torque.
a. Define angular acceleration. Write down the kinematic equations of rotational
motion for constant angular acceleration.
b. Differentiate between angular displacement, angular velocity, and angular
acceleration.
c. A disc starts from rest and accelerates uniformly to 20 rad/s in 5 seconds. Find (i)
angular acceleration, and (ii) angle turned during this time. .( Ans: 4 rads-2, , 50 rad)
19. Suppose a flywheel is rotating about its center. Think that masses are symmetrically
attached at different distances.
a. Define center of mass. How does its position affect the rotational stability?
b. Explain the concept of rotational kinetic energy. How is it different from
translational kinetic energy?
c. Refer to the given figure:
[Illustration: Fig 1.17]
PHYSICS, Grade 12 22
Two masses of 1 kg and 2 kg are placed at 0.2 m and 0.1 m from the center of a
massless rod. Find the total moment of inertia of the system about the center. (Ans:
0.06 kgm2)
19. Long rods rotating about different axes behave differently based on how the mass
is spread.
a. Why is moment of inertia considered the rotational analogue of mass?
b. Derive the expression for the moment of inertia of a uniform rod about an axis
through its center and perpendicular to its length.
c. A rod of mass 5 kg and length 1 m rotates about its center. Calculate its moment of
inertia and its rotational kinetic energy when spinning at 12 rads-1.\ .( Ans: 0.4167
kgm2, 30 J)
20. In the figure aside, a uniform rod of length L and mass M is pivoted at its center and
can rotate freely in a horizontal plane. Two small masses, m1 and m2 , are attached at
equal distances from the center.
[Illustration: Fig 1.18]
(Assume the figure shows a horizontal rod with equal masses at both ends, pivoted at
the center.)
a. Write the expression for the total moment of inertia of the system about the central
pivot. Mention each contributing term.
b. How will the moment of inertia change if one mass is moved closer to the center?
Justify your answer conceptually.
c. Given:
- Rod: mass M=3 kg, length L=1.2 , Point masses: m1=m2=1kg
- Each point mass is at a distance 0.6 m from the center
Calculate the total moment of inertia of the system about the center. .
(Ans: 1.08 kgm2)
21. A figure skater demonstrates how angular momentum is conserved by changing her
body position mid-spin.
a. Define angular momentum for a rotating body. State its SI unit.
b. State and explain the principle of conservation of angular momentum. Give one
real-life example other than a skater.
c. A skater of moment of inertia 3 kg·m² spins at 4 rad/s. She pulls in her arms and
reduces her moment of inertia to 1.5 kg·m². Find her new angular velocity. What does
this tell us about angular momentum? ( Ans: 8 rads-1.Angular momentum conserves
during the processes)
b. Derive the formula for work done by a torque in producing angular displacement.
c. A constant torque of 8 N·m is applied to a wheel. If it turns through an angle of 5
radians in 2 seconds, calculate the work done and average power delivered.( Ans:40 J,
20 W)
PHYSICS, Grade 12 23
22. A ballet dancer performs a spin with arms extended, then pulls her arms in to spin
faster.
(a) Name and state the physical principle that explains why her angular velocity
increases when she pulls her arms in.
(b) Derive the relation between angular momentum and moment of inertia for a
rotating body.
(c) If the dancer’s initial moment of inertia is 4 kgm2 and angular velocity is 2rads-1,
and she reduces her moment of inertia to 2 kgm2, calculate her new angular velocity.(
Ans: 4 rads-1)
23. A uniform rod of length L and mass M is free to rotate about an axis perpendicular to
its length and passing through its center. Two small masses m1 and m2 are attached at
distances d1 and d2 respectively from the center of the rod.
a. Give the significance of moment of inertia in rotational motion.
b. Write the formula for the moment of inertia of the uniform rod about the given
axis.
c. State the principle of conservation of angular momentum with a real-life example.
d. Calculate the total moment of inertia of the system (rod + masses) about the axis of
rotation. If the system is released from rest and rotates under a constant torque τ, find:
i. The angular acceleration α.
ii. The angular velocity after rotating through an angle θ.
Given:
M=6 kg, L=2 m , m1=2 kg at d1=0.5 m
m2 = 3kg at d2=0.8 m
τ=12 N⋅m, θ=π rad
1
(Take Irod = 12ML2 (5 marks) (Ans: 4.42 kgm2, 2.7 rads-2, 4.13 rads-1)
24. A figure skater of moment of inertia I spins at angular velocity ω1 with her arms
extended. She then pulls her arms in, reducing her moment of inertia to I2.
a. Define angular momentum and state its SI unit.
b. State the principle of conservation of angular momentum.
c. Explain why the skater spins faster after pulling her arms inward.
d. Given I1 =5 kg⋅m2, ω1=2 rads-1, and I2 = 2 kg⋅m2, calculate:
i. The new angular velocity ω2.
ii. The change in rotational kinetic energy of the skater. (Take rotational kinetic
1
energy K= 2Iω2 (Ans:5 rads-1, 15 J)
PHYSICS, Grade 12 24
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