Centre of Mass>Mathematical Understanding of Centre of Mass

Newton's laws of motion are designed for point masses, but in real life, all objects have measurable sizes. The concept of centre of mass helps us treat these finite-sized objects as if they were point objects located at a specific position. This makes it possible to apply Newton's laws to real objects. The centre of mass is a very important concept in physics that simplifies the study of the motion of extended bodies.

"Centre of mass is a point about which the summation of moments of masses in the system is zero."

When we have n particles with masses m₁, m₂, ..., m_n, the total mass is:

M = \[\sum_{i=1}^nm_i\]

If these particles are located at position vectors \[\vec{r}_1,\vec{r}_2,...,\vec{r}_n\] from origin O, then the position vector \[\vec r\] of their centre of mass is:

\[\vec{r}=\frac{\sum_{i=1}^nm_i\vec{r}_i}{\sum_{i=1}^nm_i}=\frac{\sum_{i=1}^nm_i\vec{r}_i}{M}\]

Special Case: If the origin is at the centre of mass, then \[\vec r\]$0$ and \[\sum_{i=1}^nm_i\vec{r}_i\] = 0

For Cartesian Coordinates:

The x-coordinate of the centre of mass:
x = \[\frac{\sum_{i=1}^nm_ix_i}{M}\]

The y-coordinate of the centre of mass:
y = \[\frac{\sum_{i=1}^nm_iy_i}{M}\]

The z-coordinate of centre of mass:
z = \[\frac{\sum_{i=1}^nm_iz_i}{M}\]​

For objects with continuous mass distribution and uniform density, we use integration instead of summation:

\[\vec{r}=\frac{\int\vec{r}dm}{\int dm}=\frac{\int\vec{r}dm}{M}\]

where M = ∫ dm is the total mass.

Cartesian coordinates become:

x = \[\frac{\int xdm}{M}\]

y = \[\frac{\int ydm}{M}\]

z = \[\frac{\int zdm}{M}\]

• For two point masses: Centre of mass divides the distance in inverse proportion to the masses
• For geometrically symmetric objects with uniform density, the Centre of mass is at the geometrical centre

Some standard results:

• Isosceles triangular plate: \[x_c=0,y_c=\frac{H}{3}\]
• Right angled triangular plate: \[x_c=\frac{P}{3},y_c=\frac{q}{3}\]
• Thin semicircular ring (radius R): \[x_c=0,y_c=\frac{2R}{\pi}\]
• Thin semicircular disc (radius R): \[x_c=0,y_c=\frac{4R}{3\pi}\]
• Hemispherical shell (radius R): \[x_c=0,y_c=\frac{R}{2}\]
• Solid hemisphere (radius R): \[x_c=0,y_c=\frac{3R}{8}\]
• Hollow right circular cone (height h): \[x_c=0,y_c=\frac{h}{3}\]
• Solid right circular cone (height h): \[x_c=0,y_c=\frac{h}{4}\]

• Allows us to apply Newton's laws to real extended objects by treating them as point masses
• Simplifies complex calculations for objects with many particles
• Helps predict the motion of rigid bodies and systems
• Essential for understanding rotational motion and stability
• Makes it easier to analyze collisions and the momentum of extended bodies
• Used in engineering and design to ensure balance and stability

Problem: Find the centre of mass of a uniform cardboard shaped like the letter 'E' made of 10 equal squares.

Solution Steps:

1. Treat each square as mass m at its center (numbered 1 to 10
2. Choose origin at central left mass (m₅)
3. Use symmetry:

• Masses m₁, m₂, m₃ combine to an effective 3m at position (1, 2)
• Masses m₈, m₉, m₁₀ combine to an effective 3m at position (1, -2)
• These two groups combine to 6m at position (1, 0)
• Mass m₆ is also at (1, 0), so a total of 7m at (1, 0)
• Masses m₄, m₅, m₇ give an effective 3m at the origin (0, 0)

4. Now we have two point masses: 3m at (0,0) and 7m at (1,0)
5. Calculate x-coordinate:
\[x_c=\frac{3\times0+7\times1}{3+7}=\frac{7}{10}=0.7\mathrm{cm}\]

Answer: Centre of mass is at 0.7 cm from the origin along the x-axis

Problem: Three hollow spheres of radii 1 cm, 2 cm, and 3 cm are at vertices of an equilateral triangle ABC (side 10 cm). Find the centre of mass.

Solution Steps:

1. Mass is proportional to surface area (proportional to r²)

• If mass at A = m, then mass at B = 4m, mass at C = 9m

2. Choose origin at C (largest mass) and B on the positive x-axis

• C coordinates: (0, 0)
• B coordinates: (10, 0)
• A coordinates: (5, \[\frac{10\sqrt{3}}{2}\])

3. Calculate x-coordinate:
\[x_c=\frac{m\times5+4m\times10+9m\times0}{m+4m+9m}=\frac{45m}{14m}=\frac{45}{14}\mathrm{cm}\]

4. Calculate y-coordinate:
\[y_c=\frac{m\times\frac{10\sqrt{3}}{2}+4m\times0+9m\times0}{14m}=\frac{10\sqrt{3}}{28}\mathrm{cm}\]

Answer: Centre of mass is at (\[\frac{45}{14},\frac{10\sqrt{3}}{28}\]) cm from origin at C

Problem: A hole of radius r is cut from a uniform disc of radius 2r. The centre of the hole is at a distance r from the disc centre. Find the centre of mass of the remaining part.

Solution Steps:

Method I - Using Entire Disc:

1. Original disc has centre of mass at its centre O (our origin)
2. Cut portion has mass m and centre at distance r from O
3. Remaining portion has mass = 4m - m = 3m (since area ratio is 4:1)
4. Use the centre of mass formula for a full disc:
0 = \[\frac{m\times r+3m\times x}{m+3m}\]
5. Solve for x: 0 = mr+3mx
x = \[-\frac{r}{3}\]

Method II - Using Negative Mass Concept:

Use formula: \[\vec{r}_{c}=\frac{M\vec{R}-m\vec{r}}{M-m}\]

With M = 4m, m = m, R = 0, r = r:
\[r_c=\frac{0-m\times r}{4m-m}=\frac{-mr}{3m}=-\frac{r}{3}\]

Answer: Centre of mass is at distance \[\frac {r}{3}\] from disc centre, on opposite side of the hole

1. Seesaw Balance: A heavier child sits closer to the pivot to balance a lighter one further out.
2. High Jump: Jumpers arch their body so the center of mass stays lower, helping them clear the bar.
3. Car Stability: Cars have a low center of mass to avoid tipping on turns.
4. Rocket Flight: Rocket stability depends on how the center of mass shifts as fuel burns.
5. Tightrope Walk: Walkers use a long pole to lower their center of mass and stay balanced.
6. Phone Balance: A phone balances on your finger when your finger is under its center of mass.