## Grade 11 Physics Note

# Rotational Dynamics

**Define moment of inertia:**

The inertness in rotational motion is called moment of inertia and is denoted by I.

The SI unit of moment of inertia is kgm^{2}.

And the unit of moment of inertia is [ML^{2}T^{o}].

**Moment of couple;**

If two equal and opposite force act on an object in opposite direction then, these two forces are called couple. Its effect is to create rotation without translation, or more generally without any acceleration of the centre of mass.The turning effect of force of couple is called torque. Let’s take a rod in which a couple F and F act on it. The torque acting on the object is T = either of the force of couple ${\rm{*}}$ perpendicular distance

= F ${\rm{*}}$ d

**Angular momentum:**

Moment of linear momentum of an object is called angular momentum. It is represented by L. It is a vector quantity.

**Principle of conservation of angular momentum:**

The principle of conservation of angular momentum states that, “if no external torque acts on a body rotating about an axis, then its angular momentum about that axis is always conserved.”

Let’s take a body of mass ‘m’ rotating an axis in which ‘P’ is its linear momentum and ‘r’ is the perpendicular distance from the axis of rotation. Then, the magnitude of angular momentum is given by

$\left| {{\rm{\vec L}}} \right| = \left| {{\rm{\vec P*\vec r}}} \right|$

${\rm{or}},{\rm{\: L}} = \left| {{\rm{\vec P}}} \right|\left| {{\rm{\vec r}}} \right|{\rm{sin}}90{\rm{*}}\left| {{\rm{\hat n}}} \right|$

$ = {\rm{P*r*}}1{\rm{*}}1$

$ = {\rm{mv*r}}$

$ = {\rm{m}}\left( {{\rm{r}}\omega } \right){\rm{*r}}$

$ = {\rm{m}}{{\rm{r}}^2}{\rm{*}}\omega $

${\rm{L}} = {\rm{I}}\omega $ ------ (i)

Differentiating eqn (i) w.r.t. time, we get,

$\frac{{{\rm{dL}}}}{{{\rm{dt}}}} = \frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{I}}\omega } \right)$

$ = {\rm{I}}\frac{{{\rm{d}}\omega }}{{{\rm{dt}}}}$

$ = {\rm{I}}\alpha $ Where, $\alpha $ is the angular acceleration of the object.

But, I $ \propto $ T

So, $\frac{{{\rm{dL}}}}{{{\rm{dt}}}} = {\rm{T}}$

Since, no external torque is acting on the rotating object, we have,

T = 0

So,

$\frac{{{\rm{dL}}}}{{{\rm{dt}}}} = 0$

Or. dL = 0

Integrating on both sides, we get,

$\mathop \smallint \nolimits^ {\rm{dL}}$ = constant

Or, L = constant ------ (ii)

Or, ${\rm{I}}\omega = {\rm{constant}}$ -------- (iii)

Or, ${{\rm{I}}_1}{\omega _1} = {{\rm{I}}_2}{\omega _2}$ ------ (iv)

Eqn (ii), (iii), (iv) are the principle of conservation of angular momentum.

**Radius of gyration:**

${\rm{K}} = \sqrt {\frac{{{\rm{r}}_1^2 + {\rm{r}}_2^2 + \ldots + {\rm{r}}_{\rm{n}}^2}}{{\rm{n}}}} $

Here K is the radius of gyration.

Hence, the radius of gyration is defined as the square root of mean of sum of square of perpendicular distance of the particles from the axis of rotation and it is the effective distance of rigid object.

**Moment of inertia of a rigid body:**

The inertness in rotational motion is called moment of inertia and is denoted by I. If an object of mass ‘m’ is moving in a straight line then this mass measures the inertia of the object in linear motion but in rotational motion, mass is not used to measure inertness or inertia.

The product of mass and square of perpendicular distance from the axis of rotation gives the magnitude of moment of inertia. If ‘m’ be the mass of the object and ‘$\mu $’ be the perpendicular distance from the axis of rotation then the moment of inertia is

${\rm{I}} = {\rm{m*}}{\mu ^2}{\rm{\: }}$

The SI unit of moment of inertia is kgm^{2}.

And the unit of moment of inertia is [ML^{2}T^{o}].

**Radius of gyration:**

We know that a moment of inertia of a rigid body containing many particles is given by

${\rm{I}} = \varepsilon {\rm{m}}{{\rm{r}}^2}$

$ = {{\rm{m}}_1}{\rm{r}}_1^2 + {{\rm{m}}_2}{\rm{r}}_2^2 + \ldots + {{\rm{m}}_{\rm{n}}}{\rm{r}}_{\rm{n}}^2$

If ${{\rm{m}}_1} = {{\rm{m}}_2} = {{\rm{m}}_3} = \ldots = {{\rm{m}}_{\rm{n}}}$

Then,

${\rm{I}} = {\rm{mr}}_1^2 + {\rm{mr}}_2^2 + \ldots + {\rm{mr}}_{\rm{n}}^2$

$ = {\rm{m}}\left( {{\rm{r}}_1^2 + {\rm{r}}_2^2 + \ldots + {\rm{r}}_{\rm{n}}^2} \right)$

$ = {\rm{mn}}\left( {\frac{{{\rm{r}}_1^2 + {\rm{r}}_2^2 + \ldots + {\rm{r}}_{\rm{n}}^2}}{{\rm{n}}}} \right)$

$ = {\rm{M}}{{\rm{K}}^2}$

Where,

${{\rm{K}}^2} = \frac{{{\rm{r}}_1^2 + {\rm{r}}_2^2 + \ldots + {\rm{r}}_{\rm{n}}^2}}{{\rm{n}}}$ and ${\rm{M}} = {\rm{m*n\: }}$

${\rm{or}},{\rm{\: K}} = \sqrt {\frac{{{\rm{r}}_1^2 + {\rm{r}}_2^2 + \ldots + {\rm{r}}_{\rm{n}}^2}}{{\rm{n}}}} $

And K is the radius of gyration.

Hence, the radius of gyration is defined as the square root of mean of sum of square of perpendicular distance of the particles from the axis of rotation and it is the effective distance of rigid object.

The moment of inertia of uniform thin rod of mass M and Length L is about an axis through its centre and perpendicular to its length is given by

${\rm{I}} - \frac{{{\rm{M}}{{\rm{L}}^2}}}{{12}}$

If K is the radius of gyration of the rod about the axis, then we have

${\rm{I}} = {\rm{M}}{{\rm{K}}^2}$

Then combining both we get

${\rm{M}}{{\rm{K}}^2} = \frac{{{\rm{M}}{{\rm{L}}^2}}}{{12}}$

Therefore ${\rm{K}} = \frac{{\rm{l}}}{{\sqrt {12} }}$

**Relation for the moment of inertia of circular disc:**

The inertness in rotational motion is called moment of inertia and is denoted by I. If an object of mass ‘m’ is moving in a straight line then this mass measures the inertia of the object in linear motion but in rotational motion, mass is not used to measure inertness or inertia.

The product of mass and square of perpendicular distance from the axis of rotation gives the magnitude of moment of inertia. If ‘m’ be the mass of the object and ‘$\mu $’ be the perpendicular distance from the axis of rotation then the moment of inertia is

${\rm{I}} = {\rm{m*}}{\mu ^2}{\rm{\: }}$

Moment of inertia for thin circular disc

Consider a thin uniform disc of mass M and radius R rotating about an axis YY’ passing through its centre O and perpendicular to its plane.

Then, area of disc = ${\rm{\pi }}{{\rm{R}}_2}$

And mass = $\frac{{\rm{M}}}{{{\rm{\pi }}{{\rm{R}}^2}}}$

${\rm{\: }}$

Consider a thin element of disc of radial distance x from the centre O as shown in fig

Area of the element = 2π * dx

Mass of the element, dm = 2π *dx. $\frac{{\rm{M}}}{{{\rm{\pi }}{{\rm{R}}^2}}}$= $\frac{{2{\rm{M}}}}{{{{\rm{R}}^2}}}$*dx

Moment of inertia of element about axis YY’ = dm x^{2}

$\left( {\frac{{2{\rm{M}}}}{{{{\rm{R}}^2}}}{\rm{*dx}}} \right){\rm{\: }}{{\rm{X}}^2} = \frac{{2{\rm{M\: }}{{\rm{x}}^3}}}{{{{\rm{R}}^2}}}{\rm{dx}}$

Moment of inertia of the whole disc about axis is given by

I = $\mathop\smallint\limits_{\rm{O}}^{\rm{R}} \begin{array}{*{20}{c}}{\frac{{2{\rm{M\: }}{{\rm{x}}^3}}}{{{{\rm{R}}^2}}}dx=\frac{{2{\rm{M}}}}{{{{\rm{R}}^2}}}\mathop\smallint\limits_{\rm{O}}^{\rm{R}} {{\rm{x}}^3}{\rm{dx}}=\frac{{2{\rm{M}}}}{{{{\rm{R}}^2}}}{{\left[{\frac{{{{\rm{x}}^4}}}{4}} \right]}_{\rm{o}}}^{\rm{R}}=\frac{{2{\rm{M}}}}{{{{\rm{R}}^2}}}.\frac{{{{\rm{R}}^4}}}{4}=\frac{{{\rm{M}}{{\rm{R}}^2}}}{2}\:}\\\:\end{array}$

${\rm{I}} = \frac{{{\rm{M}}{{\rm{R}}^2}}}{2}{\rm{\: \: }}$

This is the required relation for the moment of inertia of circular disc.

**Angular momentum and obtain the relation between them:**

The turning effect of force is called torque. It is denoted by T (tou). The product of force and perpendicular distance from the axis of rotation gives the magnitude of torque.

i.e. ${\rm{T}} = \left| {{\rm{\vec F*}}\vec \mu } \right|$

Moment of linear momentum of an object is called angular momentum. It is represented by L. It is a vector quantity. If P be the linear momentum and r be the perpendicular distance from the axis of rotation then, the angular momentum is given by,

${\rm{\vec L}} = {\rm{\vec P*\vec r}}$

Relation between angular momentum and torque

The angular momentum of rigid body rotating about an axis with a velocity$\omega $ is

${\rm{L}} = {\rm{I}}\omega $………..i

Where, I is the moment of inertia of the body

Differentiating with respect to time we get,

$\frac{{{\rm{dl}}}}{{{\rm{dt}}}} = \frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{I}}\omega } \right)$

Moment of inertia I remain constant only when the axis of rotation is stationary.

In this case, we have

$\frac{{{\rm{dl}}}}{{{\rm{dt}}}} = {\rm{I}}\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{I}}\omega } \right)$

$\frac{{{\rm{dl}}}}{{{\rm{dt}}}} = {\rm{I\: }}\alpha $

Where α = $\frac{{{\rm{d}}\omega }}{{{\rm{dt}}}}$

But torque acting on body is $\Gamma = {\rm{I}}\alpha $

We get,

$\Gamma = \frac{{{\rm{dl}}}}{{{\rm{dt}}}}$

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

**Kinetic energy of a rotating body:**

The kinetic energy of a rotating object is analogous to linear kinetic energy and can be expressed in terms of the moment of inertia and angular velocity. The total kinetic energy of an extended object can be expressed as the sum of the translational kinetic energy of the center of mass and the rotational kinetic energyabout the center of mass.

Let v be the velocity of the centre of mass and T is the period of rotation of the body. During this time, it describes an angle of 2π radians about the axis. If the body covers a linear distance x in one revolution, the distance covered in one revolution is

x=2π R

The angular velocity of body is

$\omega = \frac{{2{\rm{\pi }}}}{{\rm{T}}}$

And the velocity of translation of its centre of mass is

${\rm{v}} = \frac{{2{\rm{\pi r}}}}{{\rm{T}}}$

Comparing these two equations, we get

V = $\omega {\rm{R}}$

Therefore kinetic energy of rotation, ${{\rm{E}}_{{\rm{rot}}}} = \frac{1}{2}{\rm{I}}{\omega ^2}$

And the kinetic energy translation E _{Trans} = $1/2{\rm{\: Mv}}\^2$

Total kinetic energy of a rotating body is given by

E =${\rm{\: \: \: }}{{\rm{E}}_{{\rm{rot}}}} + $ E _{Trans}

${\rm{E}} = \frac{1}{2}{\omega ^2} + \frac{1}{2}{\rm{m}}{{\rm{v}}^2}$…..This is required expression

**Couple and work done by a couple:**

If two equal and opposite force act on an object in opposite direction then, these two forces are called couple.

The turning effect of force of couple is called torque.

Let’s take a rod in which a couple F and F act on it. The torque acting on the object is

T = either of the force of couple ${\rm{*}}$ perpendicular distance

= F ${\rm{*}}$ d

Hence, the product of either of the force of couple and perpendicular distance between them gives the magnitude of torque. The SI unit of torque is Nm.

To find the work done by a torque or couple, let’s take wheel in which a couple F and F act at the ends of the diameter AB.

When the wheel rotates through angle $\theta $ then, the new diameter becomes A’, B’. Let $\mu $ be the radius of the wheel then, the work done in turning the wheel from A to A’ is

${{\rm{W}}_1} = {\rm{F*AA'}}$ -------- (i)

Similarly, work done in turning the wheel from B to B’ is

${{\rm{W}}_1} = {\rm{F*BB'}}$ -------- (ii)

The total work done is

${\rm{W}} = {{\rm{W}}_1} + {{\rm{W}}_2}$

$ = {\rm{F*A}}{{\rm{A}}^{\rm{'}}} + {\rm{F*BB'}}$

But, from the figure we have,

AA’ = BB’ = $\mu {\rm{*}}\theta $

Putting the value of AA’ and BB’ in eqn (iii) we get,

${\rm{W}} = {\rm{F*}}\mu \theta + {\rm{F*}}\mu \theta $

= 2F$\mu \theta $

$ = \left( {{\rm{F*}}2\mu } \right){\rm{*}}\theta $

${\rm{or}},{\rm{\: W}} = {\rm{T*}}\theta $ -------- (iv)

Where, T = F ${\rm{*}}$ 2$\mu $ and it is the torque action on the wheel.

Eqn (iv) give the expression work done produced by torque on couple.

The power produced by the couple or torque is given by

${\rm{P}} = \frac{{{\rm{Work}}}}{{{\rm{Time}}}} = \frac{{\rm{W}}}{{\rm{t}}} = {\rm{T*}}\frac{\theta }{{\rm{t}}} = {\rm{T*}}\omega $ ------------ (v); Where, $\omega $ is the angular velocity.

**Expression for total energy of an object rolling a plane:**

Let v be the velocity of the centre of mass and T is the period of rotation of the body. During this time, it describes an angle of 2π radians about the axis. If the body covers a linear distance x in one revolution, the distance covered in one revolution is

X =2π R

The angular velocity of body is

$\omega = \frac{{2{\rm{\pi }}}}{{\rm{T}}}$

And the velocity of translation of its centre of mass is

${\rm{v}} = \frac{{2{\rm{\pi r}}}}{{\rm{T}}}$

Comparing these two equations, we get

V = $\omega {\rm{R}}$

Therefore kinetic energy of rotation, ${{\rm{E}}_{{\rm{rot}}}} = \frac{1}{2}{\rm{I}}{\omega ^2}$

And the kinetic energy translation E _{Trans} = $1/2{\rm{\: Mv}}\^2$

Total kinetic energy of a rotating body is given by

E =${\rm{\: \: \: }}{{\rm{E}}_{{\rm{rot}}}} + $ E _{Trans}

${\rm{E}} = \frac{1}{2}{\omega ^2} + \frac{1}{2}{\rm{m}}{{\rm{v}}^2}$…..This is required expression

**Theorem of parallel and perpendicular axes:**

Theorem of parallel axis:

The moment of inertia of a body about any axis is equal to the sum of its moment of inertia about the parallel axis passing through the centre of mass and the product of the mass of the body and the square of the distance between axes.

Proof;

Consider a particle of mass, m at a point Q in the body at distance x from the axis CD. The moment of inertia of a particle about

AB = ${\rm{m}}{\left( {{\rm{x}} = {\rm{r}}} \right)^2}$

Then the moment of inertia of whole body about the axis AB is given by,

I = $\mathop \sum \nolimits^ {\rm{m}}{\left( {{\rm{x}} = {\rm{r\: }}} \right)^2}{\rm{\: }}$

$ = \mathop \sum \nolimits^ {\rm{m}}{{\rm{x}}^2} + {\rm{\: }}\mathop \sum \nolimits^ {\rm{m}}{{\rm{r}}^2} + 2\mathop \sum \nolimits^ {\rm{mxr}}$

Since, ${{\rm{I}}_{{\rm{cm}}}} = {\rm{\: }}\mathop \sum \nolimits^ {\rm{m}}{{\rm{r}}^2} + 2\mathop \sum \nolimits^ {\rm{mxr}}$. . 1

Then it can be written as

${\rm{I}} = {{\rm{I}}_{{\rm{cm}}}} + {\rm{M}}{{\rm{r}}^2}$

This proves the Theorem of parallel axis.

**Theorem of perpendicular axes:**

The moment of inertia of a plane lamina (a plane body) about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two mutually perpendicular axes in the plane of the lamina which intersect the axis.

Proof;

I_{z} = $\mathop \sum \nolimits^ {\rm{m}}{{\rm{r}}^2}$

Then, the moment of inertia of body about the axis OX is given by,

I_{x} = $\mathop \sum \nolimits^ {\rm{m}}{{\rm{y}}^2}$

And the moment of inertia of body about the axis OY is given by,

I_{y} = $\mathop \sum \nolimits^ {\rm{m}}{{\rm{x}}^2}$

Now,

${{\rm{I}}_{\rm{x}}} + {{\rm{I}}_{\rm{y}}} = \mathop \sum \nolimits^ {\rm{m\: }}\left( {{{\rm{x}}^2} + {{\rm{y}}^2}} \right)$

Or, ${{\rm{I}}_{\rm{x}}} + {{\rm{I}}_{\rm{y}}} = \mathop \sum \nolimits^ {\rm{m}}{{\rm{r}}^2}$

Where ${{\rm{r}}^2} = {{\rm{x}}^2} + {{\rm{y}}^2}$

Therefore we get,

${{\rm{I}}_{\rm{x}}} = {{\rm{I}}_{\rm{y}}} + {{\rm{I}}_{\rm{z}}}$

This proves the Theorem of perpendicular axes.