LECTURE: 05

COURSE CODE: PHY 109

COURSE TITLE: ENGINEERING PHYSICS-1

Course Instructor:

Dr. Sonia Akter Ema

Assistant Professor
Department of Mathematical & Physical Sciences
East West University
Email: sonia.ema@ewubd.edu
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 Contents

❖ Rotational motion and circular motion

❖ Rotational dynamics
❖ Moment of inertia
❖ Torque
❖ Angular momentum
.

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 Rotational Dynamics

▪ We studied motion but along a line i.e. a linear motion.

▪ Here we will study rotational motion but with its causes. Hence the name Rotational Dynamics.

▪ Before we go ahead, it is important to understand the Rotational and Circular motion.

Rotational motion: When an object is turning Circular motion: When an object moving in a
around itself. circle around a point or another object.
Example: Rotation of Earth around itself.

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• In circular motion, the body rotates around a fixed axis that is outside the body.

• But in rotational motion, the body's rotating axes are inside the body.

• In rotational motion, the axis of rotation could change.

• Whereas in a circular motion, the axis of rotation does not change.

Examples of rotational motion:

• The motion of the wheel, gears, motors, etc. is rotational motion
• The motion of the blades of the helicopter is also a rotational motion
• A spinning top, the motion of a Ferris wheel in an amusement park
• Rotation of Earth around itself

Examples of circular motion:

• When a car moves around a circular track at some speed
• When a spacecraft orbiting the Earth
• When you swing an object tied with a rope in a circular path.

To know more: https://www.khanacademy.org/science/bridge-course-class-9th-science/x6eb88c2e9caec515:week-

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3/x6eb88c2e9caec515:motion/v/circular-and-rotational-motion
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What Is the Moment of Inertia?

The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration, which is the
sum of the product of the mass of every particle with its square of the distance from the axis of rotation.

❑ The moment of inertia is also known as the angular mass or rotational inertia.
❑ The SI unit of moment of inertia is kg m².

The moment of inertia is usually specified with respect to a chosen axis of rotation. It mainly depends on the
distribution of mass around an axis of rotation. MOI varies depending on the axis that is chosen.

What Are the Factors on Which the Moment of Inertia Depends?

The moment of inertia depends on the following factors:

• The density of the material

• Shape and size of the body
• Axis of rotation (distribution of mass relative to the axis)

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Moment of Inertia Formula

In general form, the moment of inertia is expressed as I = m × r²

Here

m = Sum of the product of the mass.

r = Distance from the axis of the rotation.

Moment of Inertia of a System of Particles

The moment of inertia of a system of particles is given by,

I = Σ mᵢrᵢ²

where rᵢ is the perpendicular distance from the axis to the 𝑖 𝑡ℎ particle, which has mass mᵢ.
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 Torque

❑ Torque is the vector cross product of the

displacement of the force with respect to an
arbitrary origin and the force that is acting.

τ=r×F
❑ Unit of torque: Nm

❑ The magnitude of torque is

τ = rFsinθ
where θ is the smallest angle between r and F.

Accounting for not only the force but

also its location on the object is torque! 8
 Direction of the Torque

To know more: https://www.youtube.com/watch?v=lwCF0khryK8

https://www.youtube.com/watch?v=T99yH_gw3p8
https://www.youtube.com/watch?v=hBXO-OyZxIU
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 RELATION BETWEEN TORQUE AND MOMENT OF INERTIA

Consider a rigid body rotating about a given axis with a uniform angular acceleration α, under the action of a
torque.

Let the body consist of particles of masses m₁, m₂, m₃, ..., mₙ at ⊥ distance r₁, r₂, r₃, ..., rₙ respectively from the
axis of rotation. (as shown in figure).

As the body is rigid, angular acceleration α of all the particles of the

body is the same. If a₁, a₂, a₃, ..., aₙ are the respective linear
accelerations of the particles, then,

a₁ = r₁α, a₂ = r₂α, a₃ = r₃α

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Force on particle of mass m₁ is:
f₁ = m₁a₁ = m₁r₁α
Moment of this force about the axis of rotation:
f₁ × r₁ = (m₁r₁α) × r₁ = m₁r₁²α

Similarly, moment of forces on other particles about the axis of rotation are m₂r₂²α, m₃r₃²α, .....mₙrₙ²α

Torque acting on the body,

τ = m₁r₁²α + m₂r₂²α + m₃r₃²α + ...mₙrₙ²α

τ = (m₁r₁² + m₂r₂² + m₃r₃² + ...mₙrₙ²)α

So, τ = (Σᵢⁿ mᵢrᵢ²) α

τ = Iα

where mᵢrᵢ² = moment of inertia of the body about the given axis of rotation.

If α = 1, τ = 1 × 1 or I = τ
τ⃗ = Iα⃗
 Radius of Gyration
❑ As we said before, theoretically you can find M.I. only if you know the mass distribution in object. But
experimentally, we can find M.I. for all objects.

❑ Also, this M.I. depends upon mass of the object and how it is distributed around the axis of rotation. So, if we
only want to know the mass distribution around the axis.

Let us write M.I. as I = MK², where M is mass of object and this mass is effectively at a distance K, from the axis
of rotation.

This K, is called as Radius of Gyration. It means that the mass of object is effectively at a distance K from the axis
of rotation.

If K is radius of gyration, I = MK², is the M.I.

❑ Radius of Gyration, can be defined as the distance from the axis of rotation to a point where the total
mass of the body is supposed to be concentrated.

So, if you find M.I. at this point it will be equal to the M.I. that you would get in regular way. 12(Simple trick
!!)
Moment of inertia for some common objects

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Combined Translational and Rotational Motion

Translational

Rotational

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 DERIVATION FOR ROTATIONAL KINETIC ENERGY

Consider a body of random shape and having N particles, of masses m₁, m₂, ..., mₙ and
distance
r₁, r₂, ..., rₙ from the axis of rotation. Let ω be its angular velocity.

It is a rigid body (distance between any two points inside the body is constant) and the axis of
rotation is perpendicular to the paper. When the body rotates around its axis, each of these
particles performs uniform circular motion around this axis with the same angular speed ω.
But, as r (distance from the axis) is different for each particle, they will have different
translational velocity (v = rω). Let these velocities be v₁, v₂, ..., vₙ= 𝑟𝑁 ω.

Let us consider translational K.E. for the first particle,

K.E.₁ = (m₁v₁²)/2
but v₁ = r₁ω therefore, K.E.₁ = (m₁r₁²ω²)/2

Similarly for all other particles, we shall have translational K.E. to be:
K.E.₂ = (m₂r₂²ω²)/2 ,
K.E.₃ = (m₃r₃²ω²)/2 ,
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So on...
We know,
Rotational K.E. = sum of translational K.E. of all particles present in the body.
Therefore,
Rotational K.E. = (m₁r₁²ω²)/2 + (m₂r₂²ω²)/2 + ... + (mₙrₙ²ω²)/2
= (1/2) (m₁r₁² + m₂r₂² + ... + mₙrₙ²) (ω²)

But mr² = I (moment of Inertia)

Rotational K.E. = (1/2) (I₁ + I₂ + ... + Iₙ) (ω²)

Let I₁ + I₂ + ... + Iₙ = m₁r₁² + m₂r₂² + ... + mₙrₙ² = I

Rotational K.E. = (1/2) (I) (ω²) = Iω²/2

K.E. = 1/2 mv² (translation K.E)

K.E. = 1/2 Iω² (rotational K.E.)

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 Angular momentum
Angular momentum can be experienced by an object in two situations. They are:
Point object: The object accelerating around a fixed point. For example, Earth revolving
around the sun. Here the angular momentum is given by:

𝐿=𝑟χ𝑝

Where, ՜ is the angular velocity

𝐿

r is the radius (distance between the object and the fixed point about which it revolves)

՜ is the linear momentum.

𝑝

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Analogies between Linear Momentuωm and Angular Momentum

Linear Momentum Angular Momentum

1. 𝑝 = m 𝑣 1. 𝐿 = I ω
2. Unit of p is kg m/s 2. Unit of L is kg m²/s
Δ𝑝 Δ𝐿
3. 𝐹𝑛𝑒𝑡 = 3. τ𝑛𝑒𝑡 =
Δ𝑡 Δ𝑡
4. If there are no external forces on a 4. If there are no external torques on
system of particles, momentum of the a system of particles, angular
system is conserved momentum of the system is
conserved
 Conservation of angular momentum

❑ If there are no external torques on a system of particles, angular momentum of

the system is conserved
So, If external net torque = 0, the sum of angular momentum of the system is zero.

If Γ=0, L = σ𝑁
𝑖=1 𝐼𝑖 ω 𝑖 = constant

❑ For a system with initial moment of inertia I₁ and initial angular velocity ω₁, its
initial angular momentum is I₁ω₁. If the system changes its moment of inertia to
I₂ and angular velocity ω₂, its final angular momentum is I₂ω₂. If there is not any
external net torque, then
I₁ω₁ = I₂ω₂

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 Conservation of Angular Momentum

Just like linear momentum of an isolated system remains

conserved in absence of an external unbalanced force, we
can show that Angular momentum also remains
conserved.
We know,
𝐿=𝑟χ𝑝

where ՜ is the position vector from the axis of rotation

𝑟
and ՜ is linear momentum. Differentiating w. r. t. time,
𝑝

Therefore, we conclude that, angular momentum remains

conserved in absence of an external torque. 23
Applications of Conservation of linear momentum

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 Examples of Angular Momentum
We knowingly or unknowingly come across this
property in many instances. Some examples are
explained below.
Gyroscope
Ice-skater
A gyroscope uses the principle of angular momentum to
When an ice-skater goes for a spin she starts off
maintain its orientation. It utilises a spinning wheel that has
with her hands and legs far apart from the centre of
3 degrees of freedom. When it is rotated at high speed it
her body. But when she needs more angular
locks on to the orientation, and it won’t deviate from its
velocity to spin, she gets her hands and legs closer
orientation. This is useful in space applications where the
to her body. Hence, her angular momentum is
attitude of a spacecraft is a really important factor to be
conserved, and she spins faster.
controlled.

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 Mathematical Problems

1. A particle executes 120 revolutions per minute along the circular path of radius 1.5 m. What are (a)
linear velocity, (b) time period, (c) angular velocity? Answers: 18.8 m/s ; 0.5 sec ; 12.57 rad/s
2. A flag spinner is trying to spin a flag 1000 revolutions in a row. The average angular speed of the flag
is 2 rad/s. How much time it will take the flag spinner to reach her goal? Answer: 314 s
3. Distance between the moon and Earth is 3.84× 105 km and the moon is rotating in a circular orbit
around the Earth and completes one rotation in 27.3 days. Calculate the linear and angular speeds of
the moon. Answers: 2.66× 10−6 𝑟𝑎𝑑𝑠 −1 ; 1.022 𝑘𝑚𝑠 −1
4. A solid cylinder of mass 15 kg is rotating on its own axis with angular velocity 50 𝑟𝑎𝑑𝑠 −1 . Radius of
the cylinder is 0.2 m. Calculate the rotational K.E. and angular momentum of the cylinder. Answers:
E= 375 J ; L= 15 𝑘𝑔𝑚2 𝑠 −1
5. Planet Mars is rotating in a circular orbit of radius of 2.28× 1011 𝑚 around the sun. Calculate its
angular momentum. Mass of Mars is 6.64× 1023 𝑘𝑔 and its time period is 5.94×107 s. Answer:
3.55×1039 𝑘𝑔𝑚2 𝑠 −1
6. If radius vector ՜ = 2 𝑖෡ + 3𝑗Ƹ + 2𝑘෠ and force vector ՜ = 2 𝑖෡ + 2𝑗Ƹ + 2𝑘෠ , then calculate the torque τ⃗.
𝑟 𝐹
Answer: τ⃗=2 𝑖෡ − 2𝑘෠ 26