In classical mechanics, real-world objects are often modeled as systems of particles. While a single particle is idealized as a point mass with no spatial dimensions, real extended bodies have finite sizes and complex internal structures. To analyze their motion, we introduce the idealized concept of a rigid body .
The centre of mass is a unique point associated with a system of particles that behaves as if the entire mass of the system were concentrated there, and all external forces were applied directly at this point.
For a system of $n$ discrete particles, the position of the centre of mass is calculated as a mass-weighted average of the individual particle positions:
For continuous rigid bodies, the discrete summations are replaced by integrals over the volume of the body:
$$\vec{R} = \frac{1}{M} \int \vec{r} \\, dm$$By utilizing geometric symmetry, we can locate the centre of mass of homogeneous bodies (bodies with uniform mass density):
Mathematical Problem Statement: Find the centre of mass of three particles located at the vertices of an equilateral triangle of side length 0.5 m . The masses of the particles are 100 g , 150 g , and 200 g respectively.
Step-by-Step Algorithmic Solution Trace:
- Set the origin $(0,0)$ at the position of the 100 g mass ($m_1$).
- The 150 g mass ($m_2$) lies on the x-axis at $(0.5, 0)$.
- The 200 g mass ($m_3$) lies at the third vertex. Its coordinates are calculated using trigonometry:$$x_3 = 0.5 \times \cos(60^\circ) = 0.25 \text{ m}$$$$y_3 = 0.5 \times \sin(60^\circ) = 0.25\sqrt{3} \text{ m}$$
- Calculate the total mass: $M = 100 + 150 + 200 = 450 \text{ g}$.
- Compute the x-coordinate of the COM ($X$):$$X = \frac{100(0) + 150(0.5) + 200(0.25)}{450} = \frac{75 + 50}{450} = \frac{125}{450} = \frac{5}{18} \text{ m} \approx 0.28 \text{ m}$$
- Compute the y-coordinate of the COM ($Y$):$$Y = \frac{100(0) + 150(0) + 200(0.25\sqrt{3})}{450} = \frac{50\sqrt{3}}{450} = \frac{1}{3\sqrt{3}} \text{ m} \approx 0.19 \text{ m}$$
Final Evaluation Output: The coordinates of the centre of mass are $(\frac{5}{18} \text{ m}, \frac{1}{3\sqrt{3}} \text{ m})$.
The physical significance of the centre of mass becomes clear when we analyze its kinematics and dynamics under the influence of external forces.
Differentiating the position vector of the centre of mass with respect to time yields:
Multiplying the acceleration equation by the total mass $M$ gives:
$$M\vec{A} = \sum m_i \vec{a}_i = \sum \vec{F}_i$$According to Newton's third law, internal forces between the particles of the system occur in equal and opposite pairs and cancel out in the summation. Thus, only external forces contribute to the acceleration of the centre of mass:
$$M\vec{A} = \vec{F}_{ext}$$This fundamental result states that the centre of mass of a system of particles moves as if the entire mass of the system were concentrated at that point, and all external forces were applied directly to it.
The total linear momentum $\vec{P}$ of a system of particles is the vector sum of the individual momenta of its constituent particles:
$$\vec{P} = \sum m_i \vec{v}_i = M\vec{V}$$Differentiating this with respect to time yields the system-wide extension of Newton's second law:
$$\frac{d\vec{P}}{dt} = \vec{F}_{ext}$$If the net external force acting on the system is zero ($\vec{F}_{ext} = 0$), the total linear momentum remains constant:
$$\vec{P} = \text{constant} \implies \vec{V} = \text{constant}$$This is the law of conservation of linear momentum for a system of particles. It \implies that in the absence of external forces, the velocity of the centre of mass remains completely unaffected by internal interactions (such as radioactive decay or internal explosions).
Rotational dynamics relies heavily on the vector product to define directional rotational quantities such as torque and angular momentum.
The vector product (or cross product ) of two vectors $\vec{a}$ and $\vec{b}$ is a vector $\vec{c}$ defined as:
$$\vec{c} = \vec{a} \times \vec{b} = (ab \sin\theta) \hat{n}$$where $\theta$ is the angle between the two vectors ($0 \le \theta \le 180^\circ$) and $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, with its direction determined by the right-hand rule.For the standard Cartesian unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$:
$$\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$$$$\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}$$$$\hat{j} \times \hat{i} = -\hat{k}, \quad \hat{k} \times \hat{j} = -\hat{i}, \quad \hat{i} \times \hat{k} = -\hat{j}$$The cross product can be computed systematically using a $3 \times 3$ determinant:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = \hat{i}(a_y b_z - a_z b_y) - \hat{j}(a_x b_z - a_z b_x) + \hat{k}(a_x b_y - a_y b_x)$$Mathematical Problem Statement: Find the vector product $\vec{a} \times \vec{b}$ of the vectors $\vec{a} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} - 3\hat{k}$.
Step-by-Step Algorithmic Solution Trace:
- Set up the determinant:$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ -2 & 1 & -3 \end{vmatrix}$$
- Expand along the first row:$$\vec{a} \times \vec{b} = \hat{i}[(-4)(-3) - (5)(1)] - \hat{j}[(3)(-3) - (5)(-2)] + \hat{k}[(3)(1) - (-4)(-2)]$$
- Simplify the components:$$\text{x-component: } 12 - 5 = 7$$$$\text{y-component: } -[-9 - (-10)] = -[-9 + 10] = -1$$$$\text{z-component: } 3 - 8 = -5$$
Final Evaluation Output: $\vec{a} \times \vec{b} = 7\hat{i} - \hat{j} - 5\hat{k}$.
To describe the rotational kinematics of a rigid body, we define angular variables that apply to the body as a whole.
The linear velocity $\vec{v}$ of a particle at position vector $\vec{r}$ relative to an origin on the axis of rotation is given by the cross product:
$$\vec{v} = \vec{\omega} \times \vec{r}$$This vector is perpendicular to both $\vec{\omega}$ and $\vec{r}$, pointing along the tangent to the circular path of the particle.
The rate of change of angular velocity is defined as the angular acceleration :
$$\vec{\alpha} = \frac{d\vec{\omega}}{dt}$$For rotation about a fixed axis, the direction of $\vec{\omega}$ is constant, so the vector equation simplifies to the scalar relation:
$$\alpha = \frac{d\omega}{dt}$$These two vector quantities form the dynamic foundation of rotational mechanics, acting as the rotational analogues of force and linear momentum.
The rotational effectiveness of a force $\vec{F}$ acting on a particle at position $\vec{r}$ relative to an origin is defined as the torque $\vec{\tau}$:
$$\vec{\tau} = \vec{r} \times \vec{F}$$The rotational analogue of linear momentum for a particle with momentum $\vec{p}$ at position $\vec{r}$ is the angular momentum $\vec{l}$:
$$\vec{l} = \vec{r} \times \vec{p}$$For a system of particles, the total angular momentum is $\vec{L} = \sum \vec{l}_i$. Summing the individual torques and canceling internal torque contributions (assuming central internal forces) yields:
$$\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$$If the net external torque acting on the system is zero ($\vec{\tau}_{ext} = 0$), the total angular momentum is conserved:
$$\vec{L} = \text{constant}$$A rigid body is in mechanical equilibrium if its state of motion does not change over time.
A rigid body is in complete mechanical equilibrium if it satisfies two independent conditions:
An ideal lever is a light rod pivoted at a point called the fulcrum . When a load $F_1$ is placed at a distance $d_1$ (load arm) and an effort $F_2$ is applied at a distance $d_2$ (effort arm), rotational equilibrium about the fulcrum requires:
$$F_1 d_1 = F_2 d_2$$The mechanical advantage (M.A.) is defined as:
$$\text{M.A.} = \frac{F_1}{F_2} = \frac{d_2}{d_1}$$The rotational analogue of mass is the moment of inertia, which measures a body's resistance to rotational acceleration.
For a rigid body rotating about a fixed axis, the total kinetic energy of rotation is the sum of the kinetic energies of its individual particles:
$$K = \sum \frac{1}{2} m_i v_i^2 = \sum \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2} \left( \sum m_i r_i^2 \right) \omega^2$$We define the moment of inertia $I$ as:
$$I = \sum_{i=1}^{n} m_i r_i^2$$This allows us to write the rotational kinetic energy in a form analogous to translational kinetic energy ($K = \frac{1}{2} m v^2$):
$$K = \frac{1}{2} I \omega^2$$The radius of gyration $k$ of a body about a given axis is the radial distance from the axis to a point where the entire mass of the body could be concentrated to yield the same moment of inertia:
$$I = M k^2 \implies k = \sqrt{\frac{I}{M}}$$| Body Shape | Axis of Rotation | Moment of Inertia ($I$) |
|---|---|---|
| Thin Circular Ring (Radius $R$) | Perpendicular to plane, through center | $M R^2$ |
| Thin Circular Ring (Radius $R$) | Along any diameter | $\frac{1}{2} M R^2$ |
| Thin Uniform Rod (Length $L$) | Perpendicular to rod, through midpoint | $\frac{1}{12} M L^2$ |
| Circular Disc (Radius $R$) | Perpendicular to disc, through center | $\frac{1}{2} M R^2$ |
| Circular Disc (Radius $R$) | Along any diameter | $\frac{1}{4} M R^2$ |
| Solid Cylinder (Radius $R$) | Axis of the cylinder | $\frac{1}{2} M R^2$ |
| Hollow Cylinder (Radius $R$) | Axis of the cylinder | $M R^2$ |
| Solid Sphere (Radius $R$) | Along any diameter | $\frac{2}{5} M R^2$ |
Rotational motion about a fixed axis can be described using equations analogous to those of linear motion.
| Linear Motion (Constant $a$) | Rotational Motion (Constant $\alpha$) |
|---|---|
| $v = v_0 + at$ | $\omega = \omega_0 + \alpha t$ |
| $x = x_0 + v_0 t + \frac{1}{2} at^2$ | $\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$ |
| $v^2 = v_0^2 + 2a(x - x_0)$ | $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$ |
Mathematical Problem Statement: A flywheel of mass 20 kg and radius 20 cm is mounted on a frictionless horizontal axle. A steady pull of 25 N is applied to a cord wound around its rim. Find the angular acceleration of the flywheel.
Step-by-Step Algorithmic Solution Trace:
- Convert the radius to standard SI units: $R = 20 \text{ cm} = 0.20 \text{ m}$.
- Calculate the torque ($\tau$) exerted by the pull:$$\tau = F \times R = 25 \text{ N} \times 0.20 \text{ m} = 5.0 \text{ N m}$$
- Model the flywheel as a solid disc to find its moment of inertia ($I$):$$I = \frac{1}{2} M R^2 = \frac{1}{2} (20 \text{ kg}) (0.20 \text{ m})^2 = 10 \times 0.04 = 0.4 \text{ kg m}^2$$
- Apply Newton's second law for rotation to find the angular acceleration ($\alpha$):$$\alpha = \frac{\tau}{I} = \frac{5.0 \text{ N m}}{0.4 \text{ kg m}^2} = 12.5 \text{ rad/s}^2$$
Final Evaluation Output: The angular acceleration of the flywheel is 12.5 rad/sĀ² .
Historical Chronicle: Leonhard Euler and Rigid Body Dynamics