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 $W=\int \vec{F} \cdot d r$.

  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/606476/2451 $\endgroup$
    – Qmechanic
    Jan 11, 2021 at 8:07
  • $\begingroup$ can you give me please your comment on my answer. if you ask a question we need feedback to our answer! $\endgroup$
    – Eli
    Jan 12, 2021 at 8:51

1 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*}


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

  • $\begingroup$ Sorry dear Eli for my delayed response, I was caught up in some issues. Thank you for your response. I was however looking for a derivation that uses only the integral of $F.ds$ that would begin with it . You broke the integral from the beginning into two integrals which is not what I was looking for. $\endgroup$
    – Kashmiri
    Jan 13, 2021 at 14:19
  • $\begingroup$ no problem, thank for your response, the last equation you see only the integral that you are looking for? $\endgroup$
    – Eli
    Jan 13, 2021 at 14:22
  • $\begingroup$ Yes , can you explain it $\endgroup$
    – Kashmiri
    Jan 13, 2021 at 15:51
  • $\begingroup$ you don't have only translation motion , you have translation plus rotation motion thus $~dr\mapsto dr_{cm}+R\,d\varphi~$ see also physics.stackexchange.com/questions/606476/… $\endgroup$
    – Eli
    Jan 13, 2021 at 16:51

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