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3.10: Kinetic energy

  • Page ID
    8382
  • [ "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 \( T = \frac{1}{2}M\overline{v}^{2}\)

    (Think about what this means.)

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