Class 12 Physics Rotational dynamics Notes

Moment of inertia of rotating body

Let us consider a rigid body which consists of particles of masses m₁, m₂, m₃, —- m_n with distance r₁, r₂, r₃, —- r_n from axis of rotation AB. The moment of inertia of the body about axis AB is defined by,

Fig: A rigid body rotating

I = m₁r₁² + m₂r₂² + m₃r₃² + —- + m_nr_n²

I = Σmr²

Thus, moment of inertia of a rigid body is defined as the sum of product of mass and square of distance from axis of rotation of individual particles.

Moment of inertia of uniform rod

1) About the axis passing through centre and perpendicular to its length

Let us consider a uniform rod of length ‘l’ and mass ‘m’. Let AB be the axis passing through centre and perpendicular to its length about which moment of inertia is to be determined.

Fig: Moment of inertia of uniform rod

Consider a small segment of length dx at distance ‘x’ from centre. Now mass per unit length of the rod = m/l. So, mass of the small segment (dx) = (m/l) dx.

Now, moment of inertia of small segment about axis AB,

dI = dm·x²

∴ dI = (m/l)x² dx — (i)

Thus, moment of inertia of whole rod about axis AB is obtained by integrating eqn (i) as,

I = ∫_−l/2^l/2 dI

= ∫_−l/2^l/2 (m/l)x² dx

= (m/l) ∫_−l/2^l/2 x² dx

= (m/l) [x³/3]_−l/2^l/2

= (m/l) [ (l/2)³/3 − (−l/2)³/3 ]

= (m/l) [ l³/24 + l³/24 ]

I = ml²/12

This is required expression for M.I of a uniform rod when the axis passing through centre and perpendicular to its length.

2) About the axis passing through one end and perpendicular to length

Let us consider an uniform rod of length ‘l’ and mass ‘m’. Let AB be the axis passing through one end and perpendicular to its length about which M.I is to be determined.

Fig: Moment of inertia of rod

Consider a small segment of length ‘dx’ at distance ‘x’ from one end as shown in figure. Now, mass per unit length of the rod = m/l. So, mass of the segment (dx) = (m/l) dx.

Now, moment of inertia of small segment about axis AB,

dI = dm·x²

= (m/l)x² dx — (i)

Thus, moment of inertia of whole rod about axis AB is obtained by integrating eqn (i) as,

I = ∫₀^l dI

= ∫₀^l (m/l)x² dx

= (m/l) ∫₀^l x² dx

= (m/l) [x³/3]₀^l

= (m/l) · l³/3

I = ml²/3

This is required expression for M.I of a uniform rod when the axis passing through one end and perpendicular to its length.

Kinetic energy of rotating body

Let us consider a rigid body which consists of particles of masses m₁, m₂, m₃, —- m_n with distance r₁, r₂, r₃, —- r_n from axis of rotation AB. Also, let ω be the angular velocity with which it is rotating and v₁, v₂, v₃, —- v_n be the linear velocity of respective particles m₁, m₂, m₃, —- m_n.

Fig: A rigid body rotating

v₁ = r₁ω, v₂ = r₂ω, —- v_n = r_nω

Now, K.E of 1st particle,

T₁ = 1/2 m₁v₁²

= 1/2 m₁r₁²ω²

Similarly, rotational K.E of other particles can be written as,

T₂ = 1/2 m₂r₂²ω²

T₃ = 1/2 m₃r₃²ω²

T_n = 1/2 m_nr_n²ω²

So, Rotational K.E of whole body = Sum of K.E of individual particles

Rotational K.E = T₁ + T₂ + T₃ + —- + T_n

= 1/2 m₁r₁²ω² + 1/2 m₂r₂²ω² + 1/2 m₃r₃²ω² + —- + 1/2 m_nr_n²ω²

= 1/2 ω²(m₁r₁² + m₂r₂² + m₃r₃² + —- + m_nr_n²)

= 1/2 ω²I [∵ I = Σmr²]

∴ Rotational K.E = 1/2 Iω²

So, the rotational K.E of a body is equal to the half of the product of the momentum of inertia of the body and the square of the angular velocity of the body about the given axis of rotation.

Torque acting on rigid body

Let us consider a rigid body which consists of particles of masses m₁, m₂, m₃, —- m_n with distance r₁, r₂, r₃, —- r_n from axis of rotation AB. Suppose torque (T) is applied on the body which produces angular acceleration (α) on the body.

Let F₁, F₂, —- F_n are the forces acting on individual particles producing acceleration a₁, a₂, —- a_n such that a₁ = r₁α, a₂ = r₂α, —- a_n = r_nα.

Rigid body used for torque derivation

Now, force acting on first particle,

F₁ = m₁a₁

= m₁r₁α

Again, torque acting on this particle about axis of rotation,

T₁ = r₁F₁

= r₁m₁r₁α

= m₁r₁²α

Similarly, torque acting on other particles,

T₂ = m₂r₂²α, T₃ = m₃r₃²α, —- T_n = m_nr_n²α

Thus, net torque on the whole body = Sum of individual torque

T = T₁ + T₂ + T₃ + —- + T_n

= m₁r₁²α + m₂r₂²α + m₃r₃²α + —- + m_nr_n²α

= α(m₁r₁² + m₂r₂² + m₃r₃² + —- + m_nr_n²)

T = αI [∵ I = Σmr²]

This is the relation between the moment of inertia of a body and torque.

Angular momentum

Moment of linear momentum of an object is called angular momentum. It is denoted by ‘L’ and given by

L = linear momentum × perpendicular distance from axis of rotation

∴ L = mvr — (i)

Since v = ωr

L = m(ωr)r

L = mωr² — (ii)

Fig: Angular momentum of particle

Eqn (i) and (ii) is an expression for angular momentum. It is a vector quantity and its unit is kg m²/s.

Angular momentum

Let us consider a rigid body which consists of particles of masses m₁, m₂, m₃, —- m_n with distance r₁, r₂, r₃, —- r_n from axis of rotation AB. Also let the body be rotating with angular velocity ‘ω’ and v₁, v₂, v₃, —- v_n be the linear velocities of respective particles m₁, m₂, m₃, —- m_n. Then,

v₁ = r₁ω

v₂ = r₂ω

v₃ = r₃ω

⋮

v_n = r_nω

Now, linear momentum of 1st particle,

p₁ = m₁v₁

= m₁r₁ω

So, angular momentum of 1st particle be,

L₁ = r₁p₁

= r₁m₁r₁ω

= m₁r₁²ω

Similarly, angular momentum of other particles are,

L₂ = m₂r₂²ω

L₃ = m₃r₃²ω

⋮

L_n = m_nr_n²ω

Thus, the total angular momentum of whole body be,

L = L₁ + L₂ + L₃ + —- + L_n

= m₁r₁²ω + m₂r₂²ω + m₃r₃²ω + —- + m_nr_n²ω

= ω(m₁r₁² + m₂r₂² + m₃r₃² + —- + m_nr_n²)

= ωΣmr²

L = ωI

This is required relation between angular momentum and moment of inertia of a body which shows that the magnitude of angular momentum of a body about given axis is equal to the product of moment of inertia ‘I’ of the body and its angular velocity ‘ω’ about that axis.

Relation between angular momentum and torque

We have, angular momentum ‘L’ of a rigid body rotating about axis with angular velocity ‘ω’ is,

L = Iω — (i)

where, I = moment of inertia of a body

Different eqn (i) with respect to time, we get

dL/dt = d/dt (Iω)

dL/dt = I dω/dt

dL/dt = Iα — (ii)

where α = dω/dt = angular acceleration

Also, torque on the body is,

τ = Iα — (iii)

Comparing eqn (ii) and (iii) we get,

τ = dL/dt

This is required relation between torque and angular momentum and this relation shows that torque acting on a body is equal to the rate of change of angular momentum of the body.

Principle of conservation of angular momentum

It states that: “If no external torques act on the system then total angular momentum remains conserved.”

i.e. If τ = 0 then L = constant

⇒ Iω = constant

In general I₁ω₁ = I₂ω₂

Proof

Since torque acting on a body is equal to the rate of change of angular momentum.

i.e. τ = dL/dt

If τ = 0 then,

dL/dt = 0

dL = 0

On integrating, we get

L = constant

Iω = constant

In general I₁ω₁ = I₂ω₂

Which proof the principle of conservation of angular momentum.

Work done by a couple

Let us consider, a wheel be acted upon by a couple of forces (F, F) at point A and B. Let the wheel turns through angle ‘θ’ in a time ‘dt’ such that points A and B are displaced to point A′ and B′ and let ‘S’ be the linear displacement.

Fig: Workdone by torque

Now, workdone by force at point A,

W_A = F·S

Also, the workdone by force at point B,

W_B = F·S

∴ Total workdone (W) = W_A + W_B

= F·S + F·S

= 2FS — (i)

Also, from figure,

θ = S/r

or, S = θr — (ii)

∴ Workdone (W) = 2Fθr

= (F·2r)θ

= Tθ [∵ T = F·2r is torque due to couple]

So, workdone by couple (W) = Tθ — (iii)

Thus, workdone by a couple is the product of torque and the angle of rotation of rigid body.

Power

Power (P) = dW/dt

= d(Tθ)/dt

= T dθ/dt

P = Tω

Radius of gyration

The perpendicular distance between centre of mass and axis of rotation of a rigid body is called radius of gyration.

If ‘m’ be the mass of body and k is its radius of gyration, then moment of inertia of body be given by,

I = mk²

Translation motion and rotational motion

Translation motion	Rotational motion
1) Linear displacement, S	Angular displacement, θ
2) Linear velocity, v = dS/dt	Angular velocity, ω = dθ/dt
3) Linear acceleration, a = dv/dt = d²S/dt²	Angular acceleration, α = dω/dt = d²θ/dt²
4) Mass, m	M.I = I
5) Linear momentum, p = mv	Angular momentum, L = Iω
6) Force, F = dp/dt = ma	Torque, τ = dL/dt = Iα
7) Workdone by force, W = FS	Workdone by torque, W = τθ
8) Translation K.E = 1/2 mv²	Rotational K.E = 1/2 Iω²
9) Eqn of translation motion (i) S = ut (ii) v = u + at (iii) S = ut + 1/2 at² (iv) v² = u² + 2aS	Eqn of rotational motion (i) θ = ωt (ii) ω = ω₀ + αt (iii) θ = ω₀t + 1/2 αt² (iv) ω² = ω₀² + 2αθ

Solved numericals

Numerical 1

A constant torque of 500 Nm turns a wheel which has a moment of inertia 20 kgm² about its centre. Find the angular velocity gained in 2 s and the K.E gained.

Given,

Torque (T) = 500 Nm

M.I (I) = 20 kgm²

t = 2 s

Angular velocity (ω) = ?

K.E = ?

Now,

T = Iα

500 = 20 × α

α = 25 m²/s

Again,

ω = ω₀ + αt

ω = 0 + 25 × 2

ω = 50

Again,

K = 1/2 Iω²

= 1/2 × 20 × (50)²

= 25000 J

Numerical 2

A constant torque of 200 Nm turns a wheel about its centre. The moment of inertia about this axis is 100 kgm². Find (i) the angular velocity gained in 4 s (ii) the K.E gained after 20 revs.

Given,

Torque (T) = 200 Nm

M.I (I) = 100 kgm²

T = Iα

200 = 100 × α

α = 2 m/s²

(i) t = 4 s

ω = ω₀ + αt

ω = 0 + 2 × 4

ω = 8 rad sec⁻¹

(ii) The K.E gained after 20 revs

θ = 2π

20θ = 20 × 2π

= 40π

Now,

ω² = ω₀² + 2αθ

= 0 + 2 × 2 × 40π

= 160π

K.E = 1/2 Iω²

= 1/2 × 100 × 160 × 3.14

= 8000 × 3.14

= 25133 J

Numerical 3

A ballet dancer spins about a vertical axis at 1 revolution per second with her arms stretched. With her arms folded her moment of inertia about the axis decrease by 40%. Calculate the new rate of revolution.

Soln,

f₁ = 1 rev/sec

I₁ = I (100%)

I₂ = 60% of I

= 0.6 I

f₂ = ?

I₁ω₁ = I₂ω₂

I₁·2πf₁ = I₂·2πf₂

I × 2 × 3.14 × 1 = 0.6 I × 2 × 3.14 × f₂

f₂ = 1.67 rps

Numerical 4

An electric fan is turned off and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4 s. (a) Find the angular acceleration and the number of revolution made by the motor in 4 s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant?

Soln,

500 rev/min — 4 sec → 200 rev/min — t = ? → 0 rev/sec

In case first

f₁ = 500 rev/min

f₂ = 200 rev/min

t = 4 s

ω = ω₀ + αt

2πf₂ = 2πf₁ + αt

2 × 3.14 × 200/60 = 2 × 3.14 × 500/60 + 4α

20.93 = 52.33 + 4α

α = −7.85

Source incomplete: The supplied scanned PDF ends here. The remaining requested calculation for the number of revolutions and part (b) is not shown in the provided source, so it has not been invented.

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Class 12 Physics Rotational dynamics Notes

Rotational Dynamics

Moment of inertia of rotating body

Moment of inertia of uniform rod

1) About the axis passing through centre and perpendicular to its length

2) About the axis passing through one end and perpendicular to length

Kinetic energy of rotating body

Torque acting on rigid body