Class 12 Rotational dynamics amazing definition and derivations

Rotational dynamics

In engineering physics, rotational dynamics is the branch of mechanics that studies the motion of objects that rotate or spin about an axis, and the forces and torques that cause or change this motion

Translation Motion And Rotational Motion: A body is said to be in translation motion if each particle of it has equal linear displacement in an equal interval of time. Similarly, a rigid body is said to be in rotational motion if each particle of it has the same angular displacement in an equal interval of time

Note!
In case of rotational motion, all particles have different linear displacement, linear velocity, linear acceleration, but the same angular displacement, angular velocity, and angular acceleration.

Rigid Bodies: A body is said to be a rigid body if it does not undergo any changes in shape and volume when an external force is applied to it. The distance between any two particles for such a body will remain the same no matter how large the external forces are applied on it.

Torque or Moment of Torque: The turning effect of force in a body is called Torque. Mathematically, it is defined as the product of force and its perpendicular distance from the axis of rotation. i.e.

τ = r F1

where f₁ is perpendicular to the moment of am(r) and also f₁ is the component of I along the axis of the rotation.

τ = r F Sinθ

in Vector form:

𝜏⃗=𝑟⃗×𝐹⃗

.τ is a vector quantity whose direction can be determined by the right-hand thumb rule from r and F. It may be clockwise or anti-clockwise. The SI unit of torque is N·m.

Moment of Inertia

The Moment of Inertia of a rigid body about a particular axis of rotation is defined as the sum of the product of the masses of all particles constituting the body and the square of their respective distances from the axis of rotation. From Newton’s second law of motion
F =ma
rF = mar
rF= mr ²α [∵a=rα]
τ = (mr ²) α

Where τ is torque, α is angular acceleration, and mr² is the moment of inertia (I). Thus:

Hence, the moment of inertia of a rigid body plays the same role as mass in linear motion. I is a scalar quantity, and its unit is kg·m² in the SI system.

Radius of Gyration: The radius of gyration is the distance from the axis of rotation to the point where the total mass of the body is assumed to be concentrated. It may also be defined as the square root of the mean square distance of the various particles of a body from the axis of rotation. It is denoted by k.

The moment of inertia of a body having mass m and radius of gyration k is given by:

I=Mk²

Moment of Inertia for a Body with n Particles

Consider a body consisting of n particles, each of mass m. Let r₁, r₂, r₃, …, rₙ be their perpendicular distances from the axis of rotation. The moment of inertia (I) of the body about the axis of rotation is:

Relation Between Torque and Angular Acceleration

Consider that a rigid body is rotating about an axis YY under the action of a constant torque (T). Also, let the body consist of n particles of masses m₁, m₂, m₃ … mₙ at particular distances r₁, r₂, r₃ … rₙ from the axis of rotation, respectively. Let the torque will produce a constant angular acceleration α in each particle.

Since the particle of mass m₁ follows a circular path of radius r₁, the magnitude of linear (or tangential) acceleration of this particle is:
α1=r1α
The net external force acting on this particle due to this acceleration is:
F1=m₁a₁=m₁r₁α
And, the magnitude of the torque acting on this particle due to this force is:
τ_i=F_ir_i=(m_ir_iα)r_i=m_ir_i²α
So, the total torque acting on the body is
τ=τ₁+τ₂+τ₃+⋯+τ_n for n particles

Relation Between Moment of Inertia and Rotational Kinetic Energy

Consider a rigid body rotating with uniform angular velocity ω about a given axis (YY′) as shown in the figure. Let m1, m2, m3,…,m_n be the masses of different particles constituting the body. Let r1, r2, r3,…,r_n be their respective distances from the axis of rotation.

Kinetic Energy of the particle of mass (m₁),
= ½ m₁v₁²
= ½ m₁r₁²ω² [∵ v = rω]
Kinetic Energy of the particle of mass (m₂),
= ½ m₂v₂²
= ½ m₂r₂²ω² and so on.
Rotational Kinetic Energy of the given body (E) is given by:
E = ½ m₁r₁²ω² + ½ m₂r₂²ω² + … + ½ mₙrₙ²ω²
= ½ (Σ mᵢrᵢ²) ω²
E = ½ Iω² ……………….1
Where Σ mᵢrᵢ² = I, moment of Inertia of the body. Equation 1 is the required relation.

Relation Between Angular Momentum and Moment of Inertia:

Consider a rigid body rotating about an axis YY’ as shown in the figure.Let m₁, m₂, m₃ … mₙ be the masses of the different particles constituting the body. Let r₁, r₂, r₃ … rₙ be their respective distance from the axis of rotation. Suppose the body be rotating with uniform angular velocity ω about the axis. Although each particle within the body has the same angular velocity ω, their linear velocity will be different.

Let v₁, v₂, v₃ … vₙ be the linear velocity of the particle of masses m₁, m₂, m₃ … mₙ respectively, then.
v₁ = r₁ω,
v₂ = r₂ω,
vₙ = rₙω
The angular momentum of the mass (m₁) is given by
L₁ = m₁v₁r₁
= m₁r₁ω.r₁
= m₁r₁²ω
Similarly, the angular momentum of the body of masses m₂, m₃ … mₙ can be expressed. Thus, the angular momentum of the body is
L = L₁ + L₂ + L₃ + … + Lₙ
= m₁r₁²ω + m₂r₂²ω + m₃r₃²ω + … + mₙrₙ²ω
= (m₁r₁² + m₂r₂² + m₃r₃² + … + mₙrₙ²)ω
= ( Σ mᵢrᵢ² )ω

∴ L = Iω … …….(1)
where Σ mᵢrᵢ² = I : is the moment of inertia of the body. Equation 1 is the required relation between angular momentum and moment of inertia.

Angular Momentum:

It is defined as the product of linear momentum and perpendicular distance between the body and the axis of rotation. If a body having mass m moves with velocity v at a distance r from the axis of rotation, then

Angular Momentum (L) = Linear Momentum (p) × perpendicular distance between the body and the axis of rotation.
= r×p
= rp sinθ n̂
= rpsin90
= rp
= rmv
= r m (rω)
= mr²ω
∴ L = mωr²

The SI unit of angular momentum is kgm²s⁻¹ and its dimension is [M L² T⁻¹]. It is a vector quantity. L=r×p

Conservation of Angular Momentum:-

If no external torque acts on a system, then the total angular momentum remains conserved or constant. Mathematically,
L = constant { dL/dt = J = dL/dt = 0 }
i.e. Iω = constant
I₁ω₁ = I₂ω₂
Proof:
The torque acting on a body about a given axis of rotation is defined as the rate of change of angular momentum.

i.e. dL/dt = J … (1)
If no external torque acts on a system then,
J = 0
so, dL/dt = 0 … (2)
Integrating equation 2 on both sides, we get
∫ dL/dt = ∫ 0
L = constant
Iω = constant
I₁ω₁ = I₂ω₂

Examples of conservation of angular momentum

Motion of a planet revolving in an elliptical orbit around the sun
Ballet Dancer

Moment of Inertia of a thin uniform rod about an axis passing through its centre and perpendicular to its length.

Consider a thin rod uniform rod having mass (m) and length (l), which is rotating about an axis passing through its centre and perpendicular to its length, as shown in the figure. Now, to find the moment of inertia, consider the small elements of the rod having mass dm, length dl, at a distance x from the axis YY’. Now,

The moment of Inertia of a small element is given by
dI = dm x² —-………….. (1)
The moment of inertia of the rod is given by
x² (2)
Now,
Mass of length $l$ of the rod =

Moment of Inertia of a thin uniform rod about an axis passing through its one end and perpendicular to its length

Let us consider a thin uniform rod having mass M and length l, which is rotating about an axis passing through its one end and perpendicular to its length, as shown in the figure.
Now,
Let us consider the small elements of the rod having mass dm, length dx at a distance x from the axis YY’.
Here, the Moment of Inertia of a small element is given by
dI = dm x² — (1)
The moment of Inertia of a small element of the rod is
I=∫dI = ∫dmx^{2 …………………(2)}

Now

Kinetic Energy of the rotating body

Let us consider a rigid body consisting a number of particles having masses at a distance from the axis YY’ as shown in the figure.

Now, the linear velocity of a particle having mass
$m_1, m_2, …, m_n$ is given by

v_n= r_nω
The K.E acting on the particle having

Kinetic Energy of a rolling body

A rolling body has two types of energies i.e. translational K.E and Rotational K.E.

Consider a wheel of mass M and radius r rolling over a plane surface as shown in figure. Let v be the velocity of body and ω be the angular velocity of the body. If I be the moment of inertia of a body about a given axis of rotation then,

K.E of a rolling body is given by

Which is the required expression for the K.E. of a Rolling body

Acceleration of a rolling body down an inclined plane (without slipping)

Consider a body having mass m and radius ‘r’ which is rolling down an inclined plane at an angle θ with the horizontal. Let the body require a linear velocity “v”. After rolling down the inclined plane of height h, the K.E. of a rolling body is given by

Loss in P.E of body = Mgh ………………………(2)

Since no slipping occurs, the mechanical energy is conserved: i.e.
K.E gain = P.E loss
Using eqn (1) and (2)

Let S be the length of the inclined plane. Then from ΔABC

Work and power in rotational Dynamics:

Q1. Define a Couple and derive the expression for the work done by a Couple.

Let us consider a wheel of radius r which is rotating about an axis passing through its centre O′ with angular velocity ω. Let θ be the angular displacement when two equal and opposite forces act at points A and B as shown in fig.

Now,
Work done by Couple (W) = work done by force at A + work done by force at B.
W=F×AA′+F×BB′……………(1)
From the figure,

AA′=BB′=rθ …………(2) (since the magnitude of force is the same)

Substituting in equation (1),

$W = F r θ + F r θ$

Since the torque due to the couple is given by,

τ=F×2r=2Fr……………….(4)

Using equation (4) in (3),

Which is the required expression for the work done by a couple.Now

Numerical Problem

Q1. A wheel starts from rest and accelerates with constant angular acceleration to an angular velocity of 15 revolutions per second in 10 seconds. Calculate the angular acceleration and the angle which the wheel has rotated at the end of 2 seconds. [2075 Set C]

Solution:
Initial angular velocity () = 0 rad/sec
Frequency () = 15 rev/sec
Time () = 10 seconds

Angular acceleration () = ?
Angular displacement () =?
at time () = 2 seconds

We know,

Now,

ω=ω0+αt
94.24=0+α×10
α=9.42rad/sec²

Again,
θ=ω×t′=94.24×2=188.48rad

2. A ballet dancer spins with 2.4 rev/s with her arms unstretched when the moment of inertia about the axis of rotation is I. With her arms folded, the moment of inertia about the same axis becomes 0.6I. Calculate the new rate of spin. [2015 CIE]

Solution:

Frequency (f₁) = 2.4 rev/s
Moment of Inertia (I₁) = I
Frequency (f₂) = ?
Moment of Inertia (I₂) = 0.6I

From conservation of angular momentum,
I₁ω₁ = I₂ω₂
or I × 2π × 2.4 = 0.6I × 2πf₂
or f₂ = 2.4 ÷ 0.6 = 4 rev/sec

3. An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4 seconds. Find the angular acceleration and the number of revolutions made by the motor in the 4 sec interval.
[2073 set D]

Solution:

Initial frequency (f₁) = 500 rev/min = = 8.33 rev/sec
Final frequency (f₂) = 200 rev/min = = 3.33 rev/sec

Time (t) = 4 sec
Angular acceleration (α) = ?
No. of revolutions (n) = ?
Now,

Compiled by: Er. Basant Kumar Yadav

Rotational Dynamics — Comparison Table

Rotational Dynamics — Side‑by‑Side with Translational Motion

A clean, compact comparison of the main quantities, formulas, and relations used in engineering physics.

S/N	Terms	Translational Motion	Rotational Motion	Relation
1	Displacement	Linear displacement s (metre)	Angular displacement θ (radian)	s = r θ
2	Velocity	Linear velocity v (m s⁻¹) initial u, final v	Angular velocity ω (rad s⁻¹) initial ω₀, final ω	v = ω r
3	Acceleration	Linear acceleration a (m s⁻²)	Angular acceleration α (rad s⁻²)	a = α r
4	Time	time (t) seconds	time (t) seconds	t = t
5	Equation of motion	v = u + a t s = u t + ½ a t² v² = u² + 2 a s	ω = ω₀ + α t θ = ω₀ t + ½ α t² ω² = ω₀² + 2 α θ	(constant a, α)
6	Mass / Inertia	Mass m (kg)	Moment of inertia I (kg m²)	I = m r² (point mass)
7	Cause	Force F (N)	Torque τ (N·m)	τ = r F_⊥
8	Momentum	Linear momentum p = m v	Angular momentum L	L = r × p (= I ω)
9	Newton's 2nd law	F = m a F = dp/dt	τ = I α τ = dL/dt	—

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