The equation of motion for a system's centre of mass: $M\vec{a}_{CM} = \vec{F}_{ext}$

Key insights:

• Internal forces (between particles in the system) cancel in pairs by Newton's third law
• Only external forces affect the acceleration of the centre of mass
• The centre of mass moves as if all the mass were concentrated at that point with all external forces applied there
• If no external forces act (Fₑₓₜ = 0), the centre of mass moves with constant velocity or remains at rest

Example: When a ball spins while moving in air (neglecting air resistance), the centre of mass follows a parabolic path despite the complex motion of other points on the ball.

For two particles with masses m₁ and m₂ separated by distance d:

• If m₁ is at origin and m₂ is at position d along x-axis: Centre of mass position = $\frac{m_2d}{m_1+m_2}$ from m₁ along the line joining them

• The centre of mass divides the line joining the particles internally in inverse ratio of their masses: $m_1(OC) = m_2(CP)$

• If m₁ ≫ m₂ (m₁ much larger than m₂), the centre of mass is very close to m₁

• If m₁ = m₂, the centre of mass is exactly at the midpoint

Example: Earth-Moon system's centre of mass lies about 4,700 km from Earth's center (still inside Earth) because Earth's mass is much larger than Moon's.

At maximum compression during collision:

• Both objects move with exactly the same velocity (v₁ = v₂ = V)
• The spring/deformable surface reaches its maximum compression
• No further compression occurs at this instant
• Kinetic energy is at minimum during the collision process
• Elastic potential energy is at maximum
• Total momentum remains conserved: m₁v₁(initial) + m₂v₂(initial) = (m₁+m₂)V
• Common velocity: V = (m₁v₁ + m₂v₂)/(m₁+m₂)

Example: When a moving billiard ball hits a stationary one, maximum deformation occurs precisely when both balls move at the same instantaneous velocity.

For two particles A (mass m) and B (mass 2m) initially at rest, with mutual attraction:

Analysis:

1. Consider the system of both particles
2. No external forces act on the system
3. Initial momentum is zero, so the centre of mass remains fixed
4. Initial position of centre of mass: $X_{CM} = \frac{mx_A + 2mx_B}{m + 2m} = \frac{x_A + 2x_B}{3}$
5. The collision must occur at the centre of mass (since both particles arrive there)

If A is at position 0 and B at position d initially: $X_{CM} = \frac{0 + 2d}{3} = \frac{2d}{3}$

Therefore, collision occurs at a distance of 2d/3 from A's initial position, or d/3 from B's initial position.

Note: This applies regardless of the nature of the attractive force between the particles.

• The trajectory of a ball's centre of mass is determined solely by external forces (gravity and air resistance)
• Spinning the ball:
  • Is produced by internal forces (from the bowler's fingers)
  • Cannot change the parabolic path of the centre of mass in a vacuum
  • Can affect air resistance and create Magnus effect in air
• Practical application: Bowlers utilize this principle by:
  • Controlling the basic trajectory through the initial velocity vector
  • Adding spin to create drift, dip, or deviation after bounce
  • Using the same fundamental delivery action while varying spin for deception
• Note: While the basic parabolic trajectory remains, air effects on a spinning ball create secondary deviations that make the ball challenging to play

To find the centre of mass of a uniform semicircular wire:

1. Set up coordinate system with origin at the centre of the semicircle
2. Choose an elemental length dℓ = R dθ along the wire
3. Express coordinates of this element as (R cosθ, R sinθ)
4. Calculate mass of element: dm = (M/πR)·(R dθ) = (M/π)dθ
5. Apply centre of mass formulas:
   • X = (1/M)∫x dm = (1/M)∫(R cosθ)(M/π)dθ = 0 (by symmetry)
   • Y = (1/M)∫y dm = (1/M)∫(R sinθ)(M/π)dθ = 2R/π

The centre of mass is located at (0, 2R/π), which is below the geometric center.

When selecting a differential element for center of mass calculations:

1. Choose elements that cover the entire body when integration variable spans its limits
2. Ensure coordinates of different parts within each element differ only infinitesimally
3. Express the mass of the element in terms of the integration variable (e.g., dm = λdx for linear density)

Key requirements:

• Elements must be definable using your chosen coordinate system
• Each element's position must be representable by a single coordinate set
• The collection of all elements must exactly constitute the entire body

For a rod example:

• Choose element as section between x and x+dx
• Express its mass as dm = (M/L)dx for uniform density
• Position is represented by coordinate x
• As x varies from 0 to L, all elements collectively form the complete rod

This approach enables proper application of the integral formula: X = (1/M)∫x·dm

Integral setup for y-coordinate of semicircular plate's centre of mass:

1. Choice of differential element:

   • Select a thin semicircular ring of radius r and thickness dr
   • This allows covering the entire plate as r varies from 0 to R
   • Each element has uniform mass distribution along its arc

2. Mass of differential element:

   • Area of ring = π·r·dr (area of semicircular ring)
   • Mass density = M/(πR²/2) (total mass divided by area)
   • Mass of element (dm) = (2M/R²)·r·dr

3. Position of element's centre of mass:

   • Each ring has its own centre of mass at y = 2r/π

4. Integration to find Y-coordinate:

   • Y = (1/M)∫y·dm
   • Y = (1/M)∫(0 to R) (2r/π)·(2M/R²)·r·dr
   • Y = (4/πR²)∫(0 to R) r²·dr
   • Y = (4/πR²)·(R³/3) = 4R/3π

This approach divides the object into elements whose individual centres of mass are known, then finds their weighted average.

Center of Mass Calculation

For a system of discrete particles:

• X-coordinate: $X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + ...}{m_1 + m_2 + m_3 + ...}$

• Y-coordinate: $Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + ...}{m_1 + m_2 + m_3 + ...}$

Where:

• $m_i$ = mass of each particle
• $(x_i, y_i)$ = coordinates of each particle
• Denominator = total mass of the system

Physical Significance

1. Mass Distribution Point: The center of mass is pulled toward regions of greater mass concentration

2. Rotational Properties:

   • Point where the entire mass can be considered concentrated for calculating gravitational torque
   • Forces applied at this point produce pure translational motion with no rotation

3. System Behavior: For external forces and gravity calculations, the entire system can be treated as a single particle located at the center of mass

Note: In asymmetric mass distributions (like particles with masses 0.5 kg, 1.0 kg, and 1.5 kg forming a triangle), the center of mass will be noticeably closer to the more massive particles.

Centre of Mass Positions

• Semicircular wire: (0, 2R/π) ≈ (0, 0.637R)
• Semicircular plate: (0, 4R/3π) ≈ (0, 0.424R)
• Geometric center of semicircle: (0, R/2) = (0, 0.5R)

Key Differences Explained

1. Dimensionality affects mass distribution:

   • Wire: 1D object with mass distributed uniformly along arc length
   • Plate: 2D object with mass proportional to area (more mass at larger radii)

2. Mathematical derivation:

   • Wire: Y = ∫R·sinθ·(M/π)·dθ from θ=0 to π
   • Plate: Y = ∫∫y·dm / M (with area element proportional to r·dr·dθ)

3. Physical interpretation:

   • Wire's CM is higher (0.637R) because mass is concentrated along the curved path
   • Plate's CM is lower (0.424R) because mass distribution extends throughout the area
   • Both differ from the geometric center (0.5R) of the semicircle

4. Mass concentration visualization:

   • Wire: Equal mass per unit arc length
   • Plate: More mass at larger radii (area increases with distance from origin)

Note: These differences highlight how the dimensionality of an object fundamentally affects its mass distribution properties, even when the boundary shape is identical.