NotesShala Class 11 · Physics

Chapter06

System of Particles and Rotational Motion

Rigid Body and Centre of Mass · Torque and Angular Momentum · Equilibrium and Centre of Gravity · Moment of Inertia and Rotational Dynamics

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Rigid Body and Centre of Mass

Till now, we treated every object as a point particle. A real body has a size. It is made of a very large number of particles. So it is called an extended body. In this chapter, we study the motion of a system of particles and of a rigid body. The main ideas are centre of mass, torque, angular momentum, moment of inertia and rolling motion. Rolling is a mix of translation and rotation.

1.1Rigid Body and Types of Motion

Definition

A rigid body is a body in which the distance between every pair of particles stays fixed, however large the force on it.

No real body is perfectly rigid. But real bodies change their shape only a little. So most real bodies can be taken as rigid bodies.

There are two basic kinds of motion of a rigid body: pure translation and pure rotation. A body can also do both together.

Pure Translational Motion

During pure translation, every particle of the body moves with the same velocity at any instant. So in a given time, all the particles move through the same displacement.

A block sliding (not rolling) down an inclined planeTranslation

A block sliding down an incline:

P_1

and

P_2

have the same velocity

\vec{v}

Rotation about a Fixed Axis

Definition

Axis of Rotation

In pure rotation, every particle of the body moves in a circle. All these circles have their centres on one fixed straight line. This line is called the axis of rotation.

Look at a rotating ceiling fan. A particle $A$ on a blade moves in a circle of radius $r_1$ . A particle $B$ moves in a bigger circle of radius $r_2$ .

The centres of both circles lie on the same fixed axis. This axis is the vertical line through the rod of the fan.

A rotating ceiling fan:

A

and

B

move in circles about the same axis

Take a particle $P_1$ at a distance $r_1$ from the axis. It moves in a circle of radius $r_1$ . The centre $C_1$ of this circle lies on the axis.

This circle lies in a plane perpendicular to the axis. In the same way, a particle $P_2$ moves in a circle of radius $r_2$ with centre $C_2$ .

A rigid body rotating about the

z

-axis

iDifferent Planes

The circles of $P_1$ and $P_2$ need not be in the same plane. Still, each of these planes is at right angles to the fixed axis.

iiOn the Axis

A particle on the axis, like $P_3$ , has $r = 0$ . So it stays at rest while the body rotates.

Note

Each particle moves in a circle whose plane is perpendicular to the fixed axis. The centre of this circle is on the axis.
A particle that lies on the axis of rotation does not move.

e.g. Potter's wheel, giant wheel, merry-go-round, ceiling fanRotation about a fixed axis

Precession

Definition

Precession is the motion of an axis of rotation that turns around another line. One point of the axis stays fixed. So the axis sweeps out a cone.

A spinning top shows precession. Its pointed tip touches the ground and stays fixed. At every instant, the axis of rotation passes through this point.
An oscillating table fan also shows precession.

A spinning top in precession

Remember

Convention

In this chapter, we mostly study rotation about a fixed axis. So take every rotation to be about a fixed axis, unless the question says something else.

1.2Combination of Translation and Rotation

A rigid body can also roll. Rolling is a combination of rotation and translation. Take a cylinder rolling down an inclined plane. It turns about its own axis. At the same time, this axis moves down the slope.

As the cylinder rolls down, its points $P_1$ , $P_2$ , $P_3$ and $P_4$ have different velocities at the same instant.

If the cylinder rolls without slipping, the point of contact $P_3$ has zero velocity at every instant.

A cylinder rolling down an incline. The contact point

P_3

is at rest

So in rolling, the body turns about an axis of rotation, and this axis itself moves in translation. So any motion of a rigid body is of one of two kinds. It is pure translation, or it is translation and rotation together.

A baseball bat thrown into the air is a good example. Its motion is not pure translation. It is not pure rotation either. It is a combination of both.

The bat keeps turning. But one special point of the bat, its centre of mass, follows a smooth parabola.

Its centre of mass (dot) follows a parabola

Note

A rigid body that is not pivoted or fixed in any wayPure translation, or translation + rotation

A rigid body that is pivoted or fixed in some wayPure rotation

1.3Centre of Mass of Two Particles

Definition

The centre of mass (CM) of a system is the point where the whole mass of the system can be taken to be concentrated.

The system moves as if all its mass were placed at this point. All the external forces also seem to act at this point.

Two particles of masses $m_1$ and $m_2$ lie on the $x$ -axis. Their distances from the origin $O$ are $x_1$ and $x_2$ .

Their centre of mass is the point $C$ . Its distance from $O$ is $x_{CM}$ .

Centre of mass

C

of two particles on the

x

-axis

Centre of Mass (Two Particles)

x_{CM} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \dfrac{m_1 x_1 + m_2 x_2}{M}

where, $M = m_1 + m_2$ = total mass of the system

iPosition Vector Form

Multiply the position vector of the CM by the total mass. The result equals the sum of each mass times its own position vector.

M\,\vec{r}_{CM} = m_1\vec{r}_1 + m_2\vec{r}_2

iiEqual Masses

If $m_1 = m_2 = m$ , then:

x_{CM} = \dfrac{x_1 + x_2}{2}

So the CM is at the mid-point of the line joining them.

x_{CM}

is the mass-weighted average of

x_1

and

x_2

. The CM lies on the line joining the two particlesOn the line

1.4Centre of Mass of n Particles

Take $n$ particles of masses $m_1, m_2, \ldots, m_n$ on the $x$ -axis. Their positions are $x_1, x_2, \ldots, x_n$ .

The $x$ -coordinate of their CM is the mass-weighted average of all these positions.

A system of particles on the

x

-axis

Centre of Mass (n Particles)

x_{CM} = \dfrac{m_1 x_1 + m_2 x_2 + \cdots + m_n x_n}{m_1 + m_2 + \cdots + m_n} = \dfrac{\sum m_i x_i}{\sum m_i}

Centre of Mass of Three Particles

Three particles of masses $m_1$ , $m_2$ and $m_3$ lie at the points $(x_1, y_1)$ , $(x_2, y_2)$ and $(x_3, y_3)$ . Their CM is at the point $(X, Y)$ :

iAny Masses

X = \dfrac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}

Y = \dfrac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}

iiEqual Masses (

m_1 = m_2 = m_3 = m

)

X = \dfrac{x_1 + x_2 + x_3}{3}

Y = \dfrac{y_1 + y_2 + y_3}{3}

In Three Dimensions

When the particles are spread in three-dimensional space, the CM has three coordinates:

x-coordinate

x_{CM} = \dfrac{1}{M}\sum m_i x_i

y-coordinate

y_{CM} = \dfrac{1}{M}\sum m_i y_i

z-coordinate

z_{CM} = \dfrac{1}{M}\sum m_i z_i

Position Vector of Centre of Mass

\vec{r}_{CM} = \dfrac{1}{M}\sum_{i=1}^{n} m_i \vec{r}_i \qquad \left[M = \sum_{i=1}^{n} m_i\right]

If all the $n$ masses are equal, the CM is just the average of their positions. For example, $X = \dfrac{x_1+x_2+x_3}{3}$ .

1.5Centre of Mass of a Continuous Body

In a real body, matter is spread out continuously. It is not a set of separate particles. So we divide it into tiny pieces of mass. Then the sum over the particles becomes an integral.

Position Vectors and the CM

iVector Form

Let $\vec{r}_i = x_i\hat{i} + y_i\hat{j} + z_i\hat{k}$ be the position vector of the $i^{th}$ particle. Let $\vec{R} = X\hat{i} + Y\hat{j} + Z\hat{k}$ be the position vector of the CM. Then:

\vec{R} = \dfrac{\sum m_i \vec{r}_i}{M}

iiOrigin at the CM

Choose the origin at the centre of mass. Then $\vec{R}_{CM} = 0$ . So:

\displaystyle\sum_{i=1}^{n} m_i \vec{r}_i = 0

Concept

CM of a Continuous Body

In a continuous body, each point mass becomes a tiny mass element $dm$ . Each sum becomes an integral. So the coordinates of the CM are:

Centre of Mass (Integral Form)

x_{CM} = \dfrac{1}{M}\int x\,dm,\quad y_{CM} = \dfrac{1}{M}\int y\,dm,\quad z_{CM} = \dfrac{1}{M}\int z\,dm

where, $M$ = total mass of the body. If the origin is at the CM, then $\displaystyle\int x\,dm = \int y\,dm = \int z\,dm = 0$ .

CM When No External Force Acts

Suppose no external force acts on a system. If its CM is at rest at first, the CM stays at the same place.

If the particles shift by $\Delta x_1$ , $\Delta x_2$ and so on, the CM shifts by:

\Delta x_{CM} = \dfrac{\sum m_i\,\Delta x_i}{\sum m_i}

1.6Centre of Mass of a Uniform Thin Rod

A thin uniform rod $AB$ has mass $M$ and length $L$ . It lies along the $x$ -axis, with end $A$ at the origin.

Take a small element of length $dx$ at a distance $x$ from $A$ .

A small element

dx

at distance

x

from

A

Derivation

As the rod is uniform, its mass per unit length is the same everywhere. This is its linear mass density, $\lambda = \dfrac{M}{L}$ . So the mass of the small element is:
$dm = \dfrac{M}{L}\,dx$
The $x$ -coordinate of the CM is:
$x_{CM} = \dfrac{1}{M}\int_0^{L} x\,dm = \dfrac{1}{M}\int_0^{L} x\left(\dfrac{M}{L}\right)dx = \dfrac{1}{L}\left[\dfrac{x^2}{2}\right]_0^{L} = \dfrac{L}{2}$
The rod lies on the $x$ -axis. So $y = 0$ for every element:
$y_{CM} = \dfrac{1}{M}\int y\,dm = 0 \qquad [\because y = 0]$

Centre of Mass of a Uniform Rod

x_{CM} = \dfrac{L}{2}, \quad y_{CM} = 0

So the CM is at $\left(\dfrac{L}{2},\,0\right)$ , exactly midway between $A$ and $B$ . For a uniform body, the CM lies at its geometric centre.

1.7Motion of Centre of Mass

How the CM moves depends on how each particle moves. Start from the equation that defines the CM. Then differentiate it with respect to time. This gives the velocity and the acceleration of the CM.

M\,\vec{r}_{CM} = m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3 + \cdots + m_n\vec{r}_n

Derivation

Differentiate once with respect to time. The masses are constant. Only the position vectors change:
$M\,\dfrac{d}{dt}\vec{r}_{CM} = m_1\dfrac{d\vec{r}_1}{dt} + m_2\dfrac{d\vec{r}_2}{dt} + \cdots + m_n\dfrac{d\vec{r}_n}{dt}$
Rate of change of position = velocity, that is $\vec{v} = \dfrac{d\vec{r}}{dt}$ . So:
$M\,\vec{v}_{CM} = m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3 + \cdots + m_n\vec{v}_n$
Differentiate again. Rate of change of velocity = acceleration, that is $\vec{a} = \dfrac{d\vec{v}}{dt}$ . So:
$M\,\vec{a}_{CM} = m_1\vec{a}_1 + m_2\vec{a}_2 + \cdots + m_n\vec{a}_n$

Velocity of Centre of Mass

\vec{v}_{CM} = \dfrac{\sum_{i=1}^{n} m_i \vec{v}_i}{M}

Acceleration of Centre of Mass

\vec{a}_{CM} = \dfrac{\sum_{i=1}^{n} m_i \vec{a}_i}{M}

Net External Force on a System

M\vec{a}_{CM} = \vec{F}_{ext}

The CM moves as if the whole mass of the system were placed at it, and all the external forces acted on it there.

Internal Forces Do Not Affect the CM

Every internal force has an equal and opposite partner. These are action–reaction pairs. So the internal forces add up to zero, and only the external forces remain in the sum.

Suppose a rocket explodes in mid-air. The forces of the explosion are internal. So they add nothing to the acceleration of the CM.

The CM of the pieces keeps moving on the same parabola. This is the path of the rocket without the explosion.

After the explosion, the CM of the pieces stays on the same parabola

Note

The translational motion of an extended body is easy to study through its CM. Imagine that the total mass sits at the CM, and that every external force acts on that point.

1.8Linear Momentum of a System of Particles

Derivation

To get the total linear momentum of a system, add the momenta of all its particles as vectors. One particle has momentum $m\vec{v}$ .

Add the momenta of all $n$ particles:
$\vec{p} = \vec{p}_1 + \vec{p}_2 + \cdots + \vec{p}_n = m_1\vec{v}_1 + m_2\vec{v}_2 + \cdots + m_n\vec{v}_n$
From the motion of the CM, $\displaystyle\sum m_i\vec{v}_i = M\vec{v}_{CM}$ . So:
$\vec{p} = M\vec{v}_{CM}$

Total Linear Momentum of a System

\vec{p} = M\,\vec{v}_{CM}

Total momentum of a system = total mass $M$ × velocity of the CM. It is the same as the vector sum $m_1\vec{v}_1 + m_2\vec{v}_2 + \cdots + m_n\vec{v}_n$ .

Newton's Second Law for a System

Derivation

Differentiate $\vec{p} = M\vec{v}_{CM}$ with respect to time. The mass $M$ is constant:
$\dfrac{d\vec{p}}{dt} = M\dfrac{d}{dt}\vec{v}_{CM} = M\,\vec{a}_{CM}$
From the motion of the CM, $M\,\vec{a}_{CM} = \vec{F}_{ext}$ , the net external force. So:
$\vec{F}_{ext} = \dfrac{d\vec{p}}{dt}$

Second Law for a System

\vec{F}_{ext} = \dfrac{d\vec{p}}{dt}

The total momentum $\vec{p}$ of a system changes at a rate equal to the net external force $\vec{F}_{ext}$ on it.

Conservation of Linear Momentum

Law

The law of conservation of linear momentum states that the total linear momentum of a system stays constant when no external force acts on it.

If $\vec{F}_{ext} = 0$ , then $\dfrac{d\vec{p}}{dt} = 0$ . So the total momentum does not change with time.

Momentum Conserved

\vec{p}_{initial} = \vec{p}_{final}

CM Velocity Constant

$\vec{F}_{ext} = 0$ and $M$ is constant. So $M\vec{v}_{CM}$ is constant. The CM moves with constant velocity: $\vec{v}_{CM}$ = constant.

Three Scalar Equations

p_x = C_1,\ p_y = C_2,\ p_z = C_3

$p_x$ , $p_y$ , $p_z$ are the components of $\vec{p}$ along the $x$ , $y$ , $z$ axes. $C_1$ , $C_2$ , $C_3$ are constants.

Part 2

Torque and Angular Momentum

2.1Torque (Moment of Force)

Definition

Torque is the turning effect of a force about a point or an axis. It is also called the moment of force.

A force can push a body along its own direction. It can also turn the body about a point. Torque measures this turning effect.

The torque about a point $O$ is the vector product of two vectors. The first is the position vector $\vec{r}$ of the point where the force acts, measured from $O$ . The second is the force $\vec{F}$ .

Torque (Vector Form)

\vec{\tau} = \vec{r} \times \vec{F}

where, $\vec{r}$ = position vector, from $O$ , of the point where $\vec{F}$ acts

Force

\vec{F}

acts at

P

. The moment arm

b

is the perpendicular distance

OA

Magnitude of Torque

|\vec{\tau}| = rF\sin\theta = F\,r_\perp = r\,F_\perp

where, $\theta$ = angle between $\vec{r}$ and $\vec{F}$

iMoment Arm

r_\perp

$r_\perp = b = r\sin\theta$ . It is the length of the perpendicular drawn from $O$ to the line of action of $\vec{F}$ . It is called the moment arm.

iiPerpendicular Force

F_\perp

$F_\perp = F\sin\theta$ is the component of $\vec{F}$ perpendicular to $\vec{r}$ . Only this part of the force turns the body.

When Is Torque Zero?

r = 0

: the force acts at

O

itself

\tau = 0

F = 0

: no force acts

\tau = 0

\theta = 0^\circ

180^\circ

\vec{F}

is parallel to

\vec{r}

\tau = 0

Direction of Torque

Torque is a vector product. So $\vec{\tau}$ is perpendicular to the plane that contains $\vec{r}$ and $\vec{F}$ . The right-hand rule gives its direction.

Body turns anticlockwise

\tau

positive

Body turns clockwise

\tau

negative

\vec{\tau}

is perpendicular to the plane of

\vec{r}

and

\vec{F}

Remember

Unit and Dimensions

The SI unit of torque is the newton metre (N m). Its dimensions are $[\mathrm{ML^2T^{-2}}]$ . Work also has these same dimensions. Still, torque and work are not the same quantity.

2.2Angular Momentum of a Particle

Definition

The angular momentum $\vec{L}$ of a particle about a point is the moment of its linear momentum about that point.

A particle of mass $m$ moves with velocity $\vec{v}$ . Its linear momentum is $\vec{p} = m\vec{v}$ . Its angular momentum about $O$ is the vector product of its position vector and its linear momentum.

Angular Momentum (Vector Form)

\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})

where, $\vec{r}$ = position vector of the particle about $O$

\vec{L}

is perpendicular to the plane of

\vec{r}

and

\vec{v}

Magnitude of Angular Momentum

L = mvr\sin\theta = rp\sin\theta = r\,p_\perp = p\,r_\perp

where, $\theta$ = angle between $\vec{r}$ and $\vec{v}$ (or $\vec{p}$ )

iPerpendicular Momentum

p_\perp

$p_\perp = p\sin\theta$ is the component of $\vec{p}$ perpendicular to $\vec{r}$ .

iiPerpendicular Distance

r_\perp

$r_\perp$ is the length of the perpendicular from the origin $O$ to the line along which $\vec{p}$ acts.

iiiDirection

$\vec{L}$ is a vector product. So it is perpendicular to the plane of $\vec{r}$ and $\vec{p}$ . The right-hand rule gives its direction, as for torque.

ivSI Unit

The SI unit of angular momentum is J s. This is the same as $\text{kg m}^2\,\text{s}^{-1}$ .

2.3Relation between Torque and Angular Momentum

Force equals how fast linear momentum changes. In rotation, torque equals how fast angular momentum changes. This result is the angular form of Newton's second law.

For a Single Particle

Derivation

Start with $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$ . Differentiate it with respect to time, using the product rule:
$\dfrac{d\vec{L}}{dt} = m\left(\dfrac{d\vec{r}}{dt} \times \vec{v} + \vec{r} \times \dfrac{d\vec{v}}{dt}\right)$
Here $\dfrac{d\vec{r}}{dt} = \vec{v}$ , and $\vec{v} \times \vec{v} = 0$ . So the first term is zero. Also, $\dfrac{d\vec{v}}{dt} = \vec{a}$ :
$\dfrac{d\vec{L}}{dt} = m(\vec{r} \times \vec{a}) = \vec{r} \times (m\vec{a}) = \vec{r} \times \vec{F}$
But $\vec{r} \times \vec{F}$ is the torque $\vec{\tau}$ . So $\dfrac{d\vec{L}}{dt}$ equals the torque on the particle.

Torque–Angular Momentum (Particle)

\vec{\tau} = \dfrac{d\vec{L}}{dt}

The angular momentum of a particle changes at a rate equal to the torque on it.

2.4Torque and Angular Momentum of a System

Add the angular momenta of all the particles as vectors. This gives the total $\vec{L}$ :

\vec{L} = \vec{L}_1 + \vec{L}_2 + \cdots + \vec{L}_n = \sum_{i=1}^{n}\vec{L}_i

Derivation

Differentiate with respect to time. For each particle, $\dfrac{d\vec{L}_i}{dt} = \vec{\tau}_i$ . So the total gives the net torque:
$\dfrac{d\vec{L}}{dt} = \displaystyle\sum_{i=1}^{n}\dfrac{d\vec{L}_i}{dt} = \vec{\tau}_{net}$
The force on each particle has two parts, external and internal. So the total torque also has two parts:
$\vec{\tau} = \sum \vec{r}_i \times \vec{F}_i = \vec{\tau}_{ext} + \vec{\tau}_{int}$
Newton's third law makes the internal torques cancel pair by pair. Hence $\vec{\tau}_{int} = 0$ , and only the external torque is left.

Torque–Angular Momentum (System)

\vec{\tau}_{ext} = \dfrac{d\vec{L}}{dt}

Here $\vec{L}$ = total angular momentum of the system about a point, and $\vec{\tau}_{ext}$ = total external torque about that same point.

Note

Internal forces come in pairs of equal and opposite forces. The torques of each pair cancel. So the total angular momentum of a system can change only through external torques.

2.5Conservation of Angular Momentum

Law

The law of conservation of angular momentum states that the total angular momentum of a system stays constant when the net external torque on it is zero.

We know that $\vec{\tau}_{ext} = \dfrac{d\vec{L}}{dt}$ . If $\vec{\tau}_{ext} = 0$ , then $\dfrac{d\vec{L}}{dt} = 0$ . So $\vec{L}$ does not change with time.

Conservation of Angular Momentum

\vec{\tau}_{ext} = 0 \;\Rightarrow\; \vec{L} = \text{constant} \;\Rightarrow\; \vec{L}_{initial} = \vec{L}_{final}

For rotation about a fixed axis, $L = I\omega$ , where $I$ is the moment of inertia (see 4.1). So the working form is $I_1\omega_1 = I_2\omega_2$ .

iAll Components Conserved

If $\vec{L}$ is constant, each of its components is also constant:

L_x = \text{const},\ L_y = \text{const},\ L_z = \text{const}

iiSpin and Moment of Inertia

$L = I\omega$ stays constant. So when $I$ decreases, $\omega$ increases. When $I$ increases, $\omega$ decreases.

Applications of the Law

Ice Skater / Ballet Dancer

A skater, a dancer or an acrobat in a circus spins faster by pulling the arms and legs close to the body. This reduces $I$ , so $\omega$ increases. Stretching the arms out does the opposite.

Diver's Somersault

A diver leaves the board with the body stretched out. In the air, the diver curls up to reduce $I$ . Since $L$ is conserved, $\omega$ increases for the somersaults. Near the water, the diver stretches out again.

Helicopter's Two Propellers

A helicopter has two propellers. With only one, its body would turn the opposite way to conserve angular momentum. The second propeller stops this.

Part 3

Equilibrium and Centre of Gravity

3.1Equilibrium of a Rigid Body

Definition

Mechanical Equilibrium

A rigid body is in mechanical equilibrium when its linear momentum and angular momentum both do not change with time. Such a body has no linear acceleration and no angular acceleration.

For this, two conditions must hold at the same time:

(i) Translational Equilibrium

\sum_{i=1}^{n}\vec{F}_i = 0

Net external force = 0. So $\sum F_{ix}=0,\ \sum F_{iy}=0,\ \sum F_{iz}=0$ . The linear momentum stays constant.

(ii) Rotational Equilibrium

\sum_{i=1}^{n}\vec{\tau}_i = 0

Net external torque = 0. So $\sum \tau_{ix}=0,\ \sum \tau_{iy}=0,\ \sum \tau_{iz}=0$ . The angular momentum stays constant.

Note

When the net force is zero, the condition of rotational equilibrium does not depend on the choice of origin.
A body can also be in partial equilibrium. Then only one of the two conditions holds.

3.2Partial Equilibrium

Definition

Partial equilibrium is the state of a body in which only one of the two conditions of equilibrium holds.

Rotational Equilibrium Only

Here $\sum \vec{\tau} = 0$ , but $\sum \vec{F} \ne 0$ . So the body has no angular acceleration, but it does have a linear acceleration.

e.g. Two equal forces $F$ act in the same direction at the ends $P$ and $Q$ of a rod. Both ends are at distance $l$ from the centre $O$ .

Net torque about

O

is zero, but the net force is not zero

Translational Equilibrium Only

Here $\sum \vec{F} = 0$ , but $\sum \vec{\tau} \ne 0$ . So the body has no linear acceleration, but it does have an angular acceleration.

e.g. Two equal and opposite forces act at the ends of the rod. They form a couple.

Net force is zero, but the couple gives a net torque

3.3Couple

Definition

A couple is a pair of equal and opposite forces whose lines of action are parallel but not the same. A couple produces rotation without translation.

Force $-\vec{F}$ acts at point $A$ , whose position vector is $\vec{r}_1$ . Force $+\vec{F}$ acts at point $B$ , whose position vector is $\vec{r}_2$ .

The total torque of the two forces is called the moment of the couple.

A couple:

-\vec{F}

A

and

+\vec{F}

B

Derivation

Add the torques of the two forces about $O$ :
$\vec{\tau} = \vec{r}_1\times(-\vec{F}) + \vec{r}_2\times\vec{F} = (\vec{r}_2 - \vec{r}_1)\times\vec{F}$
By the triangle law, $\vec{r}_1 + \vec{AB} = \vec{r}_2$ . So $\vec{AB} = \vec{r}_2 - \vec{r}_1$ . The moment depends only on $\vec{AB}$ . So it does not depend on the origin.

Moment of a Couple

\vec{\tau} = \vec{AB} \times \vec{F} \qquad (\text{magnitude } \tau = F\,d)

where, $d$ = perpendicular gap between the lines of the two forces. A couple has the same moment about every point.

3.4Principle of Moments

Principle

The principle of moments states that the algebraic sum of the torques on a body in rotational equilibrium is zero.

Take anticlockwise torques as positive and clockwise torques as negative.

A light rod resting on a pivot is a lever. The pivot is called the fulcrum. A load $F_1$ acts at distance $d_1$ from the fulcrum. An effort $F_2$ acts at distance $d_2$ on the other side.

The lever is in equilibrium. So both conditions hold:

Translational equilibrium

R = F_1 + F_2

Rotational equilibrium (about the fulcrum)

F_1 d_1 = F_2 d_2

A lever: load

F_1

, effort

F_2

and reaction

R

at the fulcrum

Principle of Moments for a Lever

F_1 d_1 = F_2 d_2 \qquad (\text{load} \times \text{load arm} = \text{effort} \times \text{effort arm})

where, $F_1$ = load, $d_1$ = load arm, $F_2$ = effort, $d_2$ = effort arm

Mechanical Advantage

\text{MA} = \dfrac{F_1}{F_2} = \dfrac{d_2}{d_1}

MA > 1

We usually want a mechanical advantage greater than one. Then a large load can be lifted with a small effort.

3.5Centre of Gravity

Definition

The centre of gravity (CG) of a body is the point about which the total gravitational torque on the body is zero. A body supported at its CG stays balanced.

Think of the body as many particles. A particle of mass $m_i$ is at position vector $\vec{r}_i$ from the CG. Its weight is $m_i\vec{g}$ .

The body balances on a sharp support placed at its CG. So the total gravitational torque about this point is zero.

A body balanced on a point support at its CG

Derivation

Total gravitational torque about the CG:
$\vec{\tau}_g = \sum \vec{r}_i \times m_i\vec{g} = \left(\sum m_i\vec{r}_i\right)\times\vec{g} = 0$
Suppose $\vec{g}$ is the same at every point of the body. Then $\sum m_i\vec{r}_i = 0$ about the CG. This is exactly the condition for the centre of mass. So the CG and the CM are the same point.

CG and CM

\sum m_i\vec{r}_i = 0 \;\Rightarrow\; \text{CG coincides with CM}

This holds when $g$ is uniform over the body.

iSmall Bodies

$g$ has the same value at every point of a small body. So its CG and CM are at the same point.

iiVery Large Bodies

$g$ is different at different parts of a very large body. So its CG and CM are at different points.

Plumb-Line Method

CG of an Irregular Body

Hang the body freely from a point $P$ . Mark the vertical through $P$ with a plumb line.
Hang it from another point $Q$ . Mark the new vertical.
The two verticals meet at the CG.

The two verticals meet at the CG,

G

Part 4

Moment of Inertia and Rotational Dynamics

4.1Moment of Inertia

Definition

The moment of inertia ( $I$ ) of a body about an axis is its property to resist any change in its rotational state. This state may be rest or uniform rotation about that axis.

It is also called rotational inertia. In rotation, it plays the same role as mass plays in translation.

A rotating body has kinetic energy. Add the kinetic energies of all its particles to get it.

All the particles turn through the same angle in the same time. So they have the same angular velocity $\omega$ . A particle at distance $r$ from the axis moves with speed $v = r\omega$ .

Each particle moves on a circle about the axis

O

with speed

v = r\omega

Derivation

Put $v_i = r_i\omega$ in the total kinetic energy of the particles:
$K = \sum \tfrac{1}{2}m_i v_i^2 = \tfrac{1}{2}\left(\sum m_i r_i^2\right)\omega^2$
The quantity $\sum m_i r_i^2$ depends on how the mass is spread about the axis. This quantity is the moment of inertia $I$ . So $K = \tfrac{1}{2}I\omega^2$ .
Compare this with the translational KE, $\tfrac{1}{2}mv^2$ . So in rotation, $I$ does the job that mass does in translation.

Moment of Inertia

I = \sum_{i=1}^{n} m_i r_i^2

Rotational Kinetic Energy

K = \dfrac{1}{2}I\omega^2

Definition

For a rigid body turning about a fixed axis, the moment of inertia is the sum of $mr^2$ over all its particles. Here $m$ is the mass of a particle and $r$ is its distance from the axis.

Moment of inertia for different kinds of mass distribution:

Single Particle

I = m r^2

System of Particles

I = \sum m_i r_i^2

Continuous Body

I = \displaystyle\int r^2\,dm

Note

Rotational KE and Moment of Inertia

Rotational KE is $K = \tfrac{1}{2}I\omega^2$ . Put $\omega = 1$ . Then $K = \tfrac{1}{2}I$ , which means $I = 2\times(\text{rotational KE})$ .

So, for a body spinning with unit angular velocity, $I$ has the same numerical value as twice its rotational KE.

Applications

Flywheel

An engine shaft carries a heavy flywheel. Its moment of inertia is large. So it resists any sudden change in speed. The speed changes only gradually, and the ride is smooth, without jerks.

Bicycle / Cart Wheel

Most of the mass of the wheel is at its rim. Spokes join the rim to the axle. This increases $I$ . So the wheel keeps turning for some time, even when the rider stops pedalling.

4.2Radius of Gyration

Definition

The radius of gyration ( $k$ ) of a body is that distance from the axis where its whole mass, if placed at one point, gives the same moment of inertia as the real body.

Moment of Inertia and Radius of Gyration

I = Mk^2 \qquad\Rightarrow\qquad k = \sqrt{\dfrac{I}{M}}

where, $M$ = total mass of the body

The whole mass

M

placed at distance

k

from the axis

Radius of Gyration as RMS Distance

A body has $n$ particles, each of mass $m$ . Their distances from the axis are $r_1$ , $r_2$ , … , $r_n$ . The total mass is $M = mn$ .

Note

The value of $k$ changes if the axis moves. It also depends on how the mass is spread around the axis. The total mass alone does not decide it.

Particles of a body at distances

r_1

r_2

, … ,

r_n

from the axis

Derivation

The moment of inertia of the particles is:
$I = m(r_1^2 + r_2^2 + \cdots + r_n^2)$
By definition, $I = Mk^2$ . Equate the two and use $M = mn$ :
$Mk^2 = m(r_1^2 + \cdots + r_n^2) \;\Rightarrow\; k^2 = \dfrac{r_1^2 + \cdots + r_n^2}{n}$

Radius of Gyration (RMS Distance)

k = \sqrt{\dfrac{r_1^2 + r_2^2 + \cdots + r_n^2}{n}}

So $k$ is the root-mean-square (RMS) value of the distances of the particles from the axis.

4.3Moment of Inertia of Standard Bodies

The cards give the moment of inertia $I$ , the radius of gyration $k$ and the ratio $k^2/R^2$ for some common bodies. Each value is about the axis shown by the dashed line.

Thin Circular Ring

I

MR^2

k

R

k^2/R^2

1

Thin Rod

I

\tfrac{1}{12}ML^2

k

\tfrac{L}{\sqrt{12}}

k^2/R^2

—

Circular Disc

I

\tfrac{1}{2}MR^2

k

\tfrac{R}{\sqrt{2}}

k^2/R^2

\tfrac{1}{2}

Hollow Cylinder

I

MR^2

k

R

k^2/R^2

1

Solid Cylinder

I

\tfrac{MR^2}{2}

k

\tfrac{R}{\sqrt{2}}

k^2/R^2

\tfrac{1}{2}

Solid Sphere

I

\tfrac{2}{5}MR^2

k

\sqrt{\tfrac{2}{5}}\,R

k^2/R^2

\tfrac{2}{5}

Note

The moment of inertia of a body depends on the axis chosen. All the values above are for the axes shown.
For the ring, the disc, the two cylinders and the sphere, the axis passes through the centre. For the rod, it passes through the centre, perpendicular to its length.
The ring and the hollow cylinder have the largest ratio, $k^2/R^2 = 1$ . All their mass is at distance $R$ from the axis.
The solid sphere has the smallest ratio, $\tfrac{2}{5}$ .

4.4Kinematics of Rotational Motion

Let a body turn about a fixed axis, and let its angular acceleration stay constant. Its equations of motion have the same form as the equations of linear motion. The rotational quantities simply take the place of the linear ones.

Linear Motion

v = u + at

s = ut + \tfrac{1}{2}at^2

v^2 - u^2 = 2as

Rotational Motion

\omega = \omega_0 + \alpha t

\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2

\omega^2 - \omega_0^2 = 2\alpha(\theta - \theta_0)

Here $\theta_0$ and $\omega_0$ are the initial angular position and the initial angular velocity of the body. If $\theta_0 = 0$ , the second equation becomes $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$ .

The linear and rotational quantities match like this:

Quantity	Linear	Rotational
Displacement	$s$	$\theta - \theta_0$ (angular displacement)
Initial velocity	$u$	$\omega_0$ (initial angular velocity)
Final velocity	$v$	$\omega$ (angular velocity)
Acceleration	$a$	$\alpha$ (angular acceleration)
Time	$t$	$t$

Note

Use these equations only if $\alpha$ is constant. The linear equations, too, need a constant acceleration.

4.5Dynamics of Rotational Motion

Each quantity in translation has a partner quantity in rotation. This partner is its rotational analogue. The table compares them for motion about a fixed axis.

Quantity	Pure Translation	Pure Rotation
Position	$x$	Angular position, $\theta$
Velocity	$v = \dfrac{dx}{dt}$	$\omega = \dfrac{d\theta}{dt}$
Acceleration	$a = \dfrac{dv}{dt}$	$\alpha = \dfrac{d\omega}{dt}$
Inertia	Mass, $m$	Moment of inertia, $I$
Newton's second law	$F = ma$	$\tau = I\alpha$
Work done	$W = \displaystyle\int F\,dx$	$W = \displaystyle\int \tau\,d\theta$
Kinetic energy	$K = \tfrac{1}{2}mv^2$	$K = \tfrac{1}{2}I\omega^2$
Power	$P = Fv$	$P = \tau\omega$
Momentum	$p = mv$	$L = I\omega$

4.6Work Done by a Torque and Power

Let a rigid body turn about a fixed axis. A force $\vec{F}_1$ acts on one of its particles at point $P_1$ . Its line of action is in a plane that is at right angles to the axis.

(a) Vector

\vec{\omega}

lies on the fixed axis

(b)

\vec{F}_1

does work as

P_1

turns through

d\theta

Derivation

$P_1$ moves on a circle of radius $r_1 = CP_1$ with centre $C$ on the axis. When the body turns through a small angle $d\theta$ , $P_1$ moves through $ds_1 = r_1\,d\theta$ . So the work done by $\vec{F}_1$ is:
$dW_1 = \vec{F}_1\cdot d\vec{s}_1 = F_1\,ds_1\cos\phi_1 = F_1(r_1\,d\theta)\sin\alpha_1$
$\phi_1$ = angle between $\vec{F}_1$ and the tangent at $P_1$ . $\alpha_1$ = angle between $\vec{F}_1$ and the radius $CP_1$ . So $\phi_1 + \alpha_1 = 90^\circ$ .
The torque of $\vec{F}_1$ about the axis is $\vec{\tau}_1 = \vec{CP}_1 \times \vec{F}_1$ . Its magnitude is $\tau_1 = r_1 F_1\sin\alpha_1$ . The part $OC$ of $\vec{OP}_1$ lies along the axis, so it gives no torque about the axis. Therefore:
$dW_1 = \tau_1\,d\theta$
Now add the work done by all the forces. Their torques add up to the net torque $\tau = \tau_1 + \tau_2 + \cdots$ on the body. So $dW = (\tau_1 + \tau_2 + \cdots)\,d\theta = \tau\,d\theta$ . Divide both sides by $dt$ :
$\dfrac{dW}{dt} = \tau\dfrac{d\theta}{dt} = \tau\omega \qquad (\because \omega = \tfrac{d\theta}{dt})$

Work Done by a Torque

W = \int \tau\,d\theta \;=\; \tau\,\theta

The last form is for a constant torque.

Instantaneous Power

P = \dfrac{dW}{dt} = \tau\omega

Note

For rotation about a fixed axis, take only the forces that lie in planes perpendicular to the axis. Also take only the parts of the position vectors perpendicular to the axis. Ignore the parts of torque that are perpendicular to the fixed axis.

4.7Relation between Torque and Moment of Inertia

A rigid body has rotational KE, $K = \tfrac{1}{2}I\omega^2$ . If the body has an angular acceleration $\alpha$ , this KE keeps changing. The power is the rate of this change.

Derivation

Power = rate of change of rotational KE. Here $I$ is constant and $\alpha = \tfrac{d\omega}{dt}$ :
$P = \dfrac{d}{dt}\left(\tfrac{1}{2}I\omega^2\right) = \tfrac{1}{2}I(2\omega)\dfrac{d\omega}{dt} = I\omega\alpha$
The power of a torque is also $P = \tau\omega$ (see 4.6). Equate the two:
$\tau\omega = I\omega\alpha \;\Rightarrow\; \tau = I\alpha$

Newton's Second Law for Rotation

\tau = I\alpha

Net torque = moment of inertia × angular acceleration. It is like Newton's second law, $F = ma$ , for translation with constant mass.

4.8Angular Momentum about a Fixed Axis

About a fixed axis, the angular momentum of a rigid body has a simple form, like $p = mv$ .

Derivation

A particle of mass $m_i$ at distance $r_i$ from the axis has speed $v_i = r_i\omega$ . Its angular momentum is $m_i v_i r_i$ . So the total is:
$L = \sum m_i v_i r_i$
Put $v_i = r_i\omega$ . All the particles have the same $\omega$ :
$L = \sum m_i (r_i\omega) r_i = \left(\sum m_i r_i^2\right)\omega$
But $\sum m_i r_i^2 = I$ , the moment of inertia about the axis. So $L = I\omega$ .

Angular Momentum about a Fixed Axis

L = I\omega

This is the rotational analogue of linear momentum, $p = mv$ .

iRelation to Torque

Differentiate, keeping $I$ constant:

\dfrac{dL}{dt} = \dfrac{d(I\omega)}{dt} = I\alpha = \tau

iiConservation

If $\tau_{ext} = 0$ , then $L = I\omega$ stays constant:

I_1\omega_1 = I_2\omega_2

Note

For one particle, $\vec{L}$ and $\vec{\omega}$ need not point the same way. Linear momentum and velocity, on the other hand, always point the same way. If a symmetric body turns about a fixed axis, $\vec{L}$ lies along the axis and $\vec{L} = I\vec{\omega}$ .

4.9Rolling Motion

Definition

Rolling motion is a mix of pure translation and pure rotation happening together. It is very common in daily life. For example, the wheels of a vehicle roll on the road.

In rolling without slipping, the point of the body that touches the surface has zero velocity at every instant.

A disc rolls on a level surface. Its centre of mass $C$ is its geometric centre. $C$ moves with velocity $v_{CM}$ parallel to the surface.

At the same time, the disc spins about the axis through $C$ . So each point has two velocities at once: $v_{CM}$ and the rotational velocity $v_r$ . Its actual velocity is their vector sum. Here $v_r = r\omega$ , and $v_r$ is perpendicular to $r$ .

A disc rolling to the right:

P_0

is at rest and the top point

P_1

moves at

2v_{CM}

Derivation

At the contact point $P_0$ , the rotational velocity $R\omega$ points opposite to $v_{CM}$ .
There is no slipping, so $P_0$ is at rest. The two velocities must cancel:
$v_{CM} - R\omega = 0 \;\Rightarrow\; v_{CM} = R\omega$

Condition for Rolling Without Slipping

v_{CM} = R\omega

where, $R$ = radius of the rolling body, $\omega$ = angular velocity of the body

Velocity of the contact point $P_0$

0

(at rest)

Velocity of the centre $C$

v_{CM} = R\omega

Velocity of the topmost point $P_1$

v_{CM} + R\omega = 2v_{CM}

Kinetic Energy of a Rolling Body

Derivation

A body rolling without slipping has translational KE and rotational KE. Its total KE is their sum:
$K = \tfrac{1}{2}mv_{CM}^2 + \tfrac{1}{2}I\omega^2$
Put $I = mk^2$ and $\omega = \dfrac{v_{CM}}{R}$ . Here $k$ is the radius of gyration:
$K = \tfrac{1}{2}mv_{CM}^2 + \tfrac{1}{2}mk^2\left(\dfrac{v_{CM}}{R}\right)^2$
Take $\tfrac{1}{2}mv_{CM}^2$ common to get the final result.

Kinetic Energy of a Rolling Body

K = \dfrac{1}{2}mv_{CM}^2\left(1 + \dfrac{k^2}{R^2}\right)

where, $\dfrac{k^2}{R^2}$ is the ratio given for each body in topic 4.3

End of Chapter 6 · notesshala.in