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The fundamental manifestation of mechanical energy, kinetic energy is the energy associated with an object's translational and/or rotational motion. Kinetic energy provides the definition of work (and hence all other forms of mechanical energy) through the Work-Kinetic Energy Theorem. |
Mathematical Definition
Translational Kinetic Energy of a Point Particle
The kinetic energy of a point particle is given by:
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}{composition-setup} {excerpt}The fundamental manifestation of [mechanical energy], kinetic energy is the energy associated with an object's translational and/or rotational motion. Kinetic energy provides the definition of [work] (and hence all other forms of [mechanical energy]) through the [Work-Kinetic Energy Theorem].{excerpt} h3. Mathematical Definition h4. Translational Kinetic Energy of a Point Particle The kinetic energy of a point particle is given by: {latex}\begin{large}\[ K = \frac{1}{2}mv^{2}\]\end{large}{latex} h4. Translational Kinetic Energy of a System Since energy is a [scalar], the kinetic energy of a system of [point particles|point particle] is the sum of the kinetic energies of the constituents: {latex} |
Translational Kinetic Energy of a System
Since energy is a scalar, the kinetic energy of a system of point particles is the sum of the kinetic energies of the constituents:
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\begin{large}\[ K^{\rm sys} = \sum_{j = 1}^{N} \frac{1}{2}m_{j}v_{j}^{2}\]\end{large}{latex}
where _N_ is the number |
where N is the number of system constituents.
Kinetic Energy of a Rigid Body
Consider a rigid body that can rotate and translate. We begin by treating the rigid body as a collection of point masses that are translating with the center of mass of the body and also rotating about it with angular velocity ω. We therefore write the velocity of each point as a sum of rotational and translational parts:
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of system constituents. h4. Kinetic Energy of a Rigid Body Consider a [rigid body] that can rotate and translate. We begin by treating the rigid body as a collection of point masses that are translating with the center of mass of the body and also rotating about it with angular velocity ω. We therefore write the velocity of each point as a sum of rotational and translational parts: {latex}\begin{large}\[ \vec{v}_{j} = \vec{v}_{\rm cm} + \vec{\omega}\times \vec{r}_{j} \]\end{large}{latex} where _r_~j~ is the position of the j{^}th^ particle measured from the body's axis of rotation passing through the center of mass. With this split, the kinetic energy of the body becomes: {latex} |
where rj is the position of the jth particle measured from the body's axis of rotation passing through the center of mass.
With this split, the kinetic energy of the body becomes:
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\begin{large}\[ K = \sum_{j} \frac{1}{2}m_{j}(v_{\rm cm}^{2} + \vec{v}_{\rm cm}\cdot(\vec{\omega}\times \vec{r}_{j}) + \omega^{2}r_{j}^{2})\]\end{large}{latex}
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The
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center
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term
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will
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equal
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zero,
...
because ω and vcm are constants, so:
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ω and _v_~cm~ are constants, so: {latex}\begin{large}\[ \sum_{j} \frac{1}{2}m_{j}\vec{v}_{\rm cm}\cdot(\vec{\omega}\times \vec{r}_{j}) = \frac{1}{2}\vec{v}_{\rm cm}\cdot\left(\vec{\omega}\times \sum_{j} m_{j}\vec{r}_{j}\right)\]\end{large}{latex} |
and
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the
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sum
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over
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mjrj is constrained to equal zero because we have assumed the center of mass is at the position r = 0 in our coordinates. With this realization, and using the definition of the moment of inertia, we have:
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_~j~_r_~j~ is constrained to equal zero because we have assumed the center of mass is at the position _r_ = 0 in our coordinates. With this realization, and using the definition of the [moment of inertia], we have: {latex}\begin{large}\[ K = \frac{1}{2}m_{\rm tot}v_{\rm cm}^{2} + \frac{1}{2}I_{\rm cm}\omega^{2}\]\end{large}{latex} |
This
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result
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shows
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that
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the
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kinetic
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energy
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of
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a
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rigid
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body
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can
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be
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broken
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into
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two
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parts,
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generally
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known
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as
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the
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translational
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part
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and
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the
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rotational
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part.
Rotational Kinetic Energy
The above formula suggests a definition for the kinetic energy of a rotating body:
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h4. Rotational Kinetic Energy The above formula suggests a definition for the kinetic energy of a rotating body: {latex}\begin{large}\[ K^{\rm rot} = \frac{1}{2}I\omega^{2} \]\end{large}{latex} |