Centre of Mass of Systems and Rigid Bodies

The concept of the centre of mass (COM) is fundamental in understanding the motion of extended objects and systems of particles. Instead of treating an object as a collection of individual particles, we can often simplify its motion by considering a single point – the centre of mass – where the entire mass of the object can be considered to be concentrated. This point moves as if all the external forces were applied to it, while the motion of the object about this point is independent of the translational motion of the COM.

1. Centre of Mass of a System of Discrete Particles

For a system of 'n' discrete particles with masses m₁, m₂, ..., m_n, located at positions given by their position vectors r₁, r₂, ..., r_n, the position vector of the centre of mass (R_CM) is defined as:

R_CM = (m₁r₁ + m₂r₂ + ... + m_nr_n) / (m₁ + m₂ + ... + m_n)

In terms of components, if r_i = x_ii + y_ij + z_k and R_CM = X_CMi + Y_CMj + Z_CMk, then:

X_CM = (m₁x₁ + m₂x₂ + ... + m_nx_n) / (m₁ + m₂ + ... + m_n)

Y_CM = (m₁y₁ + m₂y₂ + ... + m_ny_n) / (m₁ + m₂ + ... + m_n)

Z_CM = (m₁z₁ + m₂z₂ + ... + m_nz_n) / (m₁ + m₂ + ... + m_n)

The total mass of the system is M = Σm_i. So, the equations can be written as:

R_CM = (1/M) Σm_ir_i

X_CM = (1/M) Σm_ix_i

Y_CM = (1/M) Σm_iy_i

Z_CM = (1/M) Σm_iz_i

Example:

Consider two particles of mass 1 kg and 2 kg placed at positions (1, 2) and (3, 4) respectively.

m₁ = 1 kg, r₁ = (1, 2)

m₂ = 2 kg, r₂ = (3, 4)

Total mass M = m₁ + m₂ = 1 + 2 = 3 kg.

X_CM = (m₁x₁ + m₂x₂) / M = (1 * 1 + 2 * 3) / 3 = (1 + 6) / 3 = 7/3.

Y_CM = (m₁y₁ + m₂y₂) / M = (1 * 2 + 2 * 4) / 3 = (2 + 8) / 3 = 10/3.

So, the centre of mass is at (7/3, 10/3).

2. Centre of Mass of a Continuous Body

For a continuous body, we can imagine dividing it into infinitesimally small mass elements dm. The position vector of the centre of mass is then given by integrating over the entire body:

R_CM = (1/M) ∫ r dm

Where M is the total mass of the body, and the integral is taken over the entire volume of the body.

In terms of components:

X_CM = (1/M) ∫ x dm

Y_CM = (1/M) ∫ y dm

Z_CM = (1/M) ∫ z dm

To evaluate these integrals, we need to express dm in terms of the coordinates. This depends on the shape and density distribution of the body.

2.1 Linear Mass Density (λ)

For a thin rod or wire of length L, where mass is distributed along its length. If dm is the mass of a small element of length dl, then dm = λ dl, where λ is the linear mass density. If the rod is uniform, λ is constant.

X_CM = (1/M) ∫ x λ dl

Y_CM = (1/M) ∫ y λ dl

Z_CM = (1/M) ∫ z λ dl

2.2 Surface Mass Density (σ)

For a thin sheet or surface of area A, where mass is distributed over its surface. If dm is the mass of a small element of area dA, then dm = σ dA, where σ is the surface mass density. If the sheet is uniform, σ is constant.

X_CM = (1/M) ∫ x σ dA

Y_CM = (1/M) ∫ y σ dA

Z_CM = (1/M) ∫ z σ dA

2.3 Volume Mass Density (ρ)

For a solid object of volume V, where mass is distributed throughout its volume. If dm is the mass of a small element of volume dV, then dm = ρ dV, where ρ is the volume mass density. If the object is uniform, ρ is constant.

X_CM = (1/M) ∫ x ρ dV

Y_CM = (1/M) ∫ y ρ dV

Z_CM = (1/M) ∫ z ρ dV

3. Centre of Mass of Some Standard Uniform Bodies

For uniform bodies with simple geometric shapes, the centre of mass is often at their geometric centre due to symmetry.

Object	Centre of Mass Location
Thin rod of length L	At the midpoint of the rod (L/2 from either end).
Rectangular plate of sides a and b	At the intersection of its diagonals (a/2, b/2 from the corner).
Circular disc of radius R	At its geometric centre.
Ring of radius R	At its geometric centre.
Sphere of radius R	At its geometric centre.
Cube of side a	At its geometric centre.
Solid cone of height H	At a distance H/4 from the base along the axis.
Hollow cone of height H	At a distance H/3 from the base along the axis.
Solid hemisphere of radius R	At a distance 3R/8 from the base along the axis of symmetry.
Hollow hemisphere of radius R	At a distance R/2 from the base along the axis of symmetry.

Example: Centre of Mass of a Uniform Rod

Consider a uniform rod of length L and mass M lying along the x-axis from x=0 to x=L. The linear mass density is λ = M/L.

dm = λ dx = (M/L) dx.

X_CM = (1/M) ∫ x dm = (1/M) ∫₀^L x (M/L) dx

X_CM = (1/L) ∫₀^L x dx = (1/L) [x²/2]₀^L = (1/L) (L²/2 - 0) = L/2.

Since the rod is along the x-axis, Y_CM = 0 and Z_CM = 0. Thus, the COM is at (L/2, 0, 0), which is the midpoint.

4. Centre of Mass of Composite Bodies

For a composite body made of simpler shapes, we can find the COM by treating each simple shape as a point mass located at its own centre of mass.

If a composite body is made of 'n' parts, with masses M₁, M₂, ..., M_n and their centres of mass at positions R₁, R₂, ..., R_n, then the COM of the composite body is:

R_CM = (M₁R₁ + M₂R₂ + ... + M_nR_n) / (M₁ + M₂ + ... + M_n)

In terms of components:

X_CM = (M₁X₁ + M₂X₂ + ... + M_nX_n) / M

Y_CM = (M₁Y₁ + M₂Y₂ + ... + M_nY_n) / M

Z_CM = (M₁Z₁ + M₂Z₂ + ... + M_nZ_n) / M

Example: Centre of Mass of a Disc with a Hole

Consider a uniform circular disc of radius R and mass M. A circular hole of radius r is cut from the disc, concentric with the disc. We need to find the COM of the remaining part.

Let the original disc be D and the hole be H. The remaining part is P = D - H.

Assume the origin (0,0) is at the centre of the disc. The COM of the original disc D is at (0,0).

The mass of the original disc is M. The mass of the hole (if it were present) is M_H. Since the density is uniform (σ = M / (πR²)), the mass of the hole is M_H = σ * (πr²) = (M / (πR²)) * (πr²) = M(r²/R²).

The mass of the remaining part P is M_P = M - M_H = M - M(r²/R²) = M(1 - r²/R²).

Let's assume the hole is cut such that its centre is at (d, 0) from the centre of the disc. The COM of the hole (considered as a mass) is at (d, 0).

We can use the principle of superposition. The COM of the remaining part P can be found using the relation:

M_D * R_D,CM = M_P * R_P,CM + M_H * R_H,CM

(0,0) * M = R_P,CM * M_P + (d,0) * M_H

0 = R_P,CM * M(1 - r²/R²) + (d,0) * M(r²/R²)

R_P,CM * M(1 - r²/R²) = - (d,0) * M(r²/R²)

R_P,CM = - (d,0) * (r²/R²) / (1 - r²/R²)

R_P,CM = - (d,0) * (r² / (R² - r²))

If the hole is cut from the edge, say its centre is at (R-r, 0), then d = R-r.

R_P,CM = - ((R-r), 0) * (r² / (R² - r²))

The X-coordinate of the COM of the remaining part will be:

X_P,CM = - (R-r) * (r² / (R² - r²))

This means the COM shifts towards the side opposite to the hole.

5. Motion of the Centre of Mass

The motion of the centre of mass of a system of particles is governed by the net external force acting on the system.

R_CM = (1/M) Σm_ir_i

Taking the first derivative with respect to time, we get the velocity of the COM:

V_CM = dR_CM/dt = (1/M) Σm_i(dr_i/dt) = (1/M) Σm_iv_i

Taking the second derivative with respect to time, we get the acceleration of the COM:

A_CM = dV_CM/dt = (1/M) Σm_i(dv_i/dt) = (1/M) Σm_ia_i

According to Newton's second law, the net force on the i-th particle is F_i = m_ia_i.

So, MA_CM = Σm_ia_i = ΣF_i.

The sum of all forces ΣF_i includes both external forces (F_ext) and internal forces (F_int) between the particles.

ΣF_i = F_ext + F_int.

By Newton's third law, the internal forces occur in equal and opposite pairs, so their vector sum is zero: ΣF_int = 0.

Therefore, MA_CM = F_ext.

This is a crucial result: The acceleration of the centre of mass of a system is equal to the net external force divided by the total mass of the system. The internal forces do not affect the motion of the centre of mass.

This means that the centre of mass of a system moves as if it were a single particle of mass M, acted upon by the net external force. If the net external force is zero, the centre of mass moves with constant velocity (or remains at rest). This principle is very useful for analysing the motion of complex systems, including explosions and collisions.

Example: Explosion of a Projectile

Suppose a projectile is fired and explodes in mid-air into several fragments. If we ignore air resistance, the net external force on the system of fragments is zero (only gravity, which is an external force, acts). The momentum of the centre of mass is conserved.

Initially, the projectile has a certain velocity. The centre of mass of the system of fragments will continue to move along the original parabolic trajectory of the projectile, even though the individual fragments may scatter in different directions.

6. Centre of Mass and Rigid Bodies

A rigid body can be considered as a continuous distribution of mass. The centre of mass is a fixed point relative to the body (unless the mass distribution changes). The motion of a rigid body can be decomposed into:

• The translational motion of its centre of mass.
• The rotational motion of the body about an axis passing through its centre of mass.

The translational motion of the COM is governed by the net external force, as discussed above. The rotational motion is governed by the net external torque.

Key Considerations for Rigid Bodies:

• Symmetry: For bodies with high symmetry, the COM often coincides with the geometric centre.
• Non-uniform Density: If the density is not uniform (ρ is a function of position), integration is necessary, and the COM will not necessarily be at the geometric centre.
• Composite Structures: Break down complex shapes into simpler ones, find the COM and mass of each part, and then use the composite body formula.
• Removal of Mass: When a part of a body is removed (like a hole being drilled), treat the removed part as a body with negative mass and find its COM.