König's theorem for kinetic energy

A system’s total kinetic energy can always be written as the sum of two contributions: one tied to the motion of the centre of mass (as if the whole mass travelled as a single block) and one internal, computed relative to the CM.

Law — König's theorem for kinetic energy

$\ev{E_{\text{cin,tot}} = \underbrace{\tfrac{1}{2}M_{\text{tot}}\,v_{\text{CM}}^2}_{E_{\text{cin,CM}}} + \underbrace{\sum_i \tfrac{1}{2}m_i\,v_i'^2}_{E_{\text{cin,int}}}}$ where $\vec{v}_i' = \vec{v}_i - \vec{v}_{\text{CM}}$ is the velocity of body $i$ in the CM reference frame.

The first term is the kinetic energy of the overall motion; the second is the “internal agitation” of the parts around the CM. There are two equivalent ways to compute $E_\text{cin,int}$, depending on which velocities are more convenient to use.

Two ways to get $E_\text{cin,int}$

$E_\text{cin,int} = \sum_i \tfrac{1}{2}m_i v_i^2 - \tfrac{1}{2}M_\text{tot} v_\text{CM}^2$ $E_\text{cin,int} = \sum_i \tfrac{1}{2}m_i\,|\vec{v}_i - \vec{v}_\text{CM}|^2$

The first uses the velocities in the lab frame (total energy minus CM energy); the second uses directly the velocities relative to the CM. The result is the same: an observer at rest and one sitting on the CM compute the same $E_\text{cin,int}$.

Example — Checking the decomposition

Two balls: $m_1 = 3\;\text{kg}$ at $v_1 = 3\;\text{m/s}$, $m_2 = 7\;\text{kg}$ at $v_2 = 2\;\text{m/s}$ (1D motion). The CM velocity is $v_{\text{CM}} = (3\cdot 3 + 7\cdot 2)/10 = 2{,}3\;\text{m/s}$.

Method 1 (difference): $E_{\text{cin,tot}} = \tfrac{1}{2}\cdot 3\cdot 9 + \tfrac{1}{2}\cdot 7\cdot 4 = 27{,}5\;\text{J}$ $E_{\text{cin,CM}} = \tfrac{1}{2}\cdot 10\cdot 2{,}3^2 = 26{,}45\;\text{J} \quad\Rightarrow\quad E_{\text{cin,int}} = 27{,}5 - 26{,}45 = 1{,}05\;\text{J}$

Method 2 (relative velocities): $v_1' = 3 - 2{,}3 = 0{,}7$, $v_2' = 2 - 2{,}3 = -0{,}3$ $E_{\text{cin,int}} = \tfrac{1}{2}\cdot 3\cdot 0{,}49 + \tfrac{1}{2}\cdot 7\cdot 0{,}09 = 0{,}735 + 0{,}315 = 1{,}05\;\text{J} \;\checkmark$

Hint

In the CMRF the total momentum is zero ($\sum_i m_i \vec{v}_i' = \vec{0}$), because the observer moves at exactly the CM’s velocity. At the point of maximum compression in a spring collision, both bodies have velocity $v_\text{CM}$: $E_\text{cin,int} = 0$ and all the internal energy is stored in the spring.

Links

Topics: Centre of mass Concepts: Kinetic energy · Internal energy Skills: Changing reference frame

Related exercises: Proof of König’s theorem · Worked exercise — Toy gas of two balls · Internal kinetic energy of two balls