Introduction to Rotational Motion

Previous chapters focused on the motion of single particles, where the body's size was considered negligible. However, real-world objects have finite size. This chapter expands our study to the motion of such extended bodies, which can be considered as a system of particles. A key concept for this is the centre of mass.

A special and important category of extended bodies is the rigid body. An ideal rigid body is one with a perfectly definite and unchanging shape. The distances between all pairs of particles within the body remain constant. While no real body is perfectly rigid, many objects like wheels, beams, and planets can be treated as rigid for most practical purposes, as their deformations are negligible.

---

What kind of motion can a rigid body have?

The motion of a rigid body can be categorized as follows:

1. Pure Translational Motion

In pure translational motion, all particles of the body have the same velocity at any given instant. The body moves as a whole without rotating. An example is a block sliding down an inclined plane without tumbling.

2. Rotational Motion

In rotational motion, the body turns around a fixed line called the axis of rotation. Every particle of the body moves in a circle. These circles lie in planes perpendicular to the axis of rotation, and their centers lie on the axis.

The rotation can be about a fixed axis (e.g., a ceiling fan) or a moving axis (e.g., a spinning top or an oscillating pedestal fan). This chapter will primarily focus on the simpler case of rotation about a fixed axis.

3. Combination of Translation and Rotation

This is the most general form of motion for a rigid body. The body's position changes (translation) while it also rotates. A common example is the rolling motion of a wheel or a cylinder down an inclined plane. At any instant, different particles of the body have different velocities.

Centre of Mass

The centre of mass (CM) is a specific point associated with a system of particles. It is the average position of all the mass in the system, weighted by each particle's mass. The motion of the system as a whole can often be described by the motion of this single point.

Centre of Mass for a System of Particles

For a system of $n$ particles with masses $m_1, m_2, \dots, m_n$ located at positions $(x_1, y_1, z_1), (x_2, y_2, z_2), \dots$ respectively, the coordinates of the centre of mass $(X, Y, Z)$ are given by:

$X = \frac{m_1x_1 + m_2x_2 + \dots + m_nx_n}{m_1 + m_2 + \dots + m_n} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} = \frac{\sum m_i x_i}{M}$

$Y = \frac{\sum m_i y_i}{M}$

$Z = \frac{\sum m_i z_i}{M}$

where $M = \sum m_i$ is the total mass of the system.

Vector Form

In vector notation, if $\textbf{r}_i$ is the position vector of the $i$-th particle and R is the position vector of the centre of mass, the formula becomes more compact:

$\textbf{R} = \frac{m_1\textbf{r}_1 + m_2\textbf{r}_2 + \dots + m_n\textbf{r}_n}{M} = \frac{1}{M} \sum_{i=1}^{n} m_i \textbf{r}_i$

If the origin of the coordinate system is chosen at the centre of mass, then $\textbf{R}=0$, which implies $\sum m_i \textbf{r}_i = 0$.

---

Centre of Mass for a Rigid Body (Continuous Mass Distribution)

For a continuous body, we imagine it to be composed of an infinite number of infinitesimal mass elements $dm$. The summation is replaced by an integral over the entire body:

$X = \frac{1}{M}\int x dm$

$Y = \frac{1}{M}\int y dm$

$Z = \frac{1}{M}\int z dm$

In vector form:

$\textbf{R} = \frac{1}{M}\int \textbf{r} dm$

Symmetry and Centre of Mass

For homogeneous bodies of regular geometric shapes, the centre of mass coincides with their geometric centre due to symmetry. For every mass element $dm$ at a position $\textbf{r}$ from the geometric centre, there is an identical element at $-\textbf{r}$, and their contributions to the integral cancel out.

• Thin Rod: Midpoint of the rod.
• Ring/Disc: Geometric centre.
• Sphere: Geometric centre.

---

Answer:

Motion of Centre of Mass

The concept of the centre of mass is powerful because it simplifies the description of the motion of a complex system. Let's start with the position vector of the centre of mass:

$M\textbf{R} = \sum m_i \textbf{r}_i$

Differentiating with respect to time gives the velocity of the centre of mass, $\textbf{V} = d\textbf{R}/dt$:

$M\frac{d\textbf{R}}{dt} = \sum m_i \frac{d\textbf{r}_i}{dt} \implies M\textbf{V} = \sum m_i \textbf{v}_i$

Differentiating again with respect to time gives the acceleration of the centre of mass, $\textbf{A} = d\textbf{V}/dt$:

$M\frac{d\textbf{V}}{dt} = \sum m_i \frac{d\textbf{v}_i}{dt} \implies M\textbf{A} = \sum m_i \textbf{a}_i$

Using Newton's second law, $\textbf{F}_i = m_i \textbf{a}_i$, where $\textbf{F}_i$ is the net force on the $i$-th particle.

$M\textbf{A} = \sum \textbf{F}_i$

The total force $\sum \textbf{F}_i$ includes both external forces (from outside the system) and internal forces (between particles within the system). By Newton's third law, internal forces occur in equal and opposite pairs, so their vector sum is zero. Thus, the sum of all forces is just the sum of all external forces, $\textbf{F}_{ext}$.

$M\textbf{A} = \textbf{F}_{ext}$

This is a profound result. It states that: The centre of mass of a system of particles moves as if the entire mass of the system were concentrated at the centre of mass and all the external forces were applied at that point.

This is why we could treat extended bodies as point particles in earlier chapters. The motion of the centre of mass (the translational component of motion) is independent of the internal forces, rotation, or vibrations of the system.

Linear Momentum of a System of Particles

The total linear momentum of a system of particles, P, is the vector sum of the linear momenta of the individual particles:

$\textbf{P} = \textbf{p}_1 + \textbf{p}_2 + \dots + \textbf{p}_n = \sum m_i \textbf{v}_i$

From the previous section, we know that $\sum m_i \textbf{v}_i = M\textbf{V}$, where V is the velocity of the centre of mass. Therefore,

$\textbf{P} = M\textbf{V}$

The total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass.

Differentiating this with respect to time:

$\frac{d\textbf{P}}{dt} = M\frac{d\textbf{V}}{dt} = M\textbf{A}$

Since we already established that $M\textbf{A} = \textbf{F}_{ext}$, we get Newton's second law for a system of particles:

$\frac{d\textbf{P}}{dt} = \textbf{F}_{ext}$

The time rate of change of the total linear momentum of a system is equal to the net external force acting on the system.

---

Conservation of Linear Momentum

If the net external force on a system is zero ($\textbf{F}_{ext}=0$), then:

$\frac{d\textbf{P}}{dt} = 0 \implies \textbf{P} = \text{Constant}$

This is the law of conservation of linear momentum for a system of particles: If the net external force on a system is zero, its total linear momentum remains constant.

Since $\textbf{P} = M\textbf{V}$, this also means that if the net external force is zero, the velocity of the centre of mass of the system remains constant. The CM moves like a free particle, regardless of the complex internal interactions.

Vector Product of Two Vectors

In addition to the scalar (dot) product, there is another way to multiply vectors called the vector product or cross product. This product results in a new vector. It is essential for defining rotational quantities like torque and angular momentum.

The vector product of two vectors a and b is a vector c, written as $\textbf{c} = \textbf{a} \times \textbf{b}$.

Definition

1. Magnitude: The magnitude of c is given by $c = ab \sin\theta$, where $\theta$ is the smaller angle between a and b.
2. Direction: The vector c is perpendicular to the plane containing both a and b. Its specific direction is given by the right-hand rule.

Right-Hand Rule

If you curl the fingers of your right hand in the direction from vector a towards vector b (through the smaller angle), your outstretched thumb points in the direction of $\textbf{a} \times \textbf{b}$.

---

Properties of the Vector Product

• Anti-commutative: Reversing the order of multiplication reverses the direction of the resultant vector.

$\textbf{a} \times \textbf{b} = -(\textbf{b} \times \textbf{a})$

• Parallel Vectors: The cross product of two parallel or anti-parallel vectors is a null vector, since $\sin 0^\circ = 0$ and $\sin 180^\circ = 0$. Thus, $\textbf{a} \times \textbf{a} = \textbf{0}$.
• Distributive Law: $\textbf{a} \times (\textbf{b} + \textbf{c}) = \textbf{a} \times \textbf{b} + \textbf{a} \times \textbf{c}$

Vector Product in Component Form

For the orthogonal unit vectors:

$\hat{\textbf{i}} \times \hat{\textbf{i}} = \hat{\textbf{j}} \times \hat{\textbf{j}} = \hat{\textbf{k}} \times \hat{\textbf{k}} = \textbf{0}$

$\hat{\textbf{i}} \times \hat{\textbf{j}} = \hat{\textbf{k}}$;   $\hat{\textbf{j}} \times \hat{\textbf{k}} = \hat{\textbf{i}}$;   $\hat{\textbf{k}} \times \hat{\textbf{i}} = \hat{\textbf{j}}$ (Cyclic Order)

$\hat{\textbf{j}} \times \hat{\textbf{i}} = -\hat{\textbf{k}}$;   $\hat{\textbf{k}} \times \hat{\textbf{j}} = -\hat{\textbf{i}}$;   $\hat{\textbf{i}} \times \hat{\textbf{k}} = -\hat{\textbf{j}}$ (Anti-Cyclic Order)

For two vectors $\textbf{a} = a_x\hat{\textbf{i}} + a_y\hat{\textbf{j}} + a_z\hat{\textbf{k}}$ and $\textbf{b} = b_x\hat{\textbf{i}} + b_y\hat{\textbf{j}} + b_z\hat{\textbf{k}}$, their cross product is:

$\textbf{a} \times \textbf{b} = (a_y b_z - a_z b_y)\hat{\textbf{i}} + (a_z b_x - a_x b_z)\hat{\textbf{j}} + (a_x b_y - a_y b_x)\hat{\textbf{k}}$

This can be easily remembered using the determinant form:

$\textbf{a} \times \textbf{b} = \begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}$

Angular Velocity and its Relation with Linear Velocity

For a rigid body rotating about a fixed axis, all particles move in circles, and every particle has the same angular velocity ($\omega$) at any instant. Angular velocity is defined as the time rate of change of angular displacement ($\theta$).

$\omega = \frac{d\theta}{dt}$

Angular velocity is a vector quantity. For rotation about a fixed axis, the vector $\boldsymbol{\omega}$ points along the axis of rotation, with its direction determined by the right-hand rule (curl fingers in the direction of rotation, thumb points in the direction of $\boldsymbol{\omega}$).

---

Relation between Linear and Angular Velocity

The magnitude of the linear velocity $v$ of a particle at a perpendicular distance $r$ from the axis of rotation is given by $v = \omega r$. The direction of the velocity is always tangent to the circular path.

This relationship can be expressed elegantly using the vector product. If r is the position vector of a particle with respect to an origin on the axis of rotation, its linear velocity v is given by:

$\textbf{v} = \boldsymbol{\omega} \times \textbf{r}$

This equation correctly gives both the magnitude ($v = \omega r \sin 90^\circ = \omega r$) and the direction (tangential to the path) of the linear velocity.

---

Angular Acceleration

In analogy with linear motion, angular acceleration ($\alpha$) is defined as the time rate of change of angular velocity.

$\boldsymbol{\alpha} = \frac{d\boldsymbol{\omega}}{dt}$

For rotation about a fixed axis, the direction of $\boldsymbol{\alpha}$ is also along the axis. If the body is speeding up, $\boldsymbol{\alpha}$ is in the same direction as $\boldsymbol{\omega}$; if it is slowing down, $\boldsymbol{\alpha}$ is in the opposite direction.

Torque and Angular Momentum

Force causes linear acceleration. What is the rotational analogue of force? This is the concept of torque or moment of force.

Torque (Moment of Force)

The ability of a force to produce rotation depends not only on its magnitude but also on where it is applied. Torque is the rotational analogue of force.

For a force F acting on a particle at a position r with respect to an origin O, the torque $\boldsymbol{\tau}$ about the origin is defined as the vector product:

$\boldsymbol{\tau} = \textbf{r} \times \textbf{F}$

• The magnitude of the torque is $\tau = rF \sin\theta = r_{\perp}F = rF_{\perp}$, where $\theta$ is the angle between r and F, $r_{\perp}$ is the perpendicular distance from the origin to the line of action of the force (called the lever arm), and $F_{\perp}$ is the component of the force perpendicular to r.
• The SI unit of torque is the newton-metre (N m).
• Torque is zero if the line of action of the force passes through the origin ($r=0$ or $\sin\theta=0$).

---

Angular Momentum

Just as torque is the rotational analogue of force, angular momentum is the rotational analogue of linear momentum.

For a particle with linear momentum p at a position r with respect to an origin O, the angular momentum l about the origin is defined as:

$\textbf{l} = \textbf{r} \times \textbf{p}$

The magnitude is $l = rp \sin\theta$.

---

Relation between Torque and Angular Momentum

There is a fundamental relationship between torque and angular momentum, which is the rotational version of Newton's second law ($\textbf{F} = d\textbf{p}/dt$).

By differentiating the definition of angular momentum with respect to time:

$\frac{d\textbf{l}}{dt} = \frac{d}{dt}(\textbf{r} \times \textbf{p}) = \left(\frac{d\textbf{r}}{dt} \times \textbf{p}\right) + \left(\textbf{r} \times \frac{d\textbf{p}}{dt}\right)$

The first term is $(\textbf{v} \times m\textbf{v})$, which is zero because v is parallel to itself. The second term is $\textbf{r} \times \textbf{F}$, which is the torque $\boldsymbol{\tau}$.

Thus, we arrive at the rotational analogue of Newton's second law:

$\boldsymbol{\tau} = \frac{d\textbf{l}}{dt}$

The time rate of change of the angular momentum of a particle is equal to the torque acting on it.

For a System of Particles

This relationship can be extended to a system of particles. The total angular momentum is $\textbf{L} = \sum \textbf{l}_i$, and the total torque is $\boldsymbol{\tau} = \sum \boldsymbol{\tau}_i$. The internal torques cancel out in pairs, so the equation becomes:

$\boldsymbol{\tau}_{ext} = \frac{d\textbf{L}}{dt}$

---

Conservation of Angular Momentum

From the above relation, if the net external torque on a system is zero ($\boldsymbol{\tau}_{ext} = 0$), then:

$\frac{d\textbf{L}}{dt} = 0 \implies \textbf{L} = \text{Constant}$

Law of Conservation of Angular Momentum: If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved.

Equilibrium of a Rigid Body

A rigid body is said to be in mechanical equilibrium if both its linear momentum and angular momentum are constant. This means the body has neither linear acceleration nor angular acceleration.

For a rigid body to be in mechanical equilibrium, two conditions must be met simultaneously:

1. Translational Equilibrium

The vector sum of all external forces acting on the body must be zero.

$\sum \textbf{F}_{ext} = 0$

This ensures that the linear momentum of the centre of mass is constant (i.e., zero linear acceleration).

---

2. Rotational Equilibrium

The vector sum of all external torques acting on the body about any point must be zero.

$\sum \boldsymbol{\tau}_{ext} = 0$

This ensures that the angular momentum of the body is constant (i.e., zero angular acceleration). If the body is in translational equilibrium, this condition holds true regardless of the point about which the torques are calculated.

These two vector equations are equivalent to a total of six independent scalar equations (three for forces, three for torques). For problems where all forces are coplanar, this reduces to three conditions: $\sum F_x = 0$, $\sum F_y = 0$, and $\sum \tau_z = 0$.

---

Principle of Moments and Levers

For a lever in equilibrium pivoted at a fulcrum, the condition for rotational equilibrium ($\sum \tau = 0$) leads to the principle of moments. If a load $F_1$ is at a distance $d_1$ (load arm) and an effort $F_2$ is at a distance $d_2$ (effort arm) from the fulcrum, then:

$F_1 d_1 = F_2 d_2$

(Anticlockwise moment = Clockwise moment)

The ratio $F_1/F_2$ is called the Mechanical Advantage (M.A.).

---

Centre of Gravity (CG)

The centre of gravity of a body is the point where the total gravitational torque on the body is zero. It is the point where the entire weight of the body can be considered to act.

For a small body in a uniform gravitational field, the centre of gravity coincides with the centre of mass. However, they are conceptually different. The CM depends only on the mass distribution, while the CG also involves gravity.

Moment of Inertia

Just as mass is a measure of a body's inertia in linear motion (resistance to change in linear velocity), the moment of inertia (I) is a measure of a body's rotational inertia (resistance to change in angular velocity).

The kinetic energy of a single particle of mass $m_i$ at a distance $r_i$ from the axis of rotation is $K_i = \frac{1}{2}m_i v_i^2$. Since $v_i = \omega r_i$, this becomes $K_i = \frac{1}{2}m_i r_i^2 \omega^2$.

The total rotational kinetic energy of the rigid body is the sum of the kinetic energies of all its particles:

$K_{rot} = \sum K_i = \sum \left(\frac{1}{2}m_i r_i^2 \omega^2\right) = \frac{1}{2}\left(\sum m_i r_i^2\right)\omega^2$

The quantity in the parenthesis is defined as the moment of inertia, $I$.

$I = \sum m_i r_i^2$

The rotational kinetic energy can then be written in a form analogous to translational kinetic energy:

$K_{rot} = \frac{1}{2}I\omega^2$

The moment of inertia depends not just on the total mass, but on how that mass is distributed relative to the axis of rotation. The farther the mass is from the axis, the larger the moment of inertia.

The SI unit of moment of inertia is kg m$^2$.

---

Radius of Gyration (k)

The radius of gyration is the distance from the axis of rotation to a point where the entire mass of the body could be concentrated without changing its moment of inertia. It is defined by the relation:

$I = Mk^2 \implies k = \sqrt{\frac{I}{M}}$

where $M$ is the total mass of the body.

Theorems of Perpendicular and Parallel Axes

These two theorems are useful for calculating the moment of inertia of a body about an axis if its moment of inertia about another related axis is known.

Theorem of Perpendicular Axes

This theorem applies only to planar bodies (laminae).

It states that the moment of inertia of a planar body about an axis perpendicular to its plane ($I_z$) is equal to the sum of its moments of inertia about two perpendicular axes lying in the plane of the body and intersecting the first axis ($I_x$ and $I_y$).

$I_z = I_x + I_y$

Answer:

---

Theorem of Parallel Axes

This theorem is applicable to a body of any shape.

It states that the moment of inertia of a body about any axis ($I$) is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass ($I_{cm}$) and the product of its mass ($M$) and the square of the perpendicular distance ($a$) between the two axes.

$I = I_{cm} + Ma^2$

Answer:

Kinematics of Rotational Motion about a Fixed Axis

There is a direct analogy between the variables and equations of linear motion and rotational motion. This makes it easy to write down the kinematic equations for rotation with constant angular acceleration.

Linear Motion	Rotational Motion
Displacement, $x$	Angular displacement, $\theta$
Velocity, $v = dx/dt$	Angular velocity, $\omega = d\theta/dt$
Acceleration, $a = dv/dt$	Angular acceleration, $\alpha = d\omega/dt$

Kinematic Equations for Uniform Angular Acceleration

For rotational motion with constant angular acceleration $\alpha$, the equations are:

$\omega = \omega_0 + \alpha t$

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

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

where $\omega_0$ and $\theta_0$ are the initial angular velocity and angular position at $t=0$.

Dynamics of Rotational Motion about a Fixed Axis

We now establish the rotational equivalent of Newton's second law, $\textbf{F} = m\textbf{a}$. The rotational analogue of force is torque, and the analogue of mass is the moment of inertia.

Work Done by a Torque

The work done by a force F for an infinitesimal displacement $d\textbf{s}$ is $dW = \textbf{F} \cdot d\textbf{s}$. For a particle in rotation, the displacement is an arc length $ds = r d\theta$. The work done by a torque $\tau$ for an infinitesimal angular displacement $d\theta$ is:

$dW = \tau d\theta$

Power in Rotational Motion

The instantaneous power is the rate of doing work, $P = dW/dt$.

$P = \frac{\tau d\theta}{dt} = \tau \omega$

Newton's Second Law for Rotation

By the work-energy theorem, the rate at which work is done on a body equals the rate of change of its kinetic energy. For a rotating body, $P = dK/dt$.

$P = \tau \omega$ and $K = \frac{1}{2}I\omega^2$

$\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}I\omega^2\right) = \frac{1}{2}I (2\omega) \frac{d\omega}{dt} = I\omega\alpha$

Equating the two expressions for power: $\tau\omega = I\omega\alpha$. This leads to:

$\tau = I\alpha$

This is Newton's second law for rotational motion. The net external torque on a body is equal to the product of its moment of inertia and its angular acceleration.

Angular Momentum in Case of Rotation about a Fixed Axis

We now consider the angular momentum for the specific case of a rigid body rotating about a fixed axis. The total angular momentum of the body L is the vector sum of the angular momenta of all its constituent particles.

For a particle of mass $m_i$ at a perpendicular distance $r_i$ from the axis, its angular momentum has a component along the axis of rotation ($L_z$) and a component perpendicular to it ($L_{\perp}$).

The component of the total angular momentum along the axis of rotation is given by:

$L_z = \sum (m_i r_i^2) \omega = \left(\sum m_i r_i^2\right) \omega$

$L_z = I\omega$

For symmetric bodies rotating about an axis of symmetry, the perpendicular component of angular momentum ($L_{\perp}$) is zero. In such cases, the total angular momentum vector lies along the axis of rotation:

$\textbf{L} = I\boldsymbol{\omega}$

Conservation of Angular Momentum (for fixed-axis rotation)

The general relation $\boldsymbol{\tau}_{ext} = d\textbf{L}/dt$ holds. For rotation about a fixed axis, this implies that the component of external torque along the axis equals the rate of change of the component of angular momentum along that axis: $\tau_z = dL_z/dt$.

$\tau_z = \frac{d(I\omega)}{dt}$

If the net external torque about the axis of rotation is zero ($\tau_{ext, z} = 0$), then:

$\frac{dL_z}{dt} = 0 \implies L_z = I\omega = \text{Constant}$

This is the principle of conservation of angular momentum for fixed-axis rotation. If the moment of inertia of a rotating system changes, its angular velocity must change inversely to keep the product $I\omega$ constant.

This principle is demonstrated by ice skaters pulling in their arms to spin faster (decreasing I, increasing $\omega$) and divers tucking their bodies to perform somersaults.

Rolling Motion

Rolling motion is a combination of translational motion of the centre of mass and rotational motion about the centre of mass.

Condition for Rolling Without Slipping

For an object to roll without slipping, the point of contact between the rolling object and the surface must be instantaneously at rest relative to the surface. This condition leads to a relationship between the velocity of the centre of mass ($v_{cm}$) and the angular velocity ($\omega$):

$v_{cm} = R\omega$

where $R$ is the radius of the rolling object.

---

Kinetic Energy of Rolling Motion

The total kinetic energy of a rolling body is the sum of its translational kinetic energy (of the centre of mass) and its rotational kinetic energy (about the centre of mass).

$K_{total} = K_{trans} + K_{rot}$

$K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

Using the condition for rolling without slipping ($\omega = v_{cm}/R$) and the radius of gyration ($I_{cm} = Mk^2$), we can write the total kinetic energy in terms of the translational velocity:

$K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}(Mk^2)\left(\frac{v_{cm}}{R}\right)^2$

$K_{total} = \frac{1}{2}Mv_{cm}^2 \left(1 + \frac{k^2}{R^2}\right)$

---

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Additional Exercises

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer: