Proof of work-energy theorem for a rigid body

Question

Could anyone show me a way to derive the work energy theorem for a rigid body whose motion is along a fixed axis ( such as of a cylinder rolling on a plane) which states that states that $W=\frac{1}{2} m V_{c m}^{2}+\frac{1}{2} I \omega^{2}$ using the basic definition that $W=\int \vec{F} \cdot d r$.

Answer 1

Starting with the EOM's: \begin{align*} &\textbf{Translation}\\ &m\,\ddot{r}_{cm}=F\\ &\dot{r}_{cm}\,m\,\ddot{r}_{cm}=\dot{r}_{cm}\, F\\ &\frac{m}{2}\frac{d}{dt}\left(\dot{r}^2_{cm}\right)=\frac{d}{dt}{r}_{cm}\, F\\ &\int \frac{m}{2}{d}\left(\dot{r}^2_{cm}\right)=\int {d}{r}_{cm}\, F\\ &\frac{m}{2}\,\left(\dot{r}^2_{cm}\right)=\int {d}{r}_{cm}\, F\\\\ &\textbf{Rotation}\\ &I\,\ddot{\varphi}_{cm}=F\,R\\\\ &\text{analog}\\ &\frac{I}{2}\,\left(\dot{\varphi}\right)^2=\int {d}{\varphi}\, F\,R\\\\ &\text{thus work for the total energy $~T=\frac{m}{2}\,\left(\dot{r}^2_{cm}\right)+\frac{I}{2}\,\left(\dot{\varphi}\right)^2~$is }\\ &\int {d}{r}_{cm}\, F+\int {d}{\varphi}\, F\,R= \int F\,\left(dr_{cm}+R\,d\varphi\right)=\int F\,dr\\ &\text{with}\\ &dr=dr_{cm}+R\,d\varphi \end{align*}

Remark:

F is the constraint force between the cylinder and the plane and $~\dot\varphi=\omega$