Problem — König's theorem for kinetic energy

Problem

Show, for a rigid body in general motion, that the total kinetic energy can be written as the sum of the translational kinetic energy of the centre of mass and the rotational kinetic energy about the CM. Apply it to a ball rolling without slipping to find the fraction of energy in rotational form.

Solution

Write the velocity of each piece of the body as the sum of the centre-of-mass velocity plus the velocity relative to the CM: $\vec{v}_i = \vec{v}_\text{CM} + \vec{v}_i'$. The total kinetic energy is $E_\text{cin} = \sum_i \tfrac{1}{2}m_i v_i^2 = \sum_i \tfrac{1}{2}m_i\left(\vec{v}_\text{CM} + \vec{v}_i'\right)\cdot\left(\vec{v}_\text{CM} + \vec{v}_i'\right)$ Expanding the square gives three terms: $E_\text{cin} = \tfrac{1}{2}\Big(\sum_i m_i\Big)v_\text{CM}^2 + \vec{v}_\text{CM}\cdot\Big(\sum_i m_i\vec{v}_i'\Big) + \sum_i \tfrac{1}{2}m_i v_i'^2$ The middle term is zero, because $\sum_i m_i\vec{v}_i' = \dfrac{d}{dt}\sum_i m_i\vec{r}_i' = 0$ (the momentum relative to the CM is zero by definition). Two terms remain: $\ev{E_\text{cin} = \tfrac{1}{2}M v_\text{CM}^2 + \tfrac{1}{2}I_\text{CM}\,\omega^2}$ Application to a rolling ball (solid sphere, $I_\text{CM} = \tfrac{2}{5}mR^2$, constraint $\omega = v/R$): $E_\text{rot} = \tfrac{1}{2}\cdot\tfrac{2}{5}mR^2\cdot\frac{v^2}{R^2} = \tfrac{1}{5}mv^2, \qquad E_\text{tot} = \tfrac{1}{2}mv^2 + \tfrac{1}{5}mv^2 = \tfrac{7}{10}mv^2$ The rotational fraction is $\dfrac{E_\text{rot}}{E_\text{tot}} = \dfrac{1/5}{7/10} = \dfrac{2}{7} \approx 29\%$.

Connections

Topics: Dinamica rotazionale Concepts: Energia cinetica rotazionale · Rotolamento Methods: Teorema di König