73 lines
3.8 KiB
Markdown
73 lines
3.8 KiB
Markdown
---
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title: 8. Centre of mass
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---
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The momentum of a closed mechanical system has different values in different (inertial) frames of reference. If a frame $K'$ moves with velocity $\v{V}$ relative to another frame $K$, then the velocities $\v{v}_a'$ and $\v{v}_a$ of the particles relative to the two frames are such that $\v{v}_a = \v{v}_a' + \v{V}$. The momenta $P$ and $P'$ in the two frames are therefore related by
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$$
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\v{P}
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=\sum_a m_a\v{v}_a
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=\sum_a m_a\v{v}'_a
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+\v{V}\sum_a m_a
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$$
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or
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```load
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1/8.1
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```
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In particular, there is always a frame of reference $K'$ in which the total
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momentum is zero. Putting $P' = 0$ in `1/8.1`, we find the velocity of this frame:
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```load
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1/8.2
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```
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If the total momentum of a mechanical system in a given frame of reference is zero, it is said to be at rest relative to that frame. This is a natural generalisation of the term as applied to a particle. Similarly, the velocity V given by `1/8.2` is the velocity of the "motion as a whole" of a mechanical system whose momentum is not zero. Thus we see that the law of conservation of momentum makes possible a natural definition of rest and velocity, as applied to a mechanical system as a whole.
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Formula `1/8.2` shows that the relation between the momentum $\v{P}$ and the velocity $V$ of the system is the same as that between the momentum and velocity of a single particle of mass $\mu = \sum m_a$, the sum of the masses of the particles in the system. This result can be regarded as expressing the additivity of mass.
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The right-hand side of formula `1/8.2` can be written as the total time derivative of the expression
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```load
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1/8.3
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```
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We can say that the velocity of the system as a whole is the rate of motion in space of the point whose radius vector is `1/8.3`. This point is called the centre of mass of the system.
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The law of conservation of momentum for a closed system can be formulated as stating that the centre of mass of the system moves uniformly in a straight line. In this form it generalises the law of inertia derived in `1/3` for a single free particle, whose "centre of mass" coincides with the particle itself.
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In considering the mechanical properties of a closed system it is natural to use a frame of reference in which the centre of mass is at rest. This eliminates a uniform rectilinear motion of the system as a whole, but such motion is of no interest.
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The energy of a mechanical system which is at rest as a whole is usually called its internal energy $E$. This includes the kinetic energy of the relative motion of the particles in the system and the potential energy of their interaction. The total energy of a system moving as a whole with velocity $V$ can be written
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```load
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1/8.4
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```
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Although this formula is fairly obvious, we may give a direct proof of it. The energies $E$ and $E'$ of a mechanical system in two frames of reference $K$ and $K'$ are related by
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\begin{align}
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E
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&= \frac{1}{2}\sum_a m_a v_a^2 + U \\
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&= \frac{1}{2}\sum_a m_a (\v{v}'_a+\v{V})^2 + U \\
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&= \frac{1}{2}\mu V^2
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+ \v{V}\sum_a m_a\v{v}'_a
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+ \frac{1}{2} \sum_a m_a {v'}_a^2 + U
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\end{align}
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```load
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1/8.5
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```
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This formula gives the law of transformation of energy from one frame to another, corresponding to formula `1/8.1` for momentum. If the centre of mass is at rest in $K'$, then $\v{P}' = 0$, $E' = E_i$, and we have `1/8.4`.
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<!-- PROBLEM -->
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<!-- Find the law of transformation of the action S from one inertial frame to another. -->
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<!-- SOLUTION. The Lagrangian is equal to the difference of the kinetic and potential energies, -->
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<!-- and is evidently transformed in accordance with a formula analogous to (8.5): -->
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<!-- Integrating this with respect to time, we obtain the required law of transformation of the -->
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<!-- action: -->
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<!-- S=S'tuV.R'+1uV2t, -->
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<!-- where R' is the radius vector of the centre of mass in the frame K'. -->
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