Introduction

Previous chapters primarily focused on the motion of a single particle, often using this idealization to describe the motion of bodies with finite size.

However, the particle model is insufficient for describing the motion of real, extended bodies (bodies with finite size and shape), especially when rotation is involved.

An extended body can be considered as a system of many particles.

This chapter introduces the concept of a system of particles and develops tools to describe the motion of such systems, including rigid bodies.

A rigid body is an idealized body where the distances between all pairs of its constituent particles remain fixed and unchanged, regardless of the forces applied.

Real bodies deform under forces, but many can be approximated as rigid bodies if deformations are negligible (e.g., wheels, tops, steel beams, planets).

What Kind Of Motion Can A Rigid Body Have?

The motion of a rigid body depends on whether it is fixed or constrained in some way.

Rigid body NOT pivoted or fixed:

• Can have pure translational motion: All particles of the body have the same velocity at any given instant. The body as a whole moves without changing its orientation.
• Can have a combination of translational and rotational motion: Different particles have different velocities at any instant. The body moves from one place to another (translation) while also changing its orientation (rotation). Rolling motion is a common example.

Rigid body pivoted or fixed:

• Can have rotational motion:
  • Rotation about a fixed axis: One line in the body is fixed. Every particle in the body moves in a circle. Each circle lies in a plane perpendicular to the axis and has its center on the axis. Particles on the axis itself are stationary. Examples: ceiling fan, potter's wheel.
  • Rotation about a fixed point: One point in the body is fixed. The axis of rotation passes through this fixed point but its direction may change over time. Examples: spinning top (point of contact fixed), oscillating table fan (pivot fixed).

This chapter primarily focuses on translational motion of systems of particles and rotational motion about a fixed axis.

Centre Of Mass

The Centre of Mass (CM) is a crucial concept for describing the overall translational motion of a system of particles or a rigid body.

For a system of two particles with masses $m_1$ and $m_2$ located at positions $x_1$ and $x_2$ along the x-axis, the position $X$ of the centre of mass is given by the mass-weighted average of their positions:

For a system of $n$ particles with masses $m_1, m_2, \dots, m_n$ located at positions $x_1, x_2, \dots, x_n$ along the x-axis, the position $X$ of the centre of mass is:

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

For a system of $n$ particles distributed in three dimensions, with the $i$-th particle of mass $m_i$ located at $(x_i, y_i, z_i)$, the coordinates $(X, Y, Z)$ of the centre of mass are:

Using position vectors, if $\mathbf{r}_i = x_i \hat{\mathbf{i}} + y_i \hat{\mathbf{j}} + z_i \hat{\mathbf{k}}$ is the position vector of the $i$-th particle, the position vector $\mathbf{R}$ of the centre of mass is:

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

For a rigid body (a continuous distribution of mass), the sums are replaced by integrals:

where $M = \int dm$ is the total mass, and $\int \mathbf{r} \, dm$ is the integral of the position vector over all infinitesimal mass elements $dm$ of the body.

In coordinates: $X = \frac{1}{M} \int x \, dm$, $Y = \frac{1}{M} \int y \, dm$, $Z = \frac{1}{M} \int z \, dm$.

If the origin is at the centre of mass, $\int \mathbf{r} \, dm = \mathbf{0}$.

For homogeneous bodies of regular shapes (uniformly distributed mass), the centre of mass is located at their geometric centre. This can be shown using symmetry arguments (e.g., for a rod with the origin at its center, for every mass element at $x$, there's one at $-x$, making the integral $\int x \, dm$ zero).

Answer:

Let the equilateral triangle be OAB with side length $l = 0.5$ m. Place the origin at vertex O, with the base OA along the x-axis.

Coordinates of the vertices:

• O: $(0, 0)$
• A: $(l, 0) = (0.5, 0)$
• B: $(l/2, l\sqrt{3}/2) = (0.25, 0.25\sqrt{3})$

Masses at the vertices:

• $m_1 = 100$ g at O $(0,0)$
• $m_2 = 150$ g at A $(0.5,0)$
• $m_3 = 200$ g at B $(0.25, 0.25\sqrt{3})$

Total mass $M = m_1 + m_2 + m_3 = 100 + 150 + 200 = 450$ g.

Coordinates of the centre of mass $(X, Y)$:

$$ X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} = \frac{(100)(0) + (150)(0.5) + (200)(0.25)}{450} \text{ m} $$ $$ X = \frac{0 + 75 + 50}{450} \text{ m} = \frac{125}{450} \text{ m} = \frac{5}{18} \text{ m} $$ $$ Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{(100)(0) + (150)(0) + (200)(0.25\sqrt{3})}{450} \text{ m} $$ $$ Y = \frac{0 + 0 + 50\sqrt{3}}{450} \text{ m} = \frac{50\sqrt{3}}{450} \text{ m} = \frac{\sqrt{3}}{9} \text{ m} $$

The centre of mass is located at $\left(\frac{5}{18} \text{ m}, \frac{\sqrt{3}}{9} \text{ m}\right)$.

Note: The geometric centre (centroid) of the triangle is at $\left(\frac{0+0.5+0.25}{3}, \frac{0+0+0.25\sqrt{3}}{3}\right) = \left(\frac{0.75}{3}, \frac{0.25\sqrt{3}}{3}\right) = \left(0.25, \frac{\sqrt{3}}{12}\right)$. The centre of mass does not coincide with the geometric centre because the masses at the vertices are not equal.

Answer:

Consider a uniform triangular lamina (a thin flat plate). We can divide the triangle into infinitesimally thin strips parallel to one of its sides, say the base MN.

Since the lamina is uniform, the mass density is constant. Each thin strip is essentially a thin rod. By symmetry, the centre of mass of each thin strip lies at its midpoint.

The locus of the midpoints of all strips parallel to the base MN is the median of the triangle drawn from the opposite vertex L to the midpoint P of MN. Thus, the centre of mass of the entire triangle must lie somewhere on the median LP.

We can repeat this argument by taking strips parallel to the other two sides, LM and LN. The centre of mass must also lie on the other two medians, MQ and NR.

Since the centre of mass must lie on all three medians, it must be located at the point where the three medians intersect. This point is the centroid of the triangle.

Thus, the centre of mass of a uniform triangular lamina is at its centroid.

Example 7.3. Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg.

Answer:

Given the dimensions and total mass $M = 3$ kg. Since the lamina is uniform, its mass density is constant. We can divide the L-shaped lamina into simpler shapes whose centres of mass are easily identifiable. The L-shape can be considered as three 1m $\times$ 1m squares.

Since the total mass is 3 kg and the area is 3 square meters, the mass of each 1m $\times$ 1m square is $3 \text{ kg} / 3 = 1$ kg. Let the mass of each square be $m_i = 1$ kg.

By symmetry, the centre of mass of each uniform square is at its geometric center. Let's define the origin at the bottom-left corner as shown in the figure.

• Square 1 (bottom-left, 0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 1): CM is at $C_1 = (0.5, 0.5)$ m. Mass $m_1 = 1$ kg.
• Square 2 (bottom-right, 1 $\le$ x $\le$ 2, 0 $\le$ y $\le$ 1): CM is at $C_2 = (1.5, 0.5)$ m. Mass $m_2 = 1$ kg.
• Square 3 (top-left, 0 $\le$ x $\le$ 1, 1 $\le$ y $\le$ 2): CM is at $C_3 = (0.5, 1.5)$ m. Mass $m_3 = 1$ kg.

Now, treat these three squares as point masses located at their respective centres of mass. The centre of mass of the L-shaped lamina is the centre of mass of this system of three point masses.

Coordinates of the centre of mass $(X, Y)$ of the L-shape:

$$ X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{(1)(0.5) + (1)(1.5) + (1)(0.5)}{1 + 1 + 1} \text{ m} $$ $$ X = \frac{0.5 + 1.5 + 0.5}{3} \text{ m} = \frac{2.5}{3} \text{ m} = \frac{5}{6} \text{ m} $$ $$ Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{(1)(0.5) + (1)(0.5) + (1)(1.5)}{1 + 1 + 1} \text{ m} $$ $$ Y = \frac{0.5 + 0.5 + 1.5}{3} \text{ m} = \frac{2.5}{3} \text{ m} = \frac{5}{6} \text{ m} $$

The centre of mass of the uniform L-shaped lamina is located at $\left(\frac{5}{6} \text{ m}, \frac{5}{6} \text{ m}\right)$.

If the three squares had different masses, say $m_1$, $m_2$, $m_3$, you would use those specific masses in the formulas for $X$ and $Y$ instead of 1 kg.

Motion Of Centre Of Mass

The motion of the centre of mass of a system of particles is particularly significant because it simplifies the description of the system's overall translational movement.

Starting with the position vector of the centre of mass $\mathbf{R} = \frac{\sum m_i \mathbf{r}_i}{M}$, we can find its velocity $\mathbf{V}$ by differentiating with respect to time (assuming masses are constant):

Thus, $M \mathbf{V} = \sum m_i \mathbf{v}_i$. The total mass times the velocity of the centre of mass is equal to the vector sum of the momenta of all individual particles in the system.

Differentiating again gives the acceleration $\mathbf{A}$ of the centre of mass:

So, $M \mathbf{A} = \sum m_i \mathbf{a}_i$.

According to Newton's Second Law for each particle, $\mathbf{F}_i = m_i \mathbf{a}_i$, where $\mathbf{F}_i$ is the net force on the $i$-th particle. Thus, $\sum m_i \mathbf{a}_i = \sum \mathbf{F}_i$.

The net force $\mathbf{F}_i$ on particle $i$ is the vector sum of all external forces $\mathbf{F}_i^{ext}$ acting on it and all internal forces $\mathbf{F}_i^{int}$ exerted by other particles in the system: $\mathbf{F}_i = \mathbf{F}_i^{ext} + \mathbf{F}_i^{int}$.

So, $M \mathbf{A} = \sum (\mathbf{F}_i^{ext} + \mathbf{F}_i^{int}) = \sum \mathbf{F}_i^{ext} + \sum \mathbf{F}_i^{int}$.

According to Newton's Third Law, internal forces between particles occur in equal and opposite pairs. When summed over the entire system, these internal forces cancel out: $\sum \mathbf{F}_i^{int} = \mathbf{0}$.

Therefore, $M \mathbf{A} = \sum \mathbf{F}_i^{ext}$. Let $\mathbf{F}_{ext} = \sum \mathbf{F}_i^{ext}$ be the total external force on the system.

This fundamental equation states that the centre of mass of a system of particles moves as if all the mass of the system were concentrated at the centre of mass, and all external forces were applied at that point.

Significance of the CM equation:

• The motion of the CM is determined solely by the total external force acting on the system. Internal forces, no matter how complex, do not affect the motion of the CM.
• We can describe the translational motion of an extended body (or any system of particles) by considering its entire mass located at the CM and applying the net external force there, just as we would for a single particle.

Example: When a projectile explodes in mid-air due to internal forces, its fragments fly in various directions. However, the centre of mass of the fragments continues to follow the original parabolic trajectory, as the external force (gravity) is unchanged by the explosion.

Linear Momentum Of A System Of Particles

The total linear momentum ($\mathbf{P}$) of a system of $n$ particles is the vector sum of the linear momenta of all individual particles:

From the relationship between CM velocity and individual particle velocities, $M \mathbf{V} = \sum m_i \mathbf{v}_i$, we see that the total linear momentum of the system is equal to the total mass times the velocity of its centre of mass:

Differentiating this with respect to time gives the relationship between the rate of change of total momentum and the acceleration of the CM:

Combining this with the CM equation of motion ($\mathbf{F}_{ext} = M \mathbf{A}$), we get Newton's Second Law for a system of particles:

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

Law of Conservation of Total Linear Momentum:

If the total external force acting on a system of particles is zero ($\mathbf{F}_{ext} = \mathbf{0}$), then $\frac{d\mathbf{P}}{dt} = \mathbf{0}$, which implies $\mathbf{P} = \text{Constant}$.

When the total external force acting on a system of particles is zero, the total linear momentum of the system remains constant (conserved).

Since $\mathbf{P} = M\mathbf{V}$, if $M$ is constant, conservation of momentum means that the velocity of the centre of mass $\mathbf{V}$ remains constant. The CM moves uniformly in a straight line.

Even if internal forces cause individual particles to move in complicated ways, the CM's motion is simple if external forces are zero.

Examples: Radioactive decay (nucleus splits, product particles move such that their CM continues the original motion), motion of binary stars (CM moves uniformly if no external forces, stars orbit around the stationary CM in the CM frame).

Vector Product Of Two Vectors

Besides the scalar product (dot product), there is another way to multiply vectors, which results in a vector quantity. This is called the vector product or cross product.

Definition Of Vector Product

The vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is denoted by $\mathbf{a} \times \mathbf{b}$ and is defined as a vector $\mathbf{c}$ with the following properties:

1. Magnitude: The magnitude of $\mathbf{c}$ is the product of the magnitudes of $\mathbf{a}$ and $\mathbf{b}$ and the sine of the angle $\theta$ between them: $$ |\mathbf{c}| = |\mathbf{a}| |\mathbf{b}| \sin \theta = ab \sin \theta $$ where $0^\circ \le \theta \le 180^\circ$. The magnitude is zero if $\mathbf{a}=\mathbf{0}$, $\mathbf{b}=\mathbf{0}$, or if $\mathbf{a}$ and $\mathbf{b}$ are parallel or antiparallel ($\theta=0^\circ$ or $180^\circ$).
2. Direction: The vector $\mathbf{c}$ is perpendicular to the plane containing both vectors $\mathbf{a}$ and $\mathbf{b}$. The specific direction is given by the right-hand rule or the right-hand screw rule:
   • Right-Hand Rule: If you point the fingers of your right hand in the direction of $\mathbf{a}$ and curl them towards the direction of $\mathbf{b}$ (through the smaller angle), your extended thumb points in the direction of $\mathbf{c} = \mathbf{a} \times \mathbf{b}$.
   • Right-Hand Screw Rule: If you place a right-handed screw perpendicular to the plane containing $\mathbf{a}$ and $\mathbf{b}$ and rotate its head from $\mathbf{a}$ to $\mathbf{b}$ (through the smaller angle), the direction the screw advances is the direction of $\mathbf{c} = \mathbf{a} \times \mathbf{b}$.

Properties of Vector Product:

• Not Commutative: The order of multiplication matters. $$ \mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a}) $$ They have the same magnitude and lie along the same line but point in opposite directions.
• Distributive Law: Vector product is distributive over vector addition. $$ \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} $$
• Scalar Multiplication: For a scalar $\lambda$: $$ \lambda (\mathbf{a} \times \mathbf{b}) = (\lambda \mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (\lambda \mathbf{b}) $$
• Parallel or Identical Vectors: The vector product of a vector with itself or a parallel vector is the null vector $\mathbf{0}$ (magnitude is 0): $$ \mathbf{a} \times \mathbf{a} = \mathbf{0} $$ This implies: $\hat{\mathbf{i}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}} \times \hat{\mathbf{k}} = \mathbf{0}$.
• Perpendicular Unit Vectors: Using the definition and right-hand rule: $$ \hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}} $$ $$ \hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}} $$ $$ \hat{\mathbf{k}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}} $$ And from non-commutativity: $\hat{\mathbf{j}} \times \hat{\mathbf{i}} = -\hat{\mathbf{k}}$, $\hat{\mathbf{k}} \times \hat{\mathbf{j}} = -\hat{\mathbf{i}}$, $\hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}}$.

Vector Product in Component Form:

If $\mathbf{a} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}} + a_z \hat{\mathbf{k}}$ and $\mathbf{b} = b_x \hat{\mathbf{i}} + b_y \hat{\mathbf{j}} + b_z \hat{\mathbf{k}}$, then $\mathbf{a} \times \mathbf{b}$ can be calculated using the distributive property and the unit vector cross products:

This can be conveniently remembered using a determinant:

Answer:

Angular Velocity And Its Relation With Linear Velocity

In rigid body rotation, different particles have different linear velocities, but they share the same angular velocity with respect to the axis of rotation.

For a rigid body rotating about a fixed axis, every particle moves in a circle perpendicular to the axis. The angular velocity $\omega$ describes how fast the body rotates.

The magnitude of angular velocity is $\omega = \frac{d\theta}{dt}$, the rate of change of angular displacement $\theta$.

Angular velocity is a vector quantity ($\mathbf{\omega}$). For rotation about a fixed axis, the angular velocity vector lies along the axis of rotation.

Its direction is given by the right-hand rule: if you curl the fingers of your right hand in the direction of rotation, your extended thumb points in the direction of $\mathbf{\omega}$. This is the direction a right-handed screw would advance if rotated with the body.

The linear velocity ($\mathbf{v}$) of a particle at a position $\mathbf{r}$ (relative to an origin on the axis of rotation) in a rigid body rotating with angular velocity $\mathbf{\omega}$ is given by the vector product:

Since $\mathbf{\omega}$ is along the axis and the component of $\mathbf{r}$ perpendicular to the axis is $r_\perp$, the magnitude of $\mathbf{v}$ is $|\mathbf{\omega} \times \mathbf{r}| = \omega r \sin \phi = \omega r_\perp$ (where $\phi$ is the angle between $\mathbf{\omega}$ and $\mathbf{r}$). $r_\perp$ is the radius of the circle the particle moves in. The direction of $\mathbf{\omega} \times \mathbf{r}$ is perpendicular to both $\mathbf{\omega}$ and $\mathbf{r}$, which is tangential to the circular path of the particle, as expected for linear velocity in circular motion.

This relation $\mathbf{v} = \mathbf{\omega} \times \mathbf{r}$ holds for rotation about a fixed axis (with origin on the axis) and for rotation about a fixed point (with origin at the fixed point).

Angular Acceleration

Analogous to linear acceleration, angular acceleration ($\mathbf{\alpha}$) is defined as the time rate of change of angular velocity:

For rotation about a fixed axis, the direction of $\mathbf{\omega}$ is fixed. So, $\mathbf{\alpha}$ is also along the axis of rotation. The vector equation simplifies to a scalar equation for the magnitudes: $\alpha = \frac{d\omega}{dt}$.

If $\omega$ is increasing, $\alpha$ is in the same direction as $\mathbf{\omega}$. If $\omega$ is decreasing, $\alpha$ is in the opposite direction to $\mathbf{\omega}$.

Torque And Angular Momentum

In rotational motion, torque is the analogue of force in linear motion, and angular momentum is the analogue of linear momentum.

Moment Of Force (Torque)

Moment of force, also called torque ($\mathbf{\tau}$), is a measure of how much a force acting on an object tends to cause that object to rotate about a pivot point or axis.

For a force $\mathbf{F}$ acting on a particle at a position vector $\mathbf{r}$ relative to a chosen origin O, the torque of this force with respect to O is defined as the vector product:

Torque is a vector quantity. Its direction is perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{F}$, given by the right-hand rule for vector products.

The magnitude of the torque is:

where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$.

Alternatively, the magnitude can be expressed as:

• $\tau = F (r \sin \theta) = F r_\perp$, where $r_\perp = r \sin \theta$ is the perpendicular distance from the origin to the line of action of the force (also called the lever arm or moment arm).
• $\tau = r (F \sin \theta) = r F_\perp$, where $F_\perp = F \sin \theta$ is the component of the force perpendicular to the position vector $\mathbf{r}$.

The SI unit of torque is Newton-metre (N m). Its dimensions are $[ML^2T^{-2}]$, same as work and energy, but it is a vector and physically distinct from work.

Torque is zero if $\mathbf{r}=\mathbf{0}$ (force acts at the origin), $\mathbf{F}=\mathbf{0}$ (no force), or if the line of action of $\mathbf{F}$ passes through the origin ($\theta=0^\circ$ or $180^\circ$).

Angular Momentum Of A Particle

Angular momentum ($\mathbf{l}$) of a particle is the rotational analogue of linear momentum.

For a particle of mass $m$ with linear momentum $\mathbf{p} = m\mathbf{v}$ at a position $\mathbf{r}$ relative to a chosen origin O, its angular momentum with respect to O is defined as the vector product:

Angular momentum is a vector quantity. Its direction is perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{p}$, given by the right-hand rule.

The magnitude of the angular momentum is:

where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$.

Alternatively, $l = r p_\perp = r_\perp p$, where $r_\perp = r \sin \theta$ is the perpendicular distance from the origin to the line of motion (direction of $\mathbf{p}$), and $p_\perp = p \sin \theta$ is the component of $\mathbf{p}$ perpendicular to $\mathbf{r}$.

Angular momentum is zero if $\mathbf{r}=\mathbf{0}$ (particle is at the origin), $\mathbf{p}=\mathbf{0}$ (particle is at rest), or if the line of motion of the particle passes through the origin ($\theta=0^\circ$ or $180^\circ$).

Relation between Torque and Angular Momentum (for a single particle):

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

This can be derived by differentiating $\mathbf{l} = \mathbf{r} \times \mathbf{p}$ with respect to time using the product rule for vector differentiation:

The first term is $\mathbf{v} \times \mathbf{p} = \mathbf{v} \times (m\mathbf{v}) = m (\mathbf{v} \times \mathbf{v}) = \mathbf{0}$ (since the cross product of parallel vectors is zero).

From Newton's Second Law for a particle, $\frac{d\mathbf{p}}{dt} = \mathbf{F}_{net}$. So the second term is $\mathbf{r} \times \mathbf{F}_{net}$, which is the definition of the net torque $\mathbf{\tau}_{net}$ on the particle.

Thus, $\frac{d\mathbf{l}}{dt} = \mathbf{\tau}_{net}$.

Angular Momentum and Torque for a System of Particles:

The total angular momentum ($\mathbf{L}$) of a system of particles about an origin is the vector sum of the angular momenta of individual particles:

The total torque ($\mathbf{\tau}$) on a system of particles about the same origin is the vector sum of the torques on individual particles:

The net force $\mathbf{F}_i$ on particle $i$ includes external forces $\mathbf{F}_i^{ext}$ and internal forces $\mathbf{F}_i^{int}$ from other particles.

Assuming Newton's Third Law holds and internal forces act along the line joining the particles, the total torque due to internal forces is zero: $\sum \mathbf{\tau}_i^{int} = \sum (\mathbf{r}_i \times \mathbf{F}_i^{int}) = \mathbf{0}$.

Thus, the total torque is equal to the sum of external torques: $\mathbf{\tau} = \mathbf{\tau}_{ext} = \sum \mathbf{\tau}_i^{ext} = \sum (\mathbf{r}_i \times \mathbf{F}_i^{ext})$.

The relationship between total external torque and total angular momentum for a system of particles is:

The time rate of change of the total angular momentum of a system of particles about a point is equal to the sum of the external torques acting on the system taken about the same point.

Conservation of Total Angular Momentum:

If the total external torque acting on a system of particles is zero ($\mathbf{\tau}_{ext} = \mathbf{0}$), then $\frac{d\mathbf{L}}{dt} = \mathbf{0}$, which implies $\mathbf{L} = \text{Constant}$.

When the total external torque acting on a system of particles is zero, the total angular momentum of the system is conserved (remains constant).

This means the magnitude and direction of the total angular momentum vector $\mathbf{L}$ are constant. This vector conservation is equivalent to the conservation of its three scalar components ($L_x, L_y, L_z$ are constant).

Examples illustrating angular momentum conservation often involve changes in moment of inertia (see Section 7.13.1).

Answer:

Answer:

Experiment with the Bicycle Rim: This classic demonstration illustrates the vector nature and conservation of angular momentum. A spinning bicycle wheel has significant angular momentum along its axle. When supported at only one end of the axle, gravity exerts a torque. Instead of falling (as it would if not spinning), the wheel's axle slowly precesses (moves sideways) around the support. The gravitational torque causes the direction of the angular momentum vector to change (precess), maintaining its magnitude if speed is constant. The rate and direction of precession are determined by the torque and the angular momentum, consistent with $\mathbf{\tau} = d\mathbf{L}/dt$.

Equilibrium Of A Rigid Body

A rigid body is in mechanical equilibrium if its state of motion (both translational and rotational) is unchanging. This means its linear momentum and angular momentum are constant, or equivalently, its linear acceleration and angular acceleration are zero.

The conditions for mechanical equilibrium of a rigid body are:

1. Translational Equilibrium: The total external force acting on the body is zero. $$ \sum \mathbf{F}_{ext} = \mathbf{0} $$ This ensures the total linear momentum $\mathbf{P}$ is constant ($\frac{d\mathbf{P}}{dt} = \mathbf{0}$). In component form, $\sum F_x = 0$, $\sum F_y = 0$, $\sum F_z = 0$.
2. Rotational Equilibrium: The total external torque acting on the body about any point is zero. $$ \sum \mathbf{\tau}_{ext} = \sum (\mathbf{r}_i \times \mathbf{F}_i^{ext}) = \mathbf{0} $$ This ensures the total angular momentum $\mathbf{L}$ is constant ($\frac{d\mathbf{L}}{dt} = \mathbf{0}$). In component form, $\sum \tau_x = 0$, $\sum \tau_y = 0$, $\sum \tau_z = 0$.

These six scalar conditions (three for force, three for torque) must be satisfied for complete mechanical equilibrium in 3D.

If all external forces are coplanar (lie in a single plane), say the x-y plane, then there are only three independent equilibrium conditions:

• $\sum F_x = 0$
• $\sum F_y = 0$
• $\sum \tau_z = 0$ (Sum of torques about any axis perpendicular to the plane, e.g., the z-axis).

Comparison with particle equilibrium: For a particle, only the translational equilibrium condition ($\sum \mathbf{F} = \mathbf{0}$) applies, as rotational motion of a point mass is not considered. For a particle, all forces must be concurrent for equilibrium.

A rigid body can be in partial equilibrium (e.g., translational but not rotational equilibrium, or vice versa).

A special case is a Couple: a pair of forces equal in magnitude, opposite in direction, and with different lines of action. The net force of a couple is zero ($\mathbf{F} + (-\mathbf{F}) = \mathbf{0}$), so it produces translational equilibrium. However, a couple produces a non-zero net torque, causing the body to undergo pure rotation (rotation without translation). Examples: turning a steering wheel, tightening a screw with a screwdriver, forces on a compass needle in a magnetic field.

Answer:

Consider a couple consisting of two forces $\mathbf{F}$ and $-\mathbf{F}$ acting at points A and B respectively on a rigid body. Let the position vector of point A relative to an arbitrary origin O be $\mathbf{r}_A$, and the position vector of point B relative to the same origin O be $\mathbf{r}_B$.

The torque (moment of force) of force $\mathbf{F}$ about origin O is $\mathbf{\tau}_B = \mathbf{r}_B \times \mathbf{F}$.

The torque of force $-\mathbf{F}$ about origin O is $\mathbf{\tau}_A = \mathbf{r}_A \times (-\mathbf{F}) = -\mathbf{r}_A \times \mathbf{F}$.

The total torque of the couple about the origin O is the vector sum of these individual torques:

$$ \mathbf{\tau}_{couple} = \mathbf{\tau}_B + \mathbf{\tau}_A = (\mathbf{r}_B \times \mathbf{F}) + (-\mathbf{r}_A \times \mathbf{F}) $$ $$ \mathbf{\tau}_{couple} = \mathbf{r}_B \times \mathbf{F} - \mathbf{r}_A \times \mathbf{F} $$

Using the distributive property of the cross product:

$$ \mathbf{\tau}_{couple} = (\mathbf{r}_B - \mathbf{r}_A) \times \mathbf{F} $$

The vector $(\mathbf{r}_B - \mathbf{r}_A)$ is the displacement vector from point A to point B, which we can denote as $\vec{AB}$.

$$ \mathbf{\tau}_{couple} = \vec{AB} \times \mathbf{F} $$

This expression for the torque of the couple depends only on the vector $\vec{AB}$ joining the points of application of the two forces and the vector $\mathbf{F}$. It does not contain the position vector $\mathbf{r}_A$ or $\mathbf{r}_B$ of the points relative to the origin O. Therefore, the torque of a couple is independent of the origin about which the moments are taken.

The moment of a couple is a vector perpendicular to the plane containing the two forces, with magnitude $|\vec{AB}| |\mathbf{F}| \sin \theta$, where $\theta$ is the angle between the vector $\vec{AB}$ and the force $\mathbf{F}$ (or $-\mathbf{F}$).

Principle Of Moments

The principle of moments is applied to objects like levers that are in rotational equilibrium.

An ideal lever is a massless rod pivoted at a fulcrum. Forces (load and effort) are applied at points along the lever.

For a lever to be in rotational equilibrium about the fulcrum, the sum of the moments of the forces about the fulcrum must be zero.

If forces $F_1$ and $F_2$ are applied perpendicular to the lever at distances $d_1$ and $d_2$ from the fulcrum, the condition for rotational equilibrium is:

If $F_1$ creates a clockwise moment and $F_2$ creates an anticlockwise moment about the fulcrum:

Principle of Moments: Load arm $\times$ Load = Effort arm $\times$ Effort. ($d_1$ is load arm, $d_2$ is effort arm).

The ratio of Load to Effort ($F_1/F_2$) is the Mechanical Advantage (M.A.) of the lever: M.A. $= F_1/F_2 = d_2/d_1$. A larger effort arm ($d_2 > d_1$) gives a mechanical advantage greater than 1, meaning a smaller effort can lift a larger load.

Centre Of Gravity

The Centre of Gravity (CG) of a body is the point where the total gravitational torque on the body is zero when it is placed in a gravitational field.

Consider an extended body made of particles $m_i$ at positions $\mathbf{r}_i$ relative to a point O. The gravitational force on particle $i$ is $m_i \mathbf{g}_i$. The total gravitational torque about O is $\mathbf{\tau}_g = \sum \mathbf{r}_i \times m_i \mathbf{g}_i$.

The CG is the point $G$ such that the total gravitational torque about $G$ is zero. If $\mathbf{r}'_i$ is the position vector of particle $i$ relative to $G$, then $\sum \mathbf{r}'_i \times m_i \mathbf{g}_i = \mathbf{0}$.

If the gravitational field $\mathbf{g}$ is uniform over the body (i.e., $\mathbf{g}_i = \mathbf{g}$ for all $i$), the condition becomes $\sum \mathbf{r}'_i \times m_i \mathbf{g} = (\sum m_i \mathbf{r}'_i) \times \mathbf{g} = \mathbf{0}$. Since $\mathbf{g} \ne \mathbf{0}$, this requires $\sum m_i \mathbf{r}'_i = \mathbf{0}$. This sum is zero if the origin is chosen at the centre of mass. Thus, in a uniform gravitational field, the Centre of Gravity coincides with the Centre of Mass.

CG and CM are different concepts: CM depends only on mass distribution, while CG depends on mass distribution and the gravitational field. They coincide in a uniform field (which is usually assumed for everyday objects near Earth's surface).

Finding the CG of an irregular shape: Suspend the object from different points. The vertical line through the suspension point will pass through the CG. The intersection of two or more such vertical lines gives the location of the CG.

Answer:

Given bar length $L = 70$ cm $= 0.70$ m. Mass of bar $M = 4.00$ kg. Knife-edges $K_1, K_2$ are 10 cm from each end, so $AK_1 = BK_2 = 10$ cm $= 0.10$ m. A load $W_1$ of mass $m_1 = 6.00$ kg is suspended at $P$, 30 cm from end A, so $AP = 30$ cm $= 0.30$ m. The bar is uniform, so its weight $W$ acts at its centre of gravity $G$, which is at the midpoint. $AG = GB = 70/2 = 35$ cm $= 0.35$ m. Weight of bar $W = Mg = 4.00g$ N. Weight of load $W_1 = m_1 g = 6.00g$ N. Reactions at knife-edges are $R_1$ and $R_2$. Assume $g = 9.8$ m/s$^2$. $W = 4.00 \times 9.8 = 39.2$ N. $W_1 = 6.00 \times 9.8 = 58.8$ N.

Calculate distances from points where forces/reactions act to the centre of gravity G:

• Distance of $R_1$ from G: $K_1G = AG - AK_1 = 0.35 \text{ m} - 0.10 \text{ m} = 0.25$ m.
• Distance of $W$ from G: 0 (W acts at G).
• Distance of $W_1$ from G: $PG = AG - AP = 0.35 \text{ m} - 0.30 \text{ m} = 0.05$ m.
• Distance of $R_2$ from G: $K_2G = BG - BK_2 = 0.35 \text{ m} - 0.10 \text{ m} = 0.25$ m.

The bar is in mechanical equilibrium, so the net force is zero and the net torque is zero.

Translational Equilibrium: Vertical forces must balance. Upward forces ($R_1, R_2$) balance downward forces ($W, W_1$).

$$ R_1 + R_2 - W - W_1 = 0 $$ $$ R_1 + R_2 = W + W_1 = 39.2 \text{ N} + 58.8 \text{ N} = 98.0 \text{ N} \quad (i) $$

Rotational Equilibrium: Net torque about any point must be zero. Let's take moments about point G (centre of gravity) for convenience (moment of W is zero). Take anticlockwise moments as positive, clockwise as negative.

• Moment of $R_1$ about G: $R_1 \times K_1G$ clockwise = $-R_1 (0.25)$.
• Moment of $W_1$ about G: $W_1 \times PG$ anticlockwise = $+W_1 (0.05) = 58.8 (0.05) = 2.94$ Nm.
• Moment of $R_2$ about G: $R_2 \times K_2G$ anticlockwise = $+R_2 (0.25)$.

Sum of moments about G = 0:

$$ -R_1 (0.25) + 2.94 + R_2 (0.25) = 0 $$ $$ 0.25 (R_2 - R_1) = -2.94 $$ $$ R_2 - R_1 = -\frac{2.94}{0.25} = -11.76 \text{ N} \quad (ii) $$

Now we have a system of two linear equations for $R_1$ and $R_2$:

(i) $R_1 + R_2 = 98.0$ (ii) $-R_1 + R_2 = -11.76$

Add (i) and (ii): $(R_1 + R_2) + (-R_1 + R_2) = 98.0 - 11.76 \implies 2R_2 = 86.24 \implies R_2 = 43.12$ N.

Substitute $R_2$ into (i): $R_1 + 43.12 = 98.0 \implies R_1 = 98.0 - 43.12 = 54.88$ N.

The reactions at the knife-edges are $R_1 \approx 54.9$ N (at 10 cm from end A) and $R_2 \approx 43.1$ N (at 10 cm from end B).

Example 7.9. A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.7.27. Find the reaction forces of the wall and the floor.

Answer:

Moment Of Inertia

In rotational motion about a fixed axis, the moment of inertia (I) is the rotational analogue of mass in linear motion. It quantifies a body's resistance to changes in its rotational motion (angular acceleration).

For a system of $n$ particles rotating about a fixed axis, the total kinetic energy of rotation is the sum of the kinetic energies of the individual particles. A particle $i$ of mass $m_i$ at a perpendicular distance $r_i$ from the axis has linear speed $v_i = r_i \omega$, where $\omega$ is the angular velocity of the body.

The kinetic energy of particle $i$ is $K_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} m_i r_i^2 \omega^2$.

The total kinetic energy of rotation is $K = \sum_{i=1}^n K_i = \sum_{i=1}^n \frac{1}{2} m_i r_i^2 \omega^2$.

Since $\omega$ is the same for all particles, $K = \frac{1}{2} (\sum_{i=1}^n m_i r_i^2) \omega^2$.

The term in the parenthesis is defined as the moment of inertia (I) of the rigid body about the given axis of rotation:

For a continuous rigid body, the summation becomes an integral: $I = \int r^2 \, dm$, where $r$ is the perpendicular distance of the mass element $dm$ from the axis.

Using this definition, the kinetic energy of rotation is:

This formula is analogous to $K = \frac{1}{2} m v^2$ in linear motion, with $I$ corresponding to $m$ and $\omega$ corresponding to $v$.

Factors affecting Moment of Inertia:

• Total mass ($M$).
• Shape and size of the body.
• Distribution of mass relative to the axis of rotation.
• Position and orientation of the axis of rotation.

Unlike mass, the moment of inertia is not a fixed property of a body; it changes with the axis of rotation chosen.

Examples of Moment of Inertia for simple homogeneous bodies (about specific axes):

Body	Axis	Figure	Moment of Inertia (I)
Thin circular ring, radius R	Perpendicular to plane, at centre		$MR^2$
Thin circular ring, radius R	Diameter		$MR^2/2$
Thin rod, length L	Perpendicular to rod, at midpoint		$ML^2/12$
Circular disc, radius R	Perpendicular to disc at centre		$MR^2/2$
Circular disc, radius R	Diameter		$MR^2/4$
Hollow cylinder, radius R	Axis of cylinder		$MR^2$
Solid cylinder, radius R	Axis of cylinder		$MR^2/2$
Solid sphere, radius R	Diameter		$2MR^2/5$

Radius of Gyration (k):

The radius of gyration is a concept used to relate the moment of inertia of a complex body to that of a simple point mass. It is defined such that the moment of inertia $I$ of a body of total mass $M$ about a given axis can be written as:

where $k$ is the radius of gyration. It has dimensions of length, $k = \sqrt{I/M}$.

The radius of gyration represents the distance from the axis at which a single point mass equal to the total mass of the body would need to be placed to have the same moment of inertia as the actual body.

Dimensions of Moment of Inertia: $[ML^2]$. SI Unit: kg m$^2$.

Flywheels are practical applications of high moment of inertia; they resist changes in angular speed, providing smooth rotation.

Theorems Of Perpendicular And Parallel Axes

These theorems provide convenient ways to calculate the moment of inertia about certain axes if the moment of inertia about other related axes is known.

Theorem of Perpendicular Axes:

• This theorem applies only to planar bodies (laminae) – flat bodies with negligible thickness.
• 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 mutually perpendicular axes lying in the plane ($I_x$ and $I_y$), provided all three axes are concurrent (pass through the same point O). $$ I_z = I_x + I_y $$

Answer:

Consider a uniform circular disc of mass $M$ and radius $R$. We know its moment of inertia about an axis perpendicular to the disc and passing through its centre is $I_z = \frac{1}{2}MR^2$ (from Table 7.1).

A disc is a planar body, so the theorem of perpendicular axes is applicable. Choose the centre of the disc as the origin O. Let the z-axis be perpendicular to the plane of the disc through O. Let the x and y axes be two mutually perpendicular diameters of the disc lying in its plane, also passing through O.

According to the theorem of perpendicular axes:

$$ I_z = I_x + I_y $$

Since the disc is uniform and circular, its moment of inertia is the same about any diameter. Therefore, $I_x = I_y$.

Substitute this into the theorem:

$$ I_z = I_x + I_x = 2I_x $$

We are given $I_z = \frac{1}{2}MR^2$.

$$ \frac{1}{2}MR^2 = 2I_x $$ $$ I_x = \frac{1}{2} \left(\frac{1}{2}MR^2\right) = \frac{1}{4}MR^2 $$

Thus, the moment of inertia of a uniform disc about any of its diameters is $\frac{1}{4}MR^2$.

Applying similarly for a ring: A ring's $I_z$ (perpendicular to plane through center) is $MR^2$. For diameters $I_x=I_y$, so $I_z = I_x + I_y = 2I_x \implies MR^2 = 2I_x \implies I_x = MR^2/2$. The moment of inertia of a uniform ring about any of its diameters is $MR^2/2$.

The theorem of perpendicular axes is applicable only to planar bodies (laminae). A solid cylinder is a three-dimensional body, so this theorem is not applicable to a solid cylinder.

Theorem Of Parallel Axes

• This theorem applies to a body of any shape (planar or 3D).
• It relates the moment of inertia about an axis to the moment of inertia about a parallel axis passing through the centre of mass (CM).
• It states that the moment of inertia ($I_{z'}$) of a body about any axis (z') is equal to the sum of its moment of inertia about a parallel axis ($I_z$) passing through its centre of mass (CM) and the product of the total mass ($M$) of the body and the square of the perpendicular distance ($a$) between the two parallel axes. $$ I_{z'} = I_z + Ma^2 $$ where z is the axis through CM and z' is the parallel axis.

Answer:

Consider a thin rod of mass $M$ and length $l$. We want to find its moment of inertia about an axis perpendicular to the rod and passing through one end. We know the moment of inertia of a thin rod about a parallel axis perpendicular to the rod and passing through its centre of mass (midpoint) is $I_{CM} = \frac{1}{12}Ml^2$ (from Table 7.1).

The axis through one end is parallel to the axis through the CM (midpoint). The perpendicular distance between these two parallel axes is half the length of the rod, $a = l/2$.

According to the theorem of parallel axes, the moment of inertia about the axis through one end ($I_{end}$) is:

$$ I_{end} = I_{CM} + Ma^2 $$ $$ I_{end} = \frac{1}{12}Ml^2 + M\left(\frac{l}{2}\right)^2 $$ $$ I_{end} = \frac{1}{12}Ml^2 + M\frac{l^2}{4} $$

Find a common denominator:

$$ I_{end} = \frac{1}{12}Ml^2 + \frac{3}{12}Ml^2 = \frac{1+3}{12}Ml^2 = \frac{4}{12}Ml^2 = \frac{1}{3}Ml^2 $$

Thus, the moment of inertia of a thin rod about an axis perpendicular to it through one end is $\frac{1}{3}Ml^2$.

Answer:

Consider a uniform ring of mass $M$ and radius $R$. We want to find its moment of inertia about a tangent to the circle of the ring. A tangent lies in the plane of the ring.

We know the moment of inertia of a uniform ring about a parallel axis passing through its centre of mass (the centre of the ring) and lying in the plane of the ring. This axis is a diameter of the ring. From Table 7.1, the moment of inertia about a diameter is $I_{diameter} = \frac{1}{2}MR^2$.

The tangent to the ring is an axis parallel to a diameter. The perpendicular distance between the diameter axis (passing through the CM) and the tangent axis is the radius of the ring, $a = R$.

According to the theorem of parallel axes, the moment of inertia about the tangent axis ($I_{tangent}$) is:

$$ I_{tangent} = I_{diameter} + Ma^2 $$ $$ I_{tangent} = \frac{1}{2}MR^2 + M(R)^2 $$ $$ I_{tangent} = \frac{1}{2}MR^2 + MR^2 = \frac{1+2}{2}MR^2 = \frac{3}{2}MR^2 $$

Thus, the moment of inertia of a uniform ring about a tangent in the plane of the ring is $\frac{3}{2}MR^2$.

Kinematics Of Rotational Motion About A Fixed Axis

Kinematics describes motion using quantities like displacement, velocity, and acceleration, without considering the forces involved. For rotation about a fixed axis, there is a direct analogy with linear motion.

Analogous Kinematic Quantities:

• Linear displacement ($x$) $\leftrightarrow$ Angular displacement ($\theta$)
• Linear velocity ($v = dx/dt$) $\leftrightarrow$ Angular velocity ($\omega = d\theta/dt$)
• Linear acceleration ($a = dv/dt$) $\leftrightarrow$ Angular acceleration ($\alpha = d\omega/dt$)

For rotation about a fixed axis, we can treat $\theta$, $\omega$, and $\alpha$ as scalar quantities, with positive/negative signs indicating the sense of rotation (e.g., anticlockwise positive, clockwise negative).

For rotational motion with uniform (constant) angular acceleration ($\alpha$), the kinematic equations are directly analogous to those for linear motion with constant linear acceleration:

1. Angular velocity-time relation: $$ \omega = \omega_0 + \alpha t $$ (Analogous to $v = v_0 + at$)
2. Angular position-time relation: $$ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 $$ (Analogous to $x = x_0 + v_0 t + \frac{1}{2} a t^2$)
3. Angular velocity-displacement relation: $$ \omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0) $$ (Analogous to $v^2 = v_0^2 + 2a (x - x_0)$)

Here, $\theta_0$ and $\omega_0$ are the initial angular position and angular velocity at time $t=0$. $\theta$ and $\omega$ are the angular position and angular velocity at time $t$.

Answer:

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Dynamics Of Rotational Motion About A Fixed Axis

Dynamics relates forces (or torques) to the resulting motion (linear or angular acceleration). For rotation about a fixed axis, we seek relationships analogous to Newton's laws of linear motion.

For rotation about a fixed axis, only the component of torque parallel to the axis causes rotation about that axis. Other components tend to change the axis orientation, which is prevented by constraints.

When calculating torque for rotation about a fixed axis:

• Consider only forces in planes perpendicular to the axis.
• Consider only components of position vectors perpendicular to the axis.

Work Done by a Torque:

Consider a force $\mathbf{F}$ acting on a particle at a perpendicular distance $r_\perp$ from the fixed axis. Over a small angular displacement $d\theta$, the particle's linear displacement is $d\mathbf{s}$ with magnitude $ds = r_\perp d\theta$ and direction tangential to the circle.

The work done by the force is $dW = \mathbf{F} \cdot d\mathbf{s} = F \cos \phi \, ds$, where $\phi$ is the angle between $\mathbf{F}$ and $d\mathbf{s}$. The component of force tangential to the circle is $F_{tan} = F \cos \phi$. So, $dW = F_{tan} ds = F_{tan} (r_\perp d\theta)$.

The magnitude of the torque due to $\mathbf{F}$ about the axis is $\tau = r_\perp F_{tan}$ (only considering the component of force perpendicular to $r_\perp$ in the plane perpendicular to the axis). So, $F_{tan} = \tau / r_\perp$.

Substitute this into the work equation: $dW = (\tau / r_\perp) (r_\perp d\theta) = \tau d\theta$.

For a rigid body, the total work done by the net external torque $\mathbf{\tau}_{ext}$ during an infinitesimal angular displacement $d\theta$ (along the axis) is:

This is analogous to $dW = F_{ext} \, dx$ for linear motion.

Power delivered by a Torque:

Instantaneous power $P = \frac{dW}{dt} = \tau_{ext} \frac{d\theta}{dt}$.

This is analogous to $P = \mathbf{F}_{ext} \cdot \mathbf{v}$ for linear motion.

Newton's Second Law for Rotation (about a fixed axis):

The rate at which work is done by the torque increases the kinetic energy of rotation: $P = \frac{dK}{dt}$.

$K = \frac{1}{2}I\omega^2$. Assuming $I$ is constant (rigid body, fixed axis), $\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}I\omega^2\right) = \frac{1}{2}I \frac{d(\omega^2)}{dt} = \frac{1}{2}I (2\omega) \frac{d\omega}{dt} = I\omega \alpha$.

Equating power and rate of change of kinetic energy:

For $\omega \ne 0$, we can divide by $\omega$:

The net external torque about the fixed axis is equal to the moment of inertia about that axis times the angular acceleration about that axis. This is the rotational analogue of $\mathbf{F}_{ext} = m\mathbf{a}$.

Comparison of Translational and Rotational Motion:

Example 7.15. A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 7.35. The flywheel is mounted on a horizontal axle with frictionless bearings. (a) Compute the angular acceleration of the wheel. (b) Find the work done by the pull, when 2m of the cord is unwound. (c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest. (d) Compare answers to parts (b) and (c).

Answer:

Angular Momentum In Case Of Rotation About A Fixed Axis

We extend the concept of angular momentum from a single particle to a system of particles, specifically for a rigid body rotating about a fixed axis (say, the z-axis).

The total angular momentum $\mathbf{L}$ of a system of particles is the vector sum $\mathbf{L} = \sum \mathbf{l}_i = \sum (\mathbf{r}_i \times \mathbf{p}_i)$.

For a particle $i$ in a rigid body rotating about the z-axis, its position vector $\mathbf{r}_i$ can be written relative to the origin as $\mathbf{r}_i = \mathbf{OC}_i + \mathbf{r}_{\perp i}$, where $\mathbf{OC}_i$ is the vector from the origin to the center of the particle's circle on the z-axis, and $\mathbf{r}_{\perp i}$ is the position vector from $C_i$ to the particle (perpendicular to the axis). The velocity is $\mathbf{v}_i = \mathbf{\omega} \times \mathbf{r}_i = \mathbf{\omega} \times \mathbf{r}_{\perp i}$ (since $\mathbf{\omega}$ is parallel to $\mathbf{OC}_i$).

The angular momentum of particle $i$ is $\mathbf{l}_i = \mathbf{r}_i \times (m_i \mathbf{v}_i) = (\mathbf{OC}_i + \mathbf{r}_{\perp i}) \times (m_i \mathbf{\omega} \times \mathbf{r}_{\perp i})$.

This expression is complex. In general, for a particle in a rotating rigid body, its angular momentum $\mathbf{l}_i$ about an origin on the axis of rotation is not parallel to the angular velocity vector $\mathbf{\omega}$.

However, the component of angular momentum along the fixed axis (z-axis) is simpler. The z-component of $\mathbf{l}_i$ is $l_{iz} = m_i r_{\perp i}^2 \omega$. Summing over all particles, the total z-component of angular momentum is:

The term in parenthesis is the moment of inertia $I$ about the z-axis.

For a rigid body rotating about a fixed axis, the component of the total angular momentum along the axis of rotation is equal to the product of the moment of inertia about that axis and the angular velocity.

For bodies that are symmetric about the axis of rotation (e.g., uniform cylinder or disc rotating about its central axis), the component of $\mathbf{L}$ perpendicular to the axis is zero ($\mathbf{L}_{\perp} = \mathbf{0}$). In this case, the total angular momentum vector $\mathbf{L}$ is parallel to the angular velocity vector $\mathbf{\omega}$ and points along the axis:

The magnitude is $L = I\omega$.

For bodies not symmetric about the axis, $\mathbf{L}$ is not parallel to $\mathbf{\omega}$ (e.g., rotating a rectangular box about an axis not passing through its center or along a symmetry axis).

From the relation $\mathbf{\tau}_{ext} = d\mathbf{L}/dt$, for a fixed z-axis, the z-component of the external torque is $\tau_{ext, z} = \frac{dL_z}{dt}$.

Since $L_z = I\omega$, if the moment of inertia $I$ about the fixed axis is constant, $\tau_{ext, z} = \frac{d(I\omega)}{dt} = I \frac{d\omega}{dt} = I\alpha$. This confirms $\tau = I\alpha$ (where $\tau$ and $\alpha$ are components along the axis).

Also, for a fixed axis, the component of angular momentum perpendicular to the axis ($\mathbf{L}_{\perp}$) is constant, $\frac{d\mathbf{L}_{\perp}}{dt} = \mathbf{\tau}_{\perp, ext}$. If $\mathbf{\tau}_{\perp, ext} = \mathbf{0}$, then $\mathbf{L}_{\perp}$ is constant. If the body is symmetric, $\mathbf{L}_{\perp}=\mathbf{0}$ always.

Conservation Of Angular Momentum

The principle of conservation of angular momentum states that if the total external torque on a system is zero, its total angular momentum is conserved ($\mathbf{L} = \text{Constant}$).

For a rigid body rotating about a fixed axis, if the component of the total external torque along the axis is zero ($\tau_{ext, z} = 0$), then the component of angular momentum along the axis is conserved:

If the external torque about the fixed axis of rotation is zero, the product of the moment of inertia about that axis and the angular velocity is conserved.

This principle is demonstrated in various situations where the moment of inertia $I$ changes, causing a corresponding change in angular velocity $\omega$ to keep the product $I\omega$ constant.

Examples:

• A figure skater or dancer performing a pirouette increases their angular speed by bringing their arms and legs closer to their body, thus decreasing their moment of inertia about the vertical axis. ($I \downarrow \implies \omega \uparrow$)
• A diver or acrobat increases their rotation speed by tucking their body (decreasing $I$).
• A person on a swivel chair rotating with weights in their outstretched arms will slow down when they extend their arms (increasing $I$) and speed up when they bring them in (decreasing $I$).

While $I\omega$ is conserved, the kinetic energy of rotation $K = \frac{1}{2}I\omega^2$ is not conserved when $I$ changes (unless work is done by internal forces to change the configuration). $K = \frac{1}{2} (I\omega) \omega = \frac{1}{2} L_z \omega$. Since $L_z$ is constant, changes in $K$ are proportional to changes in $\omega$. If $I$ decreases, $\omega$ increases, and $K$ increases. This extra kinetic energy comes from the work done by the person's muscles in changing their body configuration (e.g., pulling arms in).

Rolling Motion

Rolling motion is a common type of motion that combines both translation and rotation. A body rolls when its points of contact with the surface are momentarily at rest.

Consider a rigid body (like a wheel, cylinder, or sphere) rolling on a surface without slipping. This means that at any instant, the point of the body in contact with the surface has zero instantaneous velocity relative to the surface.

Let $v_{cm}$ be the velocity of the centre of mass (which is the translational velocity of the body) and $\omega$ be the angular velocity of rotation about the axis passing through the CM (for a symmetric body rolling on a flat surface, the axis of rotation is fixed relative to the body and passes through the CM).

The linear velocity of any point on the body is the vector sum of the translational velocity $v_{cm}$ and the linear velocity due to rotation $v_r = r\omega$, where $r$ is the distance from the CM and $v_r$ is perpendicular to $\mathbf{r}$ relative to CM.

For the point of contact $P_0$ on a disc of radius $R$, the velocity due to rotation relative to CM is $R\omega$ directed backward (opposite to $v_{cm}$). The instantaneous velocity of $P_0$ is $v_{cm} - R\omega$. The condition for rolling without slipping (point of contact is at rest relative to the surface) is:

This is the condition for pure rolling (without slipping) for a symmetric body rolling on a flat surface.

Kinetic Energy Of Rolling Motion

The total kinetic energy ($K$) of a rolling body can be expressed as the sum of its translational kinetic energy and its rotational kinetic energy about the CM:

where $m$ is the total mass, $v_{cm}$ is the speed of the centre of mass, and $I_{cm}$ is the moment of inertia about the axis passing through the CM (for rolling on a flat surface, this is the axis of rotation).

Using the condition for rolling without slipping, $v_{cm} = R\omega$, so $\omega = v_{cm}/R$. Substitute this into the KE equation:

Using the radius of gyration $I_{cm} = mk^2$, where $k$ is the radius of gyration about the CM axis:

The ratio $k^2/R^2$ depends on the shape and mass distribution of the body relative to its radius.

• Ring or Hollow Cylinder: $I_{cm} = MR^2$, so $k^2 = R^2$, $k^2/R^2 = 1$. $K = \frac{1}{2}mv_{cm}^2 (1+1) = mv_{cm}^2$.
• Solid Cylinder or Disc: $I_{cm} = \frac{1}{2}MR^2$, so $k^2 = R^2/2$, $k^2/R^2 = 1/2$. $K = \frac{1}{2}mv_{cm}^2 (1+1/2) = \frac{3}{4}mv_{cm}^2$.
• Solid Sphere: $I_{cm} = \frac{2}{5}MR^2$, so $k^2 = 2R^2/5$, $k^2/R^2 = 2/5$. $K = \frac{1}{2}mv_{cm}^2 (1+2/5) = \frac{7}{10}mv_{cm}^2$.

Example 7.16. Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

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Question 7.8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

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Question 7.22. As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s$^2$)

(Hint: Consider the equilibrium of each side of the ladder separately.)

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Question 7.28. A disc rotating about its axis with angular speed $\omega_o$ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in the direction indicated ?

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