Centre of mass of particle systems and rigid bodies, basic concepts of rotational motion

Centre of Mass of a System of Particles

The centre of mass (CM) of a system of particles is a unique point where the weighted average position of all the particles in the system is located. It's as if the entire mass of the system is concentrated at this single point. This concept is crucial for understanding the motion of extended objects and systems of bodies.

For a system of 'n' particles with masses $m_1, m_2, \dots, m_n$ and position vectors $\mathbf{r}_1, \mathbf{r}_2, \dots, \mathbf{r}_n$ respectively, the position vector of the centre of mass, $\mathbf{R}_{CM}$, is given by:

$$ \mathbf{R}_{CM} = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 + \dots + m_n \mathbf{r}_n}{m_1 + m_2 + \dots + m_n} $$

This can be written more compactly using summation notation:

$$ \mathbf{R}_{CM} = \frac{\sum_{i=1}^{n} m_i \mathbf{r}_i}{\sum_{i=1}^{n} m_i} $$

If we consider the components in a Cartesian coordinate system (x, y, z), the position of the centre of mass ($X_{CM}, Y_{CM}, Z_{CM}$) can be calculated as:

$$ X_{CM} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} $$

$$ Y_{CM} = \frac{\sum_{i=1}^{n} m_i y_i}{\sum_{i=1}^{n} m_i} $$

$$ Z_{CM} = \frac{\sum_{i=1}^{n} m_i z_i}{\sum_{i=1}^{n} m_i} $$

The total mass of the system is $M = \sum_{i=1}^{n} m_i$. So, the formula simplifies to:

$$ \mathbf{R}_{CM} = \frac{1}{M} \sum_{i=1}^{n} m_i \mathbf{r}_i $$

Example: Centre of Mass of Two Particles

Consider two particles of masses $m_1$ and $m_2$ placed at positions $x_1$ and $x_2$ along the x-axis. The position of their centre of mass is:

$$ X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} $$

If $m_1 = 1$ kg at $x_1 = 0$ m and $m_2 = 2$ kg at $x_2 = 3$ m, then:

$$ X_{CM} = \frac{(1 \text{ kg})(0 \text{ m}) + (2 \text{ kg})(3 \text{ m})}{1 \text{ kg} + 2 \text{ kg}} = \frac{0 + 6}{3} \text{ m} = 2 \text{ m} $$

The centre of mass is located at 2 meters from the origin, closer to the heavier mass.

Centre of Mass of a Continuous Body

For a continuous body, we can imagine dividing it into infinitesimally small mass elements, $dm$. Each mass element can be treated as a point particle. The position vector of the centre of mass is then found by integrating over the entire body:

$$ \mathbf{R}_{CM} = \frac{\int \mathbf{r} \, dm}{\int dm} $$

The denominator, $\int dm$, is simply the total mass $M$ of the body. In Cartesian coordinates:

$$ X_{CM} = \frac{1}{M} \int x \, dm $$

$$ Y_{CM} = \frac{1}{M} \int y \, dm $$

$$ Z_{CM} = \frac{1}{M} \int z \, dm $$

To perform these integrals, we need to express $dm$ in terms of coordinates. This involves using the density of the body.

Density and Mass Element ($dm$)

Linear Density ($\lambda$): For a thin rod or wire, mass per unit length. $dm = \lambda \, dl$.

Surface Density ($\sigma$): For a thin sheet or surface, mass per unit area. $dm = \sigma \, dA$.

Volume Density ($\rho$): For a solid object, mass per unit volume. $dm = \rho \, dV$.

If the density is uniform, $\lambda, \sigma, \rho$ are constants. If not, they are functions of position.

Examples of CM for Continuous Bodies

1. Centre of Mass of a Uniform Rod of Length L

Let the rod lie along the x-axis from $x=0$ to $x=L$. The mass per unit length is $\lambda = M/L$ (uniform).

Consider a small element of length $dx$ at position $x$. Its mass is $dm = \lambda \, dx = (M/L) \, dx$.

$$ X_{CM} = \frac{1}{M} \int_0^L x \, dm = \frac{1}{M} \int_0^L x \left( \frac{M}{L} \right) dx $$

$$ X_{CM} = \frac{1}{L} \int_0^L x \, dx = \frac{1}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{1}{L} \left( \frac{L^2}{2} - 0 \right) = \frac{L}{2} $$

Since the rod is uniform and lies along the x-axis, $Y_{CM} = 0$ and $Z_{CM} = 0$. The CM is at the geometric centre of the rod.

2. Centre of Mass of a Uniform Semicircular Ring of Radius R

Let the ring lie in the xy-plane, with its centre at the origin and the semicircle in the upper half-plane ($y \ge 0$). The mass per unit length is $\lambda = M/( \pi R)$.

Using polar coordinates, $x = R \cos \theta$ and $y = R \sin \theta$. An element of arc length is $dl = R \, d\theta$.

The mass element is $dm = \lambda \, dl = \frac{M}{\pi R} (R \, d\theta) = \frac{M}{\pi} \, d\theta$.

$$ X_{CM} = \frac{1}{M} \int x \, dm = \frac{1}{M} \int_0^\pi (R \cos \theta) \left( \frac{M}{\pi} \, d\theta \right) $$

$$ X_{CM} = \frac{R}{\pi} \int_0^\pi \cos \theta \, d\theta = \frac{R}{\pi} [\sin \theta]_0^\pi = \frac{R}{\pi} (0 - 0) = 0 $$

$$ Y_{CM} = \frac{1}{M} \int y \, dm = \frac{1}{M} \int_0^\pi (R \sin \theta) \left( \frac{M}{\pi} \, d\theta \right) $$

$$ Y_{CM} = \frac{R}{\pi} \int_0^\pi \sin \theta \, d\theta = \frac{R}{\pi} [-\cos \theta]_0^\pi = \frac{R}{\pi} (-(-1) - (-1)) = \frac{R}{\pi} (1 + 1) = \frac{2R}{\pi} $$

The CM is at $(0, \frac{2R}{\pi})$.

3. Centre of Mass of a Uniform Solid Cone of Height H and Base Radius R

Consider the cone with its vertex at the origin and its axis along the z-axis. The base is at $z=H$.

It's easier to integrate using thin disks. Consider a disk of thickness $dz$ at height $z$. Let its radius be $r$. By similar triangles:

$$ \frac{r}{z} = \frac{R}{H} \implies r = \frac{Rz}{H} $$

The volume of this disk is $dV = \pi r^2 \, dz = \pi \left( \frac{Rz}{H} \right)^2 dz = \frac{\pi R^2}{H^2} z^2 \, dz$.

The mass of this disk is $dm = \rho \, dV = \rho \frac{\pi R^2}{H^2} z^2 \, dz$, where $\rho$ is the uniform volume density. The total mass $M = \frac{1}{3} \pi R^2 H \rho$.

The CM of a thin disk is at its centre. So, the CM of this disk is at $(0, 0, z)$. We only need to calculate $Z_{CM}$.

$$ Z_{CM} = \frac{1}{M} \int z \, dm = \frac{1}{M} \int_0^H z \left( \rho \frac{\pi R^2}{H^2} z^2 \right) dz $$

$$ Z_{CM} = \frac{\rho \pi R^2}{M H^2} \int_0^H z^3 \, dz = \frac{\rho \pi R^2}{M H^2} \left[ \frac{z^4}{4} \right]_0^H = \frac{\rho \pi R^2}{M H^2} \frac{H^4}{4} $$

Substitute $M = \frac{1}{3} \pi R^2 H \rho$:

$$ Z_{CM} = \frac{\rho \pi R^2 H^4}{4} \frac{3}{\rho \pi R^2 H} = \frac{3}{4} H $$

The CM of a uniform solid cone is at $\frac{3}{4}$ of its height from the vertex.

Motion of the Centre of Mass

The motion of the centre of mass of a system of particles is independent of the internal forces between the particles. It only depends on the external forces acting on the system.

Let the total external force on the system be $\mathbf{F}_{ext}$. The equation of motion for the centre of mass is:

$$ \mathbf{F}_{ext} = M \mathbf{a}_{CM} $$

where $M$ is the total mass of the system and $\mathbf{a}_{CM}$ is the acceleration of the centre of mass.

This is Newton's second law applied to the system as a whole, treating it as a single particle of mass $M$ located at the CM.

This implies that if no net external force acts on a system ($\mathbf{F}_{ext} = 0$), the centre of mass either remains at rest or moves with uniform velocity ($\mathbf{a}_{CM} = 0$). This is the principle of conservation of linear momentum for a system of particles.

Example: Explosion of a Projectile If a projectile explodes in mid-air into several fragments, the CM of all the fragments will continue to follow the original parabolic path of the projectile, provided no external forces (like air resistance) are considered. The individual fragments will move in complex ways due to the internal explosive forces, but their combined CM will behave as if the explosion never happened.

Basic Concepts of Rotational Motion

Rotational motion describes the movement of an object around a fixed axis or a point. Unlike translational motion where all particles move in the same direction, in rotational motion, particles move in circles around the axis of rotation.

1. Angular Displacement ($\theta$)

Angular displacement is the angle through which an object rotates. It is the change in angular position. It is usually measured in radians.

If a particle moves along an arc of length $s$ on a circle of radius $r$, its angular displacement $\theta$ (in radians) is given by:

$$ \theta = \frac{s}{r} $$

Angular displacement is a vector quantity. Its direction is given by the right-hand rule: if the fingers curl in the direction of rotation, the thumb points in the direction of the angular displacement vector.

2. Angular Velocity ($\omega$)

Angular velocity is the rate of change of angular displacement with respect to time.

Average angular velocity: $ \omega_{avg} = \frac{\Delta \theta}{\Delta t} $

Instantaneous angular velocity: $ \omega = \frac{d\theta}{dt} $

It is measured in radians per second (rad/s).

If an object completes $f$ revolutions per second, its frequency is $f$. One revolution is $2\pi$ radians. So, the angular velocity is:

$$ \omega = 2\pi f $$

The time period of rotation is $T = 1/f$. So,

$$ \omega = \frac{2\pi}{T} $$

Angular velocity is also a vector quantity, along the axis of rotation, following the right-hand rule.

3. Angular Acceleration ($\alpha$)

Angular acceleration is the rate of change of angular velocity with respect to time.

Average angular acceleration: $ \alpha_{avg} = \frac{\Delta \omega}{\Delta t} $

Instantaneous angular acceleration: $ \alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2} $

It is measured in radians per second squared (rad/s²).

Angular acceleration is also a vector quantity. If the angular velocity is changing, there is angular acceleration.

4. Relation between Linear and Angular Variables

For a particle moving in a circle of radius $r$:

Linear displacement $s = r\theta$

Linear velocity $v = \frac{ds}{dt} = r \frac{d\theta}{dt} = r\omega$. The magnitude of linear velocity is $v = r\omega$.

Linear acceleration $a = \frac{dv}{dt} = r \frac{d\omega}{dt} = r\alpha$. The magnitude of linear acceleration is $a = r\alpha$.

Note that the linear velocity is tangential to the circle, while angular velocity is along the axis of rotation.

5. Uniform Circular Motion

If an object moves in a circle with constant speed, it has uniform circular motion. In this case, the magnitude of angular velocity $\omega$ is constant, so the angular acceleration $\alpha = 0$.

However, the direction of the linear velocity vector is continuously changing, so there is a linear acceleration. This acceleration is directed towards the centre of the circle and is called centripetal acceleration ($a_c$).

$$ a_c = \frac{v^2}{r} = r\omega^2 $$

This acceleration is responsible for changing the direction of velocity, keeping the object moving in a circle.

6. Non-uniform Circular Motion

If the speed of the object changes, the angular velocity $\omega$ also changes, resulting in a non-zero angular acceleration $\alpha$.

In this case, the linear acceleration has two components:

• Tangential acceleration ($a_t$): Responsible for changing the speed. $a_t = r\alpha$. It is tangential to the circle.
• Centripetal acceleration ($a_c$): Responsible for changing the direction. $a_c = r\omega^2$. It is directed towards the centre.

The total linear acceleration is the vector sum of these two components: $\mathbf{a} = \mathbf{a}_t + \mathbf{a}_c$.

The magnitude of the total acceleration is $a = \sqrt{a_t^2 + a_c^2} = \sqrt{(r\alpha)^2 + (r\omega^2)^2}$.

7. Torque ($\tau$)

Torque is the rotational analogue of force. It is a measure of the turning effect of a force about an axis. A force applied at a distance from the axis of rotation produces a torque.

Torque is defined as the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force (lever arm).

$$ \tau = r F \sin \phi $$

where $r$ is the distance from the pivot to the point where the force is applied, $F$ is the magnitude of the force, and $\phi$ is the angle between the position vector $\mathbf{r}$ and the force vector $\mathbf{F}$.

Torque is also a vector quantity, given by the cross product:

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

The direction of torque is also given by the right-hand rule. Its unit is Newton-meter (N m).

8. Moment of Inertia ($I$)

Moment of inertia is the rotational analogue of mass. It is a measure of an object's resistance to changes in its rotational motion. A larger moment of inertia means more resistance to angular acceleration.

For a system of particles, the moment of inertia about an axis is the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis of rotation.

$$ I = \sum_{i=1}^{n} m_i r_i^2 $$

For a continuous body, it is calculated by integration:

$$ I = \int r^2 \, dm $$

where $r$ is the perpendicular distance of the mass element $dm$ from the axis of rotation.

The moment of inertia depends on:

• The mass of the object.
• The distribution of mass relative to the axis of rotation.
• The orientation of the axis of rotation.

Units of moment of inertia are kg m².

9. Newton's Second Law for Rotation

The rotational analogue of Newton's second law ($F=ma$) relates torque, moment of inertia, and angular acceleration:

$$ \tau_{net} = I \alpha $$

This equation states that the net external torque acting on a rigid body is equal to the product of its moment of inertia and its angular acceleration.

10. Angular Momentum ($L$)

Angular momentum is the rotational analogue of linear momentum. It is a measure of the quantity of rotation of an object.

For a single particle of mass $m$ moving with velocity $\mathbf{v}$ at a position $\mathbf{r}$ relative to the origin:

$$ \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times (m\mathbf{v}) $$

For a system of particles or a rigid body rotating about a fixed axis:

$$ L = I \omega $$

Angular momentum is a vector quantity. Its direction is along the axis of rotation.

11. Conservation of Angular Momentum

If the net external torque acting on a system is zero ($\tau_{net} = 0$), then the total angular momentum of the system remains constant.

$$ L_{initial} = L_{final} $$

$$ I_{initial} \omega_{initial} = I_{final} \omega_{final} $$

This principle is very important and has many applications, such as:

• A figure skater pulling their arms in to spin faster.
• A diver tucking to increase rotation speed.
• The formation of galaxies and planetary systems.

Centre of Mass of Rigid Bodies

A rigid body is an object where the distance between any two points on the body remains constant. For such bodies, the centre of mass is a fixed point relative to the body. The methods discussed for continuous bodies apply directly.

The location of the CM for symmetric, uniform rigid bodies is at their geometric centre. For asymmetric or non-uniform bodies, integration using density is required.

Understanding the CM of a rigid body is essential for analyzing its motion. When a rigid body is acted upon by external forces, its CM moves as if it were a single particle subject to the net external force, according to $\mathbf{F}_{ext} = M \mathbf{a}_{CM}$.

If the external forces are such that they cause rotation, then the net torque about the CM is related to the angular acceleration of the body by $\boldsymbol{\tau}_{CM} = I_{CM} \boldsymbol{\alpha}$, where $I_{CM}$ is the moment of inertia about an axis passing through the CM.