Systems and Center of Mass
Introduced| English | Meaning |
|---|---|
| centre of mass/ˈsentə ɒv mæs/ | The point representing an object's or system's mass-weighted average position. |
| system/ˈsɪstəm/ | The object or collection of objects chosen for analysis, separated conceptually from its surroundings. |
| external force/ekˈstɜːnl fɔːs/ | A force exerted on a chosen system by something outside that system. |
| internal force/ɪnˈtɜːnl fɔːs/ | A force between two parts of the same system; internal forces cannot change the system's total momentum. |
| particle/ˈpɑːtɪkl/ | An idealised object whose size and shape are ignored so its mass can be treated as concentrated at a point. |
A wrench tumbles — yet one point glides straight
- Slide a spinning wrench across ice and every part traces a messy loop.
- But one special point drifts in a perfectly straight line — the centre of mass 质心.
- Treat a complicated object (or a group of objects) as if all its mass sat there.
- That single idea turns a whole system 系统 into one easy-to-track point.
What a system is
- A system is any group of objects you choose to study together.
- Everything else is the surroundings; forces from outside are external forces 外力.
- Forces between members of the system are internal forces 内力.
- By Newton's third law, internal forces come in canceling pairs — they never move the centre of mass.
By Newton's third law, the internal forces of a system always ____ in pairs.
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Internal forces are action–reaction pairs, so they sum to zero and cannot move the centre of mass.
Locating the centre of mass
- The centre of mass is the mass-weighted average of every part's position.
- For point masses on a line: $\displaystyle x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots}$.
- It sits closer to the heavier mass — the big mass "pulls" the average toward it.
- It need not be inside the object: a ring's centre of mass is in the empty hole.
Find the balance point
Change the two masses and their distances, and see where the system balances — the centre of mass.
A $2\ \text{kg}$ mass is at $x = 0$ and a $6\ \text{kg}$ mass at $x = 8\ \text{m}$. Where is the centre of mass, in metres?
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$x_{\text{cm}} = \dfrac{2(0) + 6(8)}{2 + 6} = \dfrac{48}{8} = 6\ \text{m}$ — closer to the heavier mass.
The centre of mass of a system is best described as the:
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It is $\frac{\sum m_i x_i}{\sum m_i}$ — each position weighted by its mass.
Two equal masses sit at $x = 2\ \text{m}$ and $x = 6\ \text{m}$. Where is their centre of mass, in metres?
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Equal masses ⇒ the centre of mass is the midpoint: $(2 + 6)/2 = 4\ \text{m}$.
Why the centre of mass matters
- External forces alone move the centre of mass: $\vec F_{\text{net,ext}} = M\,\vec a_{\text{cm}}$.
- No matter how the parts spin or collide inside, the centre of mass obeys this simple law.
- A diver somersaults, but their centre of mass follows a smooth parabola.
- This is why we can model an extended object as a single particle 质点.
Select all true statements about the centre of mass.
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It is mass-weighted (closer to heavy parts), moved only by external forces, and lets us treat the body as a point — but it is not always the geometric centre.
The centre of mass is not always the geometric centre, and it need not lie within the material. For a doughnut or a boomerang it sits in empty space. Only for a uniform, symmetric object do the two coincide.
The centre of mass must always lie inside the material of the object.
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A ring or boomerang has its centre of mass in empty space. Only symmetric solid bodies keep it inside.
A $2\ \text{kg}$ ball sits at $x = 0$ and a $6\ \text{kg}$ ball at $x = 8\ \text{m}$.
- $x_{\text{cm}} = \dfrac{(2)(0) + (6)(8)}{2 + 6} = \dfrac{48}{8} = 6\ \text{m}$.
The centre of mass is at $6\ \text{m}$ — much closer to the heavier $6\ \text{kg}$ ball.
A system is a chosen set of objects; its centre of mass is the mass-weighted average position, $x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}$, lying closer to heavier parts. Internal forces cancel in pairs; only external forces move the centre of mass ($\vec F_{\text{net,ext}} = M\vec a_{\text{cm}}$).
Write from a centre of mass at 3 m
- A 3 kg mass is at x = 1 m. A 1 kg mass is at x = 9 m.
- Find the centre of mass.
A 3 kg mass sits at x = 1 m and a 1 kg mass sits at x = 9 m. Where is the centre of mass, in metres?
12 divided by 4 is 3.
Where is the centre of mass?
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The mass-weighted average is 3 m, closer to the 3 kg mass. Other options — 6 m (the lesson centre); 5 m (the midpoint of 1 and 9).