Scooter replacement mass

The roller substitute mass is a calculation variable that has to be added to the real physical mass of a rotationally symmetrical rigid body in order to mathematically replace its rotational energy with additional translational kinetic energy . ${\displaystyle m_{RE}}$

The kinetic energy of a body that translates and rotates (e.g. a wheel rolls along a surface) corresponds to the kinetic energy of a body with a greater mass that only translates: ${\displaystyle m_{2}>m_{1}}$

${\displaystyle {\begin{alignedat}{2}E_{\text{kin1}}&&&=E_{\text{kin2}}\\\Leftrightarrow E_{\text{trans1}}&+E_{\text{rot}}&&=E_{\text{trans2}}\\\Leftrightarrow {\frac {1}{2}}\cdot m_{1}\cdot v^{2}&+{\frac {1}{2}}\cdot J\cdot \omega ^{2}&&={\frac {1}{2}}\cdot m_{2}\cdot v^{2}\end{alignedat}}}$

With a known moment of inertia and by replacing the angular velocity (since the external path velocity when rolling corresponds exactly to the translational velocity ), the roller substitute mass is obtained: ${\displaystyle J}$ ${\displaystyle \omega ={\frac {v}{r}}}$ ${\displaystyle v}$

${\displaystyle E_{\text{rot}}={\frac {1}{2}}\cdot J\cdot \left({\frac {v}{r}}\right)^{2}={\frac {1}{2}}\cdot m_{RE}\cdot v^{2}}$

${\displaystyle \Rightarrow }$ ${\displaystyle m_{RE}={\frac {J}{r^{2}}}}$

From this it follows for the kinetic energy:

${\displaystyle \Rightarrow {\frac {1}{2}}\cdot m_{1}\cdot v^{2}+{\frac {1}{2}}\cdot m_{RE}\cdot v^{2}={\frac {1}{2}}\cdot m_{2}\cdot v^{2}}$

or for the calculated total mass:

${\displaystyle \Rightarrow m_{1}+m_{RE}=m_{2}}$

example

Using the example of a sphere, it looks like this: ${\displaystyle \left(J={\frac {2}{5}}\cdot m\cdot r^{2}\right)}$

${\displaystyle {\begin{aligned}E_{\text{rot}}&={\frac {1}{2}}\cdot J\cdot \omega ^{2}\\&={\frac {1}{2}}\cdot \left({\frac {2}{5}}\cdot m\cdot r^{2}\right)\cdot \left({\frac {v}{r}}\right)^{2}\\&={\frac {1}{2}}\cdot \left({\frac {2}{5}}\cdot m\right)\cdot v^{2}\end{aligned}}}$

so is

${\displaystyle \Rightarrow m_{\text{Ersatz}}={\frac {2}{5}}\cdot m}$

The kinetic energy of the sphere is thus:

${\displaystyle {\begin{aligned}\Rightarrow E_{\text{kin}}&={\frac {1}{2}}\cdot m\cdot v^{2}+{\frac {1}{2}}\cdot \left({\frac {2}{5}}\cdot m\right)\cdot v^{2}\\&={\frac {1}{2}}\cdot \left(m+{\frac {2}{5}}\cdot m\right)\cdot v^{2}\\&={\frac {1}{2}}\cdot \left({\frac {7}{5}}\cdot m\right)\cdot v^{2}\\&={\frac {7}{10}}\cdot m\cdot v^{2}\end{aligned}}}$