Rotational energy

Rotational energy is the kinetic energy of a rigid body (example: flywheel ) that rotates around a fixed point or its (movable) center of mass . In these two cases, the kinetic energy of the body can be broken down into a translational and a rotational component. This energy is dependent on the moment of inertia and the angular velocity of the body: the further the mass is away from the axis of rotation, the more energy the body releases when its rotation is stopped.

This can be illustrated by the following experiment: Two balls of the same weight with identical radii are placed on an inclined plane and roll down, see wheel rolling down an inclined plane . A ball consists of a light material such as plastic and is made solid. The other ball, however, is hollow, but made of a denser and therefore heavier material than plastic. The hollow ball will roll more slowly, because its entire mass is distributed on a thin shell at a certain distance from the axis of rotation. The massive ball with the same mass rolls faster because there is more mass near the axis of rotation and therefore has to move more slowly on the circular path. Therefore less of its positional energy is converted into rotational energy and more into translational energy and it rolls faster.

Rotational energy is important for: turbines , generators , wheels and tires , shafts , propellers .

Moment of inertia

A body that rotates around the x-axis with angular velocity has the rotational energy ${\displaystyle \omega }$

${\displaystyle E_{\mathrm {rot} }={\frac {1}{2}}\cdot J_{x}\cdot \omega ^{2}}$

With

• ${\displaystyle J_{x}}$: Moment of inertia of the body around the x-axis
• ${\displaystyle \omega }$: Angular velocity .

This can be expressed broadly as:

${\displaystyle {\begin{aligned}E_{\mathrm {rot} }&={\frac {1}{2}}\;{\vec {\omega }}^{T}\,J\;{\vec {\omega }}\\&={\frac {1}{2}}\sum _{\alpha ,\beta =1}^{3}\,J_{\alpha \beta }\;\omega _{\alpha }\;\omega _{\beta }\end{aligned}}}$

With

• ${\displaystyle J_{\alpha \beta }}$: Inertia tensor .

To specify the energy of a body that rotates around any axis ( unit vector with ), the angular velocity is expressed in each case by its vector components in the x, y and z directions: ${\displaystyle {\vec {n}}}$ ${\displaystyle \left|{\vec {n}}\right|=1}$

${\displaystyle {\vec {\omega }}=\omega \,{\vec {n}}=\omega \cdot {\begin{pmatrix}n_{1}\\n_{2}\\n_{3}\\\end{pmatrix}}}$

The following applies to the rotational energy:

${\displaystyle {\begin{aligned}\Rightarrow E_{\mathrm {rot} }&={\frac {1}{2}}\sum _{\alpha ,\beta =1}^{3}J_{\alpha \beta }\;n_{\alpha }\;n_{\beta }\;\omega ^{2}\\&={\frac {1}{2}}\cdot J_{n}\cdot \omega ^{2}\end{aligned}}}$

with the moment of inertia with respect to any axis : ${\displaystyle J_{n}}$ ${\displaystyle {\vec {n}}}$

${\displaystyle J_{n}=\sum _{\alpha ,\beta =1}^{3}J_{\alpha \beta }\;n_{\alpha }\;n_{\beta }}$

Examples

• A sphere with a radius has the moment of inertia . If it rolls on the plane at speed , its angular velocity, and consequently its total kinetic energy, is: ${\displaystyle r}$ ${\displaystyle J={\tfrac {2}{5}}\,mr^{2}}$ ${\displaystyle v}$ ${\displaystyle \omega ={\tfrac {v}{r}}}$

${\displaystyle {\begin{aligned}E_{\mathrm {kin} }&=E_{\mathrm {trans} }+E_{\mathrm {rot} }\\&={\frac {1}{2}}mv^{2}+{\frac {1}{2}}\cdot {\frac {2}{5}}mr^{2}\cdot \omega ^{2}\\&={\frac {7}{10}}mv^{2}\end{aligned}}}$

• A body that rotates around the diagonal through its xy surface has the angular velocity:

${\displaystyle {\vec {\omega }}=\omega \,{\vec {n}}}$ With ${\displaystyle {\vec {n}}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\1\\0\\\end{pmatrix}}}$

From this it follows for the moment of inertia with respect to this axis of rotation:

${\displaystyle {\begin{aligned}J_{n}&={\vec {n}}^{T}J\,{\vec {n}}\\&={\frac {1}{\sqrt {2}}}\left(1,1,0\right)\left({\begin{matrix}J_{11}&J_{12}&J_{13}\\J_{12}&J_{22}&J_{23}\\J_{13}&J_{23}&J_{33}\\\end{matrix}}\right){\frac {1}{\sqrt {2}}}\left({\begin{matrix}1\\1\\0\\\end{matrix}}\right)\\&={\frac {1}{2}}\cdot (J_{11}+J_{12}+J_{12}+J_{22})\end{aligned}}}$

The rotational energy is obtained from:

${\displaystyle {\begin{aligned}E_{\mathrm {rot} }&={\frac {1}{2}}\cdot J_{n}\cdot \omega ^{2}\\&={\frac {1}{4}}\cdot (J_{11}+J_{12}+J_{12}+J_{22})\cdot \omega ^{2}\end{aligned}}}$

Angular momentum

The rotational energy can also be expressed by the angular momentum : ${\displaystyle {\vec {L}}}$

${\displaystyle {\begin{aligned}E_{\mathrm {rot} }&={\frac {1}{2}}\cdot {\vec {L}}\cdot {\vec {\omega }}\\&={\frac {1}{2}}\cdot {\frac {{\vec {L}}^{2}}{J}}\end{aligned}}}$

With ${\displaystyle {\vec {L}}=J\cdot {\vec {\omega }}}$

It should be noted that the angular momentum and the angular velocity are generally not parallel to each other (except for rotation around a main axis of inertia ); see also ellipsoid of inertia .

Derivation

Let the rigid body be given by individual mass points with masses m _i at the locations relative to the origin of a fixed reference system that is located at the location in the inertial system . For the general motion of rigid bodies , Euler's equation of velocity applies : ${\displaystyle {\vec {r}}_{i}}$ ${\displaystyle {\vec {b}}(t)}$

${\displaystyle {\vec {v}}({\vec {x}},t)={\dot {\vec {b}}}(t)+{\vec {\omega }}(t)\times {\vec {x}}\,.}$

This is the angular velocity of the rigid body (including the reference system fixed to the body), the velocity of and both may depend on the time t . The speed at time t is the speed of the mass point at the location in the body-fixed reference system. ${\displaystyle {\vec {\omega }}}$ ${\displaystyle {\dot {\vec {b}}}}$ ${\displaystyle {\vec {b}}}$ ${\displaystyle {\vec {v}}({\vec {x}},t)}$ ${\displaystyle {\vec {x}}}$

The kinetic energy of the body is then given by

${\displaystyle {\begin{aligned}E_{\rm {kin}}=&{\frac {1}{2}}\sum _{i}m_{i}{\vec {v}}({\vec {r}}_{i},t)^{2}={\frac {1}{2}}\sum _{i}m_{i}({\dot {\vec {b}}}+{\vec {\omega }}\times {\vec {r}}_{i})^{2}\\=&{\frac {1}{2}}\sum _{i}m_{i}{\dot {\vec {b}}}^{2}+{\frac {1}{2}}\sum _{i}m_{i}2{\dot {\vec {b}}}\cdot ({\vec {\omega }}\times {\vec {r}}_{i})+{\frac {1}{2}}\sum _{i}m_{i}({\vec {\omega }}\times {\vec {r}}_{i})^{2}\\=&\underbrace {{\frac {1}{2}}m{\dot {\vec {b}}}^{2}} _{E_{\rm {trans}}}+\underbrace {{\frac {1}{2}}\sum _{i}m_{i}({\vec {\omega }}\times {\vec {r}}_{i})^{2}} _{E_{\rm {rot}}}+m{\vec {\omega }}\cdot ({\vec {s}}\times {\dot {\vec {b}}})\end{aligned}}}$

This is the total mass of the body, E _{trans is} its translational energy, E _{rot is} its rotational energy, its center of mass and it was exploited that in the late product of three vectors their order may be cyclically reversed. The third term vanishes under four conditions: ${\displaystyle \textstyle m:=\sum _{i}m_{i}}$ ${\displaystyle \textstyle {\vec {s}}:={\frac {1}{m}}\sum _{i}m_{i}{\vec {r}}_{i}}$ ${\displaystyle m{\vec {\omega }}\cdot ({\vec {s}}\times {\dot {\vec {b}}})}$

1. If the center of mass lies at the origin ( ) or on the axis of rotation ( ), i.e. the rotation takes place around the center of mass. ${\displaystyle {\vec {s}}={\vec {0}}}$ ${\displaystyle {\vec {\omega }}\parallel {\vec {s}}}$
2. When the body-fixed system is at rest ( ) or moves along the axis of rotation ( ), which can always be established by a suitable choice of the reference point. ${\displaystyle {\dot {\vec {b}}}={\vec {0}}}$ ${\displaystyle {\vec {\omega }}\parallel {\dot {\vec {b}}}}$
3. When the reference point moves towards the center of mass ( ), which amounts to a balancing act . ${\displaystyle {\dot {\vec {b}}}\parallel {\vec {s}}}$
4. The trivial case is not considered further here. ${\displaystyle m{\vec {\omega }}={\vec {0}}}$

In the first three cases the kinetic energy is split into translational and rotational energy, but only the first two cases are of interest for gyroscopic theory . With the Lagrange identity , using the properties of the dyadic product, the rotational energy is calculated as: ${\displaystyle ({\vec {\omega }}\times {\vec {r}}_{i})^{2}=({\vec {\omega }}\cdot {\vec {\omega }})({\vec {r}}_{i}\cdot {\vec {r}}_{i})-({\vec {\omega }}\cdot {\vec {r}}_{i})^{2}}$ ${\displaystyle \otimes }$

${\displaystyle {\begin{aligned}2E_{\rm {rot}}=&\sum _{i}m_{i}({\vec {\omega }}\times {\vec {r}}_{i})^{2}=\sum _{i}m_{i}[({\vec {\omega }}\cdot {\vec {\omega }})({\vec {r}}_{i}\cdot {\vec {r}}_{i})-({\vec {\omega }}\cdot {\vec {r}}_{i})^{2}]\\=&{\vec {\omega }}\cdot \sum _{i}m_{i}[({\vec {r}}_{i}\cdot {\vec {r}}_{i}){\vec {\omega }}-({\vec {\omega }}\cdot {\vec {r}}_{i}){\vec {r}}_{i}]={\vec {\omega }}\cdot \sum _{i}m_{i}[({\vec {r}}_{i}\cdot {\vec {r}}_{i})\mathbf {1} -{\vec {r}}_{i}\otimes {\vec {r}}_{i}]\cdot {\vec {\omega }}\\=&{\vec {\omega }}\cdot \mathbf {\Theta } _{b}\cdot {\vec {\omega }}\\\Rightarrow E_{\rm {rot}}=&{\frac {1}{2}}{\vec {\omega }}\cdot \mathbf {\Theta } _{b}\cdot {\vec {\omega }}={\frac {1}{2}}{\vec {\omega }}\cdot {\vec {L}}\end{aligned}}}$

Therein is the inertia tensor of the rigid body with respect to , its own angular momentum , and 1 of the unit tensor . In the body-fixed system the inertia tensor is constant, but in the inertial system it is not when the body rotates. ${\displaystyle \textstyle \mathbf {\Theta } _{b}:=\sum _{i}m_{i}[({\vec {r}}_{i}\cdot {\vec {r}}_{i})\mathbf {1} -{\vec {r}}_{i}\otimes {\vec {r}}_{i}]}$ ${\displaystyle {\vec {b}}}$ ${\displaystyle {\vec {L}}:=\mathbf {\Theta } _{b}\cdot {\vec {\omega }}}$

See also

• Potential energy
• Scooter replacement mass
• Flywheel storage
• Flywheel storage power plant

Footnotes

1. ↑ ^{a b} Institute for Physics at the University of Rostock (ed.): Theoretical Physics II - Theoretical Mechanics . Chapter 5 - Rigid Body and Top Theory. ( uni-rostock.de [PDF; accessed on June 6, 2017]).
2. ↑ The dyadic product is defined with any three vectors by ${\displaystyle {\vec {a}},{\vec {b}},{\vec {c}}}$ ${\displaystyle ({\vec {a}}\otimes {\vec {b}})\cdot {\vec {c}}:=({\vec {b}}\cdot {\vec {c}}){\vec {a}}}$