Motion of System of Particles and Rigid Body

The Centre of Mass (CM) of a system of particles is defined as a unique point where the entire mass of the system can be considered to be concentrated for the description of its translational motion under the action of external forces.

For a discrete system of N particles having masses $m_1, m_2, \dots, m_N$ and position vectors $\vec{r}_1, \vec{r}_2, \dots, \vec{r}_N$ relative to an arbitrary origin, the position vector of the centre of mass $\vec{R}_{\text{cm}}$ is defined by:

where $M = \sum_{i=1}^N m_i$ is the total mass of the system. In Cartesian coordinates:

For continuous extended bodies, the summation is replaced by integration over infinitesimal mass elements $dm$:

Key standard centre of mass locations for uniform symmetric bodies:

• Uniform thin rod of length $L$: At the midpoint ($x_{\text{cm}} = L/2$).
• Semicircular ring of radius $R$: On the axis of symmetry at distance $y_{\text{cm}} = \frac{2R}{\pi}$ from the center.
• Semicircular disc of radius $R$: On the axis of symmetry at distance $y_{\text{cm}} = \frac{4R}{3\pi}$ from the center.
• Hemispherical shell of radius $R$: At distance $y_{\text{cm}} = \frac{R}{2}$ from the center of the base.
• Solid hemisphere of radius $R$: At distance $y_{\text{cm}} = \frac{3R}{8}$ from the center of the base.
• Solid cone of height $h$: At distance $y_{\text{cm}} = \frac{h}{4}$ from the center of its circular base.

Differentiating the position vector with respect to time yields the velocity of the centre of mass:

Differentiating once more gives the acceleration of the centre of mass:

According to Newton's Third Law, internal mutual interaction forces between particles occur in equal and opposite collinear pairs and cancel out vectorially: $\sum \vec{F}_{\text{int}} = 0$. Therefore:

Law of Conservation of Linear Momentum of CM: If the net external force on a system is zero ($\sum \vec{F}_{\text{ext}} = 0$), then $\vec{A}_{\text{cm}} = 0$ and the velocity of the centre of mass remains strictly constant ($\vec{V}_{\text{cm}} = \text{constant}$). For example, when a projectile explodes mid-flight due to internal chemical forces, its fragments scatter in all directions, but its centre of mass continues along the identical parabolic trajectory as if no explosion had occurred.

The Centre of Mass is the unique mass-weighted mean position of all matter in a body. The Centre of Gravity is the point through which the resultant gravitational force (weight) acts on the body, about which the net torque due to gravity is zero. In a uniform gravitational field where $g$ is constant across the entire body, CM and CG coincide identically. In non-uniform gravitational fields (such as near supermassive astronomical bodies or in structures spanning kilometers), the centre of gravity lies slightly below the centre of mass because the lower sections experience a greater gravitational field.

The vector or cross product of two vectors $\vec{A}$ and $\vec{B}$ inclined at an angle $\theta$ ($0 \le \theta \le \pi$) is defined as a vector $\vec{C} = \vec{A} \times \vec{B}$ whose magnitude is:

and whose direction is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$, given by the right-hand thumb rule. Formally:

Core Mathematical Properties:

• Anti-commutative: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
• Collinear vectors: If $\vec{A} \parallel \vec{B}$ ($\theta = 0^\circ$ or $180^\circ$), then $\vec{A} \times \vec{B} = 0$. In particular, $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$.
• Orthogonal unit vectors: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$.
• Determinant Form: In rectangular Cartesian components:
$$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = \hat{i}(A_y B_z - A_z B_y) - \hat{j}(A_x B_z - A_z B_x) + \hat{k}(A_x B_y - A_y B_x)$$
• Geometrical Significance: The magnitude $|\vec{A} \times \vec{B}|$ represents the area of the parallelogram formed by $\vec{A}$ and $\vec{B}$ as adjacent sides. The area of a triangle formed by $\vec{A}$ and $\vec{B}$ is $\frac{1}{2}|\vec{A} \times \vec{B}|$.

Torque ($\vec{\tau}$) represents the turning or rotational effect of a force about a given pivot point or axis of rotation. For a force $\vec{F}$ acting at a point whose position vector is $\vec{r}$ relative to the origin:

The magnitude is $\tau = r F \sin\theta = F (r \sin\theta) = F \cdot r_\perp$, where $r_\perp = r\sin\theta$ is known as the lever arm or perpendicular distance of the line of action of the force from the axis of rotation. Dimensions: $[ML^2T^{-2}]$, SI unit: $\text{N}\cdot\text{m}$.

The Angular Momentum ($\vec{L}$) of a particle of mass $m$ moving with linear momentum $\vec{p} = m\vec{v}$ about a reference point is defined as:

Magnitude: $L = r p \sin\theta = m v r \sin\theta = p \cdot r_\perp$. Dimensions: $[ML^2T^{-1}]$, SI unit: $\text{kg}\cdot\text{m}^2/\text{s}$ or $\text{J}\cdot\text{s}$ (identical to Planck's constant $h$).

Differentiating $\vec{L}$ with respect to time $t$ using the product rule:

Since $\frac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$, the first term is $\vec{v} \times (m\vec{v}) = m(\vec{v} \times \vec{v}) = 0$. By Newton's Second Law, $\frac{d\vec{p}}{dt} = \vec{F}_{\text{net}}$. Hence:

This is the fundamental rotational analogue of Newton's Second Law: The time rate of change of total angular momentum of a particle equals the net external torque acting on it.

A rigid body is in complete mechanical equilibrium if both its linear momentum and angular momentum are constant in time. This imposes two independent vector conditions:

1. Translational Equilibrium: The vector sum of all external forces acting on the body must be zero:$$\sum \vec{F}_{\text{ext}} = 0 \iff \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$$This guarantees that the linear acceleration of the centre of mass is zero ($\vec{a}_{\text{cm}} = 0$).
2. Rotational Equilibrium: The vector sum of external torques acting on the body about any reference point must be zero:$$\sum \vec{\tau}_{\text{ext}} = 0 \iff \sum \tau_x = 0, \quad \sum \tau_y = 0, \quad \sum \tau_z = 0$$This guarantees that the angular acceleration of the body is zero ($\vec{\alpha} = 0$).

A couple consists of a pair of equal and opposite parallel forces whose lines of action do not coincide. Since the forces are equal in magnitude and opposite in direction, the net translational force of a couple is identically zero:

However, the torque produced by a couple is non-zero. If the forces act at points with position vectors $\vec{r}_1$ and $\vec{r}_2$, the net torque about an arbitrary origin $O$ is:

where $\vec{r}_{12} = \vec{r}_1 - \vec{r}_2$ is the relative displacement vector between the points of application. The magnitude of the torque is $\tau = F \cdot d$, where $d$ is the perpendicular distance between the lines of action of the two forces, known as the arm of the couple.

For a body pivoted at a fulcrum to remain in rotational equilibrium, the sum of clockwise moments must balance the sum of counter-clockwise moments about the fulcrum:

In a mechanical lever with an applied effort $E$ acting at distance $d_E$ (effort arm) and lifting a load $L$ acting at distance $d_L$ (load arm):

If the effort arm is longer than the load arm ($d_E > d_L$), the mechanical advantage is greater than 1, meaning a smaller effort can lift a substantially heavier load.

The Moment of Inertia ($I$) is the quantitative measure of rotational inertia of a rigid body about a specified axis of rotation. It represents the resistance of a body to changes in its rotational state, exactly analogous to inertial mass $M$ in translational motion.

For a system of particles of masses $m_i$ at perpendicular distances $r_i$ from the axis of rotation:

Dimensions: $[ML^2]$, SI unit: $\text{kg}\cdot\text{m}^2$. It is a tensor quantity, but for rotation about a fixed axis, it acts as a positive scalar.

Factors determining Moment of Inertia: (1) Total mass of the body, (2) Shape and size of the body, (3) Distribution of mass relative to the axis of rotation, and (4) Position and orientation of the axis of rotation. Note: $I$ is independent of angular velocity $\omega$ and applied torque $\tau$.

The radius of gyration ($k$) of a rigid body about a given axis is defined as the radial distance from the axis of rotation where, if the entire mass $M$ of the body were concentrated as a single point, the moment of inertia would remain unchanged:

For $n$ identical particles of mass $m$ distributed at distances $r_1, r_2, \dots, r_n$ from the axis:

Thus, the radius of gyration is the root-mean-square (RMS) distance of the constituent particles from the axis of rotation. Its SI unit is the metre ($\text{m}$).

1. Theorem of Parallel Axes (Steiner's Theorem): The moment of inertia of a rigid body about any arbitrary axis ($I$) equals the moment of inertia of the body about a parallel axis passing through its centre of mass ($I_{\text{cm}}$), plus the product of the total mass $M$ of the body and the square of the perpendicular distance $d$ between the two parallel axes:

Scope: Valid for bodies of ANY arbitrary shape and dimension (1D, 2D, or 3D).

2. Theorem of Perpendicular Axes: For any planar lamina (a two-dimensional flat body) lying in the $x\text{--}y$ plane, the moment of inertia about an axis $z$ perpendicular to its plane equals the sum of its moments of inertia about two mutually perpendicular axes $x$ and $y$ lying within its plane and intersecting at the point where the $z$-axis passes through:

Scope: Valid strictly for planar 2D laminas only; completely invalid for 3D solid bodies.

When a rigid body rotates about a fixed axis with constant angular acceleration $\alpha$, its kinematic motion is described by equations directly analogous to linear motion:

• $\omega = \omega_0 + \alpha t$ (analogous to $v = u + at$)
• $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ (analogous to $s = ut + \frac{1}{2}at^2$)
• $\omega^2 = \omega_0^2 + 2\alpha \theta$ (analogous to $v^2 = u^2 + 2as$)
• $\theta_n = \omega_0 + \frac{\alpha}{2}(2n - 1)$ (angular displacement in $n^{\text{th}}$ second)

Every linear dynamic variable possesses an exact rotational counterpart:

According to the rotational analogue of Newton's Second Law, $\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}$. If the net external torque acting on a rotating system is zero ($\sum \vec{\tau}_{\text{ext}} = 0$):

Statement: When no net external torque acts on a system, its total angular momentum remains strictly conserved in magnitude and direction.

Classic Physical Demonstrations:

• Ballet Dancer / Figure Skater: When a spinning skater draws their outstretched arms close to their chest, their mass shifts closer to the rotation axis, decreasing the radius of gyration $k$ and moment of inertia $I$. Because $I\omega = \text{constant}$, their angular spin velocity $\omega$ increases substantially. Their rotational kinetic energy $K = \frac{L^2}{2I}$ increases because internal muscular work is performed to pull the arms inward against centrifugal reaction.
• Springboard Diver: A diver tucks their arms and legs into a ball after launching to reduce $I$, completing multiple rapid somersaults, then stretches out just before hitting the water to increase $I$ and slow down rotation for a smooth vertical entry.
• Kepler's Second Law of Planetary Motion: The gravitational attraction of the Sun is a central force directed along the position vector $\vec{r}$, so $\vec{\tau} = \vec{r} \times \vec{F}_g = 0$. Consequently, the planet's orbital angular momentum is conserved: $L = m r v_\perp = \text{constant}$, implying the areal velocity $\frac{dA}{dt} = \frac{L}{2m}$ is strictly constant. When a planet approaches perihelion (closest distance $r_p$), its orbital velocity $v_p$ reaches its maximum.

Pure rolling motion is a simultaneous superposition of two distinct motions: (1) Pure translation of the centre of mass with linear velocity $v_{\text{cm}}$, and (2) Pure rotation about an axis through the centre of mass with angular velocity $\omega$.

For a round body of radius $R$ rolling without slipping on a stationary horizontal surface, the velocity of any point on the body at distance $r$ from the centre is given vectorially by $\vec{v} = \vec{v}_{\text{cm}} + (\vec{\omega} \times \vec{r})$.

• Point of contact ($P$): The rotational velocity is directed backward ($-R\omega$), while the translational velocity is directed forward ($v_{\text{cm}}$). For rolling without slipping, the relative velocity at the instantaneous contact point must be zero:$$v_P = v_{\text{cm}} - \omega R = 0 \implies v_{\text{cm}} = \omega R, \quad a_{\text{cm}} = \alpha R$$
• Topmost point ($T$): Both translational and rotational velocities point forward in the same direction: $v_T = v_{\text{cm}} + \omega R = 2v_{\text{cm}}$.
• Centre of mass ($C$): $v_C = v_{\text{cm}}$.

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:

Expressing $I_{\text{cm}} = M k^2$ and using $\omega = v_{\text{cm}}/R$:

The ratio of translational to total kinetic energy is $\frac{K_{\text{trans}}}{K_{\text{total}}} = \frac{1}{1 + k^2/R^2}$, and the ratio of rotational to total kinetic energy is $\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{k^2/R^2}{1 + k^2/R^2}$.

Consider a round body of mass $M$, radius $R$, and radius of gyration $k$ released from rest from the top of an inclined plane of inclination $\theta$ and vertical height $h$.

1. Linear Velocity at the Bottom: By conservation of mechanical energy (potential energy converted into total rolling kinetic energy without dissipative slipping loss):

2. Linear Acceleration Down the Incline: Differentiating $v^2 = 2as$ with distance along the incline $s = h/\sin\theta$ gives:

3. Static Friction Force Required for Rolling Without Slipping: The force of static friction acts up the incline to provide the clockwise torque $\tau = f_s R = I\alpha$ necessary for rotation:

4. Minimum Coefficient of Static Friction: To prevent slipping, $f_s \le \mu_s N = \mu_s M g \cos\theta$, which yields:

Since both acceleration $a$ and bottom velocity $v$ are inversely proportional to $(1 + k^2/R^2)$, the body with the smallest ratio of $k^2/R^2$ will experience the largest acceleration, reach the bottom in the shortest time, and achieve the highest final speed:

• Solid Sphere: $k^2/R^2 = 2/5 = 0.40 \implies a = \frac{5}{7}g\sin\theta \approx 0.714\,g\sin\theta$ (1st place, fastest)
• Solid Cylinder / Disc: $k^2/R^2 = 1/2 = 0.50 \implies a = \frac{2}{3}g\sin\theta \approx 0.667\,g\sin\theta$ (2nd place)
• Spherical Shell: $k^2/R^2 = 2/3 \approx 0.67 \implies a = \frac{3}{5}g\sin\theta = 0.600\,g\sin\theta$ (3rd place)
• Thin Circular Ring / Hoop: $k^2/R^2 = 1.00 \implies a = \frac{1}{2}g\sin\theta = 0.500\,g\sin\theta$ (4th place, slowest)