Systems and Center of Mass
Introduced| English |
|---|
| center of mass/ˈsentə ɒv mæs/ |
| system/ˈsɪstəm/ |
| internal forces/ɪnˈtɜːnl ˈfɔːsɪz/ |
| external forces/ekˈstɜːnl ˈfɔːsɪz/ |
| point particle/pɔɪnt ˈpɑːtɪkl/ |
| integral/ˈɪntɪɡrəl/ |
| net external force/net ekˈstɜːnl fɔːs/ |
Throw a spinning wrench
- Toss a wrench across the room and it tumbles wildly -- yet one single point traces a smooth, simple arc.
- That special point is the center of mass 质心.
- No matter how an object spins or wobbles, its center of mass moves as if all the mass were squeezed into that one spot.
- Find it, and a messy tumbling problem becomes a simple point-mass problem.
Treat a system as one point
- A system 系统 is any group of objects we choose to study together.
- Forces inside the system are internal forces 内力; forces from outside are external forces 外力.
- The whole system moves as if it were a single point particle 质点 located at its center of mass.
Select all true statements about the center of mass.
The first two are the core rules. The third is false -- it is the mass-weighted center, not the shape's midpoint.
Center of mass of discrete masses
- For separate point masses along a line, the center of mass is the mass-weighted average position:
- Heavier masses pull the center of mass toward themselves.
A $1\ \text{kg}$ mass is at $x = 0$ and a $3\ \text{kg}$ mass is at $x = 4\ \text{m}$. Where is the center of mass (in m)?
$x_{cm} = \dfrac{1(0) + 3(4)}{1 + 3} = \dfrac{12}{4} = 3\ \text{m}$ -- close to the heavier mass.
Move one mass farther from the others. The center of mass...
The mass-weighted average follows the mass -- moving a mass out pulls $x_{cm}$ toward it.
Masses $2\ \text{kg}$ at $x=0$, $2\ \text{kg}$ at $x=2$, and $4\ \text{kg}$ at $x=5$. Find $x_{cm}$ (in m).
$x_{cm} = \dfrac{2(0)+2(2)+4(5)}{2+2+4} = \dfrac{24}{8} = 3\ \text{m}$.
Center of mass by integration
- For a continuous object, replace the sum with an integral 积分 over its mass:
- Slice the object into tiny bits $dm$, weight each by its position, and add them up.
Centre of mass of a system
Tap the parts to see where the centre of mass of a two-body system sits.
For a continuous body the sum in the center-of-mass formula becomes an ____.
$x_{cm} = \tfrac{1}{M}\int x\,dm$ -- an integral over the mass distribution.
Only external forces move it
- The center of mass accelerates only in response to the net external force 净外力.
- Internal forces -- however violent -- can never shift the center of mass.
- That is why the tumbling wrench's center of mass follows the same smooth path a simple thrown ball would.
A hidden explosion inside a floating system (no outside forces) can shift its center of mass.
Explosion forces are internal, so they cannot move the center of mass. Only a net external force can.
A $2\ \text{kg}$ mass sits at $x = 0$ and a $4\ \text{kg}$ mass at $x = 3\ \text{m}$.
- $x_{cm} = \dfrac{2(0) + 4(3)}{2 + 4} = \dfrac{12}{6} = 2\ \text{m}$.
- The center of mass sits closer to the heavier $4\ \text{kg}$ mass, as expected.
The center of mass is not always the geometric center. For a uniform rod it is the middle, but load one end and the center of mass shifts toward the heavy end -- it is the mass-weighted average, not the shape's midpoint.
The center of mass is the mass-weighted average position: $x_{cm} = \dfrac{\sum m_i x_i}{\sum m_i}$ (or $\tfrac{1}{M}\int x\,dm$ for a continuous body). A whole system moves as if all its mass were at that point, and only the net external force can accelerate it.