The study of rotational motion of a rigid body is said to be rotational dynamics. A solid body in which the particles are compactly arranged and their positions are not disturbed by any external force applied on it is called rigid body. There are three types of motion:

a) Translational motion

Eg. – A motion of stone when kicked, A motion of moving bus except tyres or passenger etc.

b) Rotational motion

Eg. moving of fan, axle of mill, motion of bike in double stand.
Here, angular displacement/ velocity / acceleration is constant but linear displacement / velocity / acceleration of each particle are different.

c) Translational + Rotational motion

Eg. motion of tyre of a moving bus, motion of football, motion of bullet fired from gun.

Equations of angular motion

1. Consider, a rigid body rotating about a fixed axis with constant angular acceleration [latex]\alpha [/latex]
   So,
   [latex]\alpha = \frac{d\omega}{dt}[/latex]
   [latex]d\omega = \alpha dt[/latex] ………….. (i)
   Using initial conditions, if t = 0, [latex]\omega = \omega_o[/latex]
   And, if t = t, [latex]\omega = \omega[/latex]
   Integrating [latex]eq^n. (i)[/latex], we get,
   [latex]\int_{\omega_o}^{\omega}d\omega = \alpha \int_{0}^{t} dt[/latex]
   Or, [latex][\omega]_{\omega_o}^{\omega} = \alpha[t]_{0}^{t}[/latex]
   Or, [latex]\omega – \omega_o = \alpha t[/latex]

2. Suppose, [latex]\omega[/latex] be the angular velocity of the rigid body at an instant t.
   So,
   [latex]\omega = \frac{d\theta}{dt}[/latex]
   Or, [latex]d\theta = \omega dt[/latex]
   Or, [latex]d\theta = (\omega_o + \alpha t)dt[/latex] ………… (ii)
   Using initial conditions, if t = 0, [latex]\theta = 0[/latex]
   And, if t = t, [latex]\theta = \theta[/latex]
   Integrating [latex]eq^n.[/latex] (ii), we get,
   [latex]\int_{\theta_o}^{\theta} d\theta = \int_{0}^{t} (\omega_o + \alpha t)dt[/latex]
   Or, [latex]\int_{\theta_o}^{\theta} d\theta = \int_{0}^{t} \omega_o dt + \alpha \int_{0}^{t} tdt[/latex]
   Or, [latex][\theta]_{\theta_o}^{\theta} = \omega_o[t]_{0}^{t} + \alpha [\frac{t^2}{2}]_{0}^{t}[/latex]
   Or, [latex]\theta – 0 = \omega_o(t-0) + \frac{1}{2}\alpha (t^2 – 0)[/latex]

3. The angular acceleration  can be expressed as:
   [latex]\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta}.\frac{d\theta}{dt}[/latex]
   
   Or, [latex]\alpha = \frac{d\omega}{d\theta}.\omega[/latex]
   
   Or, [latex]\omega.d\omega = \alpha.d\theta[/latex] …………….. (iii)
   Using initial conditions, if t = 0, [latex]\theta = 0, \omega = \omega_o[/latex]
   
                                       And, if [latex]t = t, \theta = \theta, \omega = \omega[/latex]
   Integrating eq^n. (iii), we get,
   [latex]\int_{\omega_o}^{\omega}\omega.d\omega = \int_{0}^{\theta}\alpha.d\theta[/latex]
   
   Or, [latex][\frac{\omega^2}{2}]_{\omega_o}^{\omega} = \alpha \int_{0}^{\theta}d\theta[/latex]
   
   Or, [latex](\frac{\omega^2 – \omega_o^2}{2}) = \alpha[\theta]_{0}^{\theta}[/latex]
   
   Or, [latex]\omega^2 – \omega_o^2 = 2\alpha(\theta – 0)[/latex]
   

Moment of Inertia

According to Newton’s first law of motion, object has no ability to change its state of rest or uniform linear motion by itself. This inability is called inertia. Similarly, a body rotating about an axis is unable to produce a change in its rotational motion. This inability of the rotating body is said to be rotational inertia.

Moment of inertia of a body about a given axis of rotation is the sum of the products of the masses of the particles of a body and square of their respective perpendicular distance from the axis of rotation.
It is analogous to mass in linear motion.
Here, in the figure, both masses are same. i.e. 5kg. But, r₁ = 1m and r₂ = 2 m. So, rotation will be easier in r_l = 1m.
This is due to more moment of inertia in r₂ = 2m than r₁ = lm. i.e. M.I. of any rigid body depends on the mass of the body as well as distance of the mass from the axis of rotation.

Moment of Inertia of a rigid body

Moment of inertia of any rigid body is defined as the resistance or inability of the body to change its state or rotation by itself.

Let us consider, a rigid body consisting of large number of small particles of masses m₁, m₂, ……., m_n at the distance r₁, r₂,……….. r_n from the axis of rotation YY’.

But, the moment of inertia of the masses m₁, m₂, ………………, m_n about the axis YY’ are given by,

I₁ = m₁r₁², I₂ = m₂r₂² and so on.

So, Total MI (I) = I₁ + I₂ + ………. + I_n

Or, I = m₁r₁² + m₂r₂² + ……….. + m_nr_n²

∴ I = [latex]\sum_{i=1}^{n}m_ir_i^2[/latex]

If m₁ = m₂ = …………. m_n = m.

I = m [latex]\sum_{i=1}^{n}r_i^2[/latex]

Its unit is Kgm².

Note: Moment of inertia of any rigid body is not constant. It depends on the axis of rotation.

Radius of Gyration (K)

We know, I = mr² = mk²

In above figure,

I = m₁r₁² + m₂r₂² + ………… + m_nr_n²

If m₁ = m₂ = …………. m_n = m.

∴ I = m (r₁² + r₂² + ……….. + r_n²)

Or, I = [latex]\frac{mn}{n}[/latex] (r₁² + r₂² + ……….. + r_n²)

= M ([latex]\frac{r_1^2+r_2^2+…………..+r_n^2}{n}[/latex])

Or, I = Mk², where, k = [latex]\sqrt{\frac{r_1^2+r_2^2+…………+r_n^2}{n}}[/latex]

Definition

It is defined as the root mean square of the distance of the particles of the body from the axis of rotation.

Moment of inertia of a thin uniform rod

a. About an axis passing through centre and perpendicular to length

Let us consider, a thin uniform rod of length ‘l’ and mass ‘M’, YY’ be the axis of rotation from the centre of rod as shown in figure.

Now, let us choose small portion of the rod of length dx and mass dm at the distance x from the axis of rotation.

So, mass per unit length of rod = [latex]\frac{M}{l}[/latex]

[latex]\therefore[/latex] mass of the strip (dm) = [latex]\frac{M}{l}.dx[/latex]

Or, [latex] dm = \frac{M}{l}.dx[/latex]

So, moment of inertia of the small strip is

[latex]dI = dm.x^2[/latex]

Or, [latex]dI = \frac{M}{l}x^2 dx[/latex] …………… (i)

Now, moment of inertia of rod about the axis YY’ is

[latex]I = \int_{-l/2}^{l/2}dl[/latex]

= [latex]\frac{M}{l}\int_{-l/2}^{l/2}x^2 dx[/latex] 

= [latex]\frac{M}{l}[\frac{x^3}{3}]_{-l/2}^{l/2}[/latex]

= [latex]\frac{M}{3l}[(\frac{l}{2})^3-(\frac{-l}{2})^3][/latex]

= [latex]\frac{M}{3l}[\frac{l^3}{8}+\frac{l^3}{8}][/latex]

= [latex]\frac{2Ml^3}{24l}[/latex]

∴ [latex]I = \frac{Ml^2}{12}[/latex] ……………… (ii)

Now, let, K be the radius of gyration of the rod about the axis YY’, then,

[latex]I = MK^2[/latex] …………… (iii)

From eq^ns. (ii) and (iii), we get,

[latex]K^2 = \frac{l^2}{12}[/latex]

Or, [latex]K = \frac{l}{\sqrt{12}}[/latex]

[latex]\therefore K = \frac{l}{2\sqrt{3}}[/latex]

b. About an axis passing through one of the edge and perpendicular to the length

Let us consider, a thin uniform rod of length ‘l’ and mass “M”. YY’ be the axis of rotation through one end of rod as shown in figure.

Now, let us choose small strip of the rod of length dx and mass dm at the distance x from the axis of rotation.

So, mass per unit length of rod = [latex]\frac{M}{l}[/latex]

Mass of small strip (dm) = [latex]\frac{M}{l}.dx[/latex]

So, moment of inertia of the small strip is dI = dmx² 

[latex]dI = \frac{M}{l}.dx.x^2[/latex] …………….… (i)

Now, moment of inertia of the rod about the axis YY’ is

[latex]I = \int_{0}^{l}\frac{M}{l}x^2 dx[/latex]

= [latex]\frac{M}{l}[\frac{x^3}{3}]_{0}^{l}[/latex]

= [latex]\frac{Ml^2}{3}[/latex]

[latex]\therefore I = \frac{Ml^2}{3}[/latex] ………. (ii)

Let, K be the radius of gyration of the rod about the axis YY’, then,

[latex]I = MK^2[/latex] ……………. (iii)

From equations (ii) and (iii), we get,

[latex]\frac{Ml^3}{3} = MK^2[/latex]

Or, [latex]K^2 = \frac{l^2}{3}[/latex]

∴ K = [latex]\frac{l}{\sqrt{3}}[/latex]

Relation between Torque ([latex]\tau[/latex]) and angular acceleration ([latex]\alpha[/latex])

Let us consider, a rigid body where YY’ be the axis of rotation as shown in figure. [latex]m_1, m_2, …………, m_n[/latex] be the masses of different particles of rigid body at the distance r₁, r₂, ……. r_n from the axis of rotation.

The torque [latex](\tau)[/latex] will produce a constant angular acceleration [latex](\alpha)[/latex]. The magnitude of linear acceleration of a particle of mass [latex]m_1[/latex] is related with angular acceleration as, [latex]a = r_1\alpha[/latex].

Since, Total torque [latex](\tau)[/latex] of rigid body = sum of torque due to all particles.

i.e. [latex]\tau = \tau_1 + \tau_2 + ……………… + \tau_n[/latex] ……. (i)

but, from definition,

[latex]\tau_1[/latex] = F₁r₁ = [latex]m_1a_1r_1 = m_1(r_1\alpha)r_1 = m_1r_1^2\alpha[/latex]

Now, from equation (i)

[latex]\tau = [m_1r_1^2 + m_2r_2^2 + ………. + m_nr_n^2]\alpha[/latex]

= [latex](\sum_{i=1}^{n}m_i r_i^2)\alpha[/latex]

                  = Iα

∴ τ = Iα.

Angular Momentum (L)

It is defined as the moment of linear momentum. It is the product of linear momentum of body or the particle and perpendicular distance from axis of rotation.

i.e. L = Pr.

Or, L = mvr.

Or, L = m(ωr)r = mωr².

Unit: Kgm²s^-1

Vectorically, [latex]\vec{L} = \vec{r}\times \vec{P}[/latex]

Relation between angular momentum (L) and moment of inertia (I):

Let us consider, a rigid body YY’ be the axis of rotation as shown in figure. m₁, m₂, …. m_n be the masses of different particles of rigid body at the distance r₁, r_{2, ……….} r_n form the axis of rotation.

Total angular momentum of rigid body = sum of angular momentum due to all particles.

i.e. L = L₁ + L₂ + ……… L_n.

but, from definition,

L₁ = P₁r₁ = m₁v₁r₁ = m₁ωr₁²   
[∵ ω₁ = ω₂ = ……….. = ω_n = ω]

Now, L = L₁ + L₂ + ……… L_n.

= [latex][m_1r_1^2 + m_2r_2^2 + ……. + m_nr_n^2]\times \omega[/latex]

= [latex][\sum_{i=1}^{n}m_i r_i^2]\times \omega[/latex]

= Iω

∴ L = Iω.

Relation between angular momentum (L) and Torque (𝛕):

Let us consider, a rigid body where YY’ be the axis of rotation as shown in figure. m₁ = m₂ = …………. m_n be the masses of different particles of rigid body at the distance r₁, r₂, …. r_n from the axis of rotation.

We know that, Total torque of rigid body

([latex]\tau[/latex]) = Iα ……… (i)

And, L = Iω. ………. (ii)

Differentiating (ii), with respect to t, we get,

= Iα ………….… (iii)

From (i) and (iii), we get,

[latex]\tau = \frac{dL}{dt}[/latex]

Principle of conservation of angular momentum:

If no external torque acts on the body or system (if the body of system is isolated), then, the total angular momentum of a body or system is always constant or conserved. i.e.

If [latex]\tau[/latex] = 0, L = Constant.

Proof:

Let us consider, a body of mass ‘m’ which is in rotational motion having angular momentum ‘L’ rotating with angular velocity ‘ω’, I is the total moment of inertia in the body.

We know,

[latex]\tau = \frac{dL}{dt}[/latex]

If external torque [latex]\tau[/latex] = 0, then,

[latex]\frac{dL}{dt}[/latex] = 0.

i.e. L = constant.

Iω = Constant.

Or, I₁ω₁ = I₂ ω₂

Hence, the total angular momentum of body is conserved.

Kinetic energy of rotating body / Rotational kinetic energy:

Let us consider, a rigid body where YY’ be the axis of rotation as shown in figure. m_1, m₂ and m_n be the masses of different particles of rigid body at the distance r₁, r₂, …….. r₂ from axis of rotation. We know that,

Total kinetic energy of rotation i.e.

KE_rot. = KE₁ + KE₂ + ….. KE_n

Or, KE_rot. 

= [latex]\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + …….. + \frac{1}{2}m_nv_n^2[/latex]

= [latex]\frac{1}{2}(m_1r_1^2 + m_2r_2^2 + …….. + m_nr_n^2)\omega^2[/latex]

[latex][\omega = \frac{v}{r}, \omega_1 = \frac{v_1}{r_1}, \omega_n = \frac{v_n}{r_n}][/latex]

= [latex]\frac{1}{2}\sum_{r=1}^{n}(m_ir_i^2)\omega^2[/latex]

KE_rot. = [latex]\frac{1}{2}I\omega^2[/latex]

∴ Rotational kinetic energy is defined as half the product of moment of inertia (I) and square of angular velocity (ω).

Kinetic energy of rolling body:

Here, total rolling kinetic energy (K.E._roll.) = Translational K.E. ([latex]K.E._{T}[/latex]) + rotational K.E. (K.E._rot.)

∴ KE_roll. = [latex]KE_T[/latex] + KE_rot.

= [latex]\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/latex]

= [latex]\frac{1}{2}m\omega^2R^2 + \frac{1}{2}mk^2\omega^2[/latex]        [∵ v = ωr]

∴ K.E._roll. = [latex]\frac{1}{2}m\omega^2 (R^2 + k^2)[/latex]

= [latex]\frac{1}{2}mv^2 + \frac{1}{2}mk^2.\frac{v^2}{R^2}[/latex]  [∵ v = ωr]

∴ K.E._roll. = [latex]\frac{1}{2}mv^2[1+\frac{k^2}{R^2}][/latex]

Translational Motion	Rotational Motion
1. S → Linear displacement.	Angular displacement ([latex]\theta[/latex]). [latex]\theta = \frac{S}{r}[/latex] ∴ S = [latex]\theta[/latex]r.
2. Linear velocity (v)	Angular velocity (ω), v = ωr.
3. Acceleration (a)	Angular acceleration (α); a = αr
4. Mass (m)	Moment of inertia (I) = mk2
5. Linear momentum (P) = mv.	Angular momentum, L = Pr = mvr = mωr2 = Iω.
6. Force (F) = [latex]\frac{dP}{dt}[/latex] = ma.	Torque (τ) = [latex]\frac{dL}{dt}[/latex] = Iα [latex]\tau[/latex] = Fr = [latex]\vec{r}\times \vec{F}[/latex]
7. Work (w) = FS	Work done by Torque = [latex]\tau \theta[/latex]
8. K.E. = [latex]\frac{1}{2}mv^2[/latex]	K.E. of rotation = [latex]\frac{1}{2}I\omega^2[/latex]
9. Equation of motion: i. v = u + at ii. S = ut + [latex]\frac{1}{2}at^2[/latex] iii. v2 = u2 + 2aS	Equation of motion: ω = ωo + αt [latex]\theta[/latex] = ωo + [latex]\frac{1}{2}at^2[/latex] ω2 = ωo + 2α[latex]\theta[/latex]

Acceleration of a rolling body on an inclined plane:

Loss of potential energy = mgh ………… (i)

∴ Gain in Kinetic energy = [latex]\frac{1}{2}mv^2 [\frac{k^2}{R^2}+1][/latex]

Now, mgh = [latex]\frac{1}{2}mv^2[\frac{k^2}{R^2}+1][/latex]

2as = v² = [latex]\frac{2gSsin\theta}{\frac{k^2}{R^2 + 1}}[/latex]                      
[∵ v² = 2as + u², u= 0 and,

Sin[latex]\theta[/latex] = h/S → h = Ssin[latex]\theta[/latex]]

∴ a = [latex]\frac{gsin\theta}{\frac{k^2}{R^2 + 1}}[/latex]

Workdone by the couple:

Suppose, a couple of forces F acting tangentially to the rim of a wheel of radius r rotates it through an angle [latex]\theta[/latex] as shown in the figure. By the definition of torque, the torque [latex]\tau[/latex] acting on the wheel is given by,

[latex]\tau = F \times AB[/latex]

Or, [latex]\tau = F \times 2r[/latex] …………….. (i)           

(∵ [latex]AB = AO + OB = r + r = 2r[/latex])

The work done by the couple of forces F is given by: 
W = F x AA’ + F x

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