System of Particles and Rotational Motion - Comprehensive Notes

Fundamentals of Particle Systems and Rigid Bodies

• Point Object vs. Extended Object:

  • Point Object: An object having a certain mass but negligible physical size. Motion of physical bodies is initially analyzed by treating them as point objects.

  • Extended Object (Real Object): An object made up of a large number of particles where these constituent particles exert forces on each other.

• System of Particles: A collection of a large number of particles interacting with each other.

• Rigid Body:

  • Definition: A body with a perfectly definite and unchanging shape, where the distances between all pairs of constituent particles do not change.

  • Ideal vs. Real Situation: In real physical situations, no body is perfectly rigid. However, for practical applications, bodies such as wheels, steel beams, metallic spheres, wooden blocks, etc., undergo deformations under force that are so small that they are considered rigid bodies.

Types of Rigid Body Motion

• General Rigid Body Motion:

  • Unpivoted or Unfixed Rigid Body: A rigid body that is not pivoted or fixed in some way exhibits either pure translation or a combination of translation and rotation.

  • Pivoted or Fixed Rigid Body: A rigid body that is pivoted or fixed in some way exhibits rotational motion. The rotation may occur about an axis that is fixed or moving.

• Translational Motion:

  • Definition: Motion of a body where, at any instant of time, all particles of the body possess the exact same velocity.

  • Example: A wooden block sliding down an inclined plane.

• Rotational Motion (Rotation):

  • Definition: Motion of a body where every particle of the body moves in a circle lying in a plane perpendicular to a fixed axis, with its circle's center positioned on the fixed axis.

  • Example: A rotating ceiling fan.

  • Behavior of Axis Particles: Any particle lying directly on the axis of rotation remains at rest while the rigid body rotates around the axis of rotation. Thus, the axis of rotation is fixed.

• Precession:

  • Definition: The movement of the axis of a rotating body around a vertical axis.

  • Condition: During precession, the point of contact of the rotating body with the ground remains fixed.

Concept and Formulations of Centre of Mass

• Definition: The centre of mass of a system of particles is the unique point where the entire mass of the system can be assumed to be concentrated.

• Physical Significance:

  • This point-like mass exhibits the same type of translational motion as the system as a whole if some net external force acts on this point mass as acting on the system.

  • The centre of mass of a body or system functions as its balancing point.

• Two-Particle System:

  • Consider two particles lying on the X-axis with masses $m_1$ and $m_2$ at distances $x_1$ and $x_2$ from the origin, respectively.

  • Distance of the centre of mass $X_{cm}$ from the origin is:     $X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

  • Position vector of the centre of mass:     $\mathbf{R} = X_{cm}\hat{i}$

  • Special Case 1 (Equal Masses): If $m_1 = m_2 = m$:     $X_{cm} = \frac{m x_1 + m x_2}{m + m} = \frac{m(x_1 + x_2)}{2m} = \frac{x_1 + x_2}{2}$     Thus, for two particles of equal mass, the centre of mass lies exactly midway between them.

  • Special Case 2 (2D Non-Linear System): If the two particles do not lie on a straight line, define X and Y axes in the plane containing the particles with positions $(x_1, y_1)$ and $(x_2, y_2)$:     $X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$     $Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$     $\mathbf{R} = X_{cm}\hat{i} + Y_{cm}\hat{j}$

• Three-Particle System:

  • Consider three particles not lying in the same straight line with masses $m_1$, $m_2$, and $m_3$ at coordinates $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, respectively.

  • Coordinates of the centre of mass $(X_{cm}, Y_{cm})$:     $X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$     $Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}$     $\mathbf{R} = X_{cm}\hat{i} + Y_{cm}\hat{j}$

  • Special Case (Equal Masses): For particles of equal mass $m_1 = m_2 = m_3 = m$:     $X_{cm} = \frac{m x_1 + m x_2 + m x_3}{m + m + m} = \frac{m(x_1 + x_2 + x_3)}{3m} = \frac{x_1 + x_2 + x_3}{3}$     $Y_{cm} = \frac{m y_1 + m y_2 + m y_3}{m + m + m} = \frac{m(y_1 + y_2 + y_3)}{3m} = \frac{y_1 + y_2 + y_3}{3}$     Thus, for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.

• System of $n$-Particles:

  • Consider a system of $n$ particles with masses $m_1, m_2, \dots, m_n$ and coordinates $X_{cm}, Y_{cm}, Z_{cm}$ for total mass $M = \sum_{i=1}^{n} m_i$:     $X_{cm} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} = \frac{1}{M} \sum_{i=1}^{n} m_i x_i$     $Y_{cm} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i} = \frac{1}{M} \sum_{i=1}^{n} m_i y_i$     $Z_{cm} = \frac{\sum_{i=1}^{n} m_i z_i}{\sum_{i=1}^{n} m_i} = \frac{1}{M} \sum_{i=1}^{n} m_i z_i$

  • Position Vector of Centre of Mass $\mathbf{R}$:     $\mathbf{R} = X_{cm}\hat{i} + Y_{cm}\hat{j} + Z_{cm}\hat{k}$     $\mathbf{R} = \left(\frac{1}{M} \sum_{i=1}^{n} m_i x_i\right)\hat{i} + \left(\frac{1}{M} \sum_{i=1}^{n} m_i y_i\right)\hat{j} + \left(\frac{1}{M} \sum_{i=1}^{n} m_i z_i\right)\hat{k}$     $\mathbf{R} = \frac{1}{M} \sum_{i=1}^{n} m_i (x_i \hat{i} + y_i \hat{j} + z_i \hat{k}) = \frac{1}{M} \sum_{i=1}^{n} m_i \mathbf{r}_i$

• Location Property: The centre of mass of a system may or may not lie inside the physical boundaries of the system.

Centre of Mass for Continuous Distributions and Rigid Bodies

• Continuous Mass Distribution Formulation:

  • In a rigid body, the number of particles is so large that summation over individual discrete particles is impossible.

  • Because particle spacing is negligible, the body is treated as a continuous mass distribution.

  • Subdivide the body into $n$ small mass elements $dm_1, dm_2, \dots, dm_n$.

  • Summation is replaced by integration:     $\sum m_i \mathbf{r}_i = \int \mathbf{r} \, dm$

  • Vector position equation for a continuous body:     $\mathbf{R} = \frac{1}{M} \int \mathbf{r} \, dm$

  • Expressing position vectors as $\mathbf{R} = X_{cm}\hat{i} + Y_{cm}\hat{j} + Z_{cm}\hat{k}$ and $\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}$, comparing components yields:     $X_{cm} = \frac{1}{M} \int x \, dm$     $Y_{cm} = \frac{1}{M} \int y \, dm$     $Z_{cm} = \frac{1}{M} \int z \, dm$

• Origin Selected at Centre of Mass:

  • If the centre of mass is selected as the origin of the coordinate system, then $\mathbf{R} = 0$.

  • This condition implies:     $\int \mathbf{r} \, dm = 0$

  • Corresponding component integrals:     $\int x \, dm = 0$     $\int y \, dm = 0$     $\int z \, dm = 0$

• Geometric Conclusions for Special Bodies:

  • Homogeneous Bodies: For homogeneous bodies such as a circular solid disc, an ice cube, a sugar cube, a solid sphere, a hollow sphere, a marble ball, a billiard ball, an iron ball, or a uniform thin rod, the centre of mass coincides with the geometric centers of the bodies.

  • Symmetric Bodies: For bodies possessing an axis of symmetry such as a solid cylinder, a hollow cylinder, or a wheel, the centre of mass lies directly on the axis of symmetry.

Derivation for a Uniform Thin Rod

• System Setup:

  • Consider a uniform rod of length $L$.

  • Define end A of the rod as origin ($x = 0$) and end B as $x = L$.

  • Since the rod is uniform, mass per unit length (linear mass density) $\lambda$ is constant:     $\lambda = \frac{M}{L}$

• Element Selection:

  • Consider a small mass element of length $dx$ located at distance $x$ from end A.

  • Mass of element $dm$:     $dm = \lambda \, dx$

• Step-by-Step Mathematical Derivation:

  • Coordinate equation for the centre of mass:     $X_{cm} = \frac{1}{M} \int x \, dm$

  • Substitute $dm = \lambda \, dx$ with integration bounds from $0$ to $L$:     $X_{cm} = \frac{1}{M} \int_0^L x \lambda \, dx$

  • Factor out constant $\lambda$:     $X_{cm} = \frac{\lambda}{M} \int_0^L x \, dx$

  • Integration step:     $X_{cm} = \frac{\lambda}{M} \left[ \frac{x^2}{2} \right]_0^L = \frac{\lambda}{M} \left( \frac{L^2}{2} \right)$

  • Substitute $\lambda = \frac{M}{L}$:     $X_{cm} = \frac{\left(\frac{M}{L}\right)}{M} \left( \frac{L^2}{2} \right)$

  • Simplifying terms:     $X_{cm} = \frac{L}{2}$

  • Conclusion: The centre of mass of a uniform thin rod lies at its geometric midpoint $x = \frac{L}{2}$.

Motion and Kinematics of the Centre of Mass

• Position Relation:

  • For a system having $n$ particles of masses $m_1, m_2, m_3, \dots, m_n$ with position vectors $\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3, \dots, \mathbf{r}_n$:     $\mathbf{R} = \frac{\sum_{i=1}^{n} m_i \mathbf{r}_i}{M}$     $M \mathbf{R} = \sum_{i=1}^{n} m_i \mathbf{r}_i$     $M \mathbf{R} = m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3 + \dots + m_n \mathbf{r}_n$

• Velocity Relation:

  • Differentiating the position vector equation with respect to time $t$:     $M \frac{d\mathbf{R}}{dt} = m_1 \frac{d\mathbf{r}_1}{dt} + m_2 \frac{d\mathbf{r}_2}{dt} + m_3 \frac{d\mathbf{r}_3}{dt} + \dots + m_n \frac{d\mathbf{r}_n}{dt}$

  • Representing derivative terms as linear velocities $\mathbf{V}$ (centre of mass velocity) and $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \dots, \mathbf{v}_n$ (particle velocities):     $M \mathbf{V} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 + m_3 \mathbf{v}_3 + \dots + m_n \mathbf{v}_n$

• Acceleration Relation:

  • Differentiating the velocity equation with respect to time $t$ again:     $M \frac{d\mathbf{V}}{dt} = m_1 \frac{d\mathbf{v}_1}{dt} + m_2 \frac{d\mathbf{v}_2}{dt} + m_3 \frac{d\mathbf{v}_3}{dt} + \dots + m_n \frac{d\mathbf{v}_n}{dt}$

  • Representing derivative terms as accelerations $\mathbf{A}$ (centre of mass acceleration) and $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \dots, \mathbf{a}_n$ (particle accelerations):     $M \mathbf{A} = m_1 \mathbf{a}_1 + m_2 \mathbf{a}_2 + m_3 \mathbf{a}_3 + \dots + m_n \mathbf{a}_n$