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# 3.10: Kinetic energy

[ "article:topic", "authorname:tatumj", "showtoc:no" ]

We remind ourselves that we are discussing particles, and that all kinetic energy is translational kinetic energy.

Notation:

• $$T_{C}$$ = kinetic energy with respect to the centre of mass C.
• $$T$$ = kinetic energy with respect to the origin O.

Theorem:

$T = T_{C} + \frac{1}{2}M\overline{v}^{2}\tag{3.10.1}\label{eq:3.10.1}$

Thus:

$$T = \frac{1}{2}\sum m_{i}{v}^{2}_{i} = \frac{1}{2} \sum m_{i} ({\bf v} ^{\prime}_{i} + \overline{{\bf v} })\cdot ({\bf v} ^{\prime}_{i} + \overline{{\bf v} })$$

$$= \frac{1}{2}\sum m{v}^{\prime 2}_{i} \times \overline{{\bf v} } \sum m{{\bf v} }^{\prime}_{i} + \frac{1}{2} v^{-2} \sum m_{i}$$.

$$\therefore \qquad T = T_{C} + \frac{1}{2}M\overline{v}^{2}$$.

Corollary:

If $$\overline{{\bf v} } = 0, T = T_{C}$$ . (Think about what this means.)

Corollary:

Corollary: For a non-rotating rigid body, $$T_{C}$$ = 0, and therefore $$T = \frac{1}{2}M\overline{v}^{2}$$