# Bead on a smooth rotating rod

Suppose a bead ($m$) is free to move on a thin rod in the otherwise empty space. The rod is made to rotate at constant angular speed $\omega$. Lets assume the initial position of the bead is $(r_o, \theta_o)$. To solve this problem I can write down the Lagrangian of the system.

$$L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)$$

Hence the equation of motion will be, (note $\dot{\theta}=\omega\rightarrow$ constant)

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right)=\frac{\partial L}{\partial r} \implies \ddot{r} = r\dot{\theta}^2 = r\omega^2 \implies r=r_o e^{\omega t}$$

And

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)=\frac{\partial L}{\partial \theta} \implies \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)= 0 \implies r^2 \omega = constant \implies r = constant$$

I don't understand where I have made any mistake.

• It appears that the two members of the Lagrangian have different units, the second containing a $rad^2$. – Energizer777 Nov 13 '15 at 17:57
• @Energizer777 Dimensionally it is correct. $r\dot{\theta}$ is a velocity. – The Imp Nov 13 '15 at 18:01
• If you are considering $\dot{\theta}$ to be constant, I dont think you should be doing $\frac{\partial L}{\partial \dot{\theta}}$ – Oswald Nov 13 '15 at 18:34
• Well, for starters, which do you think is the right one? – Kyle Kanos Nov 13 '15 at 18:35
• First one?, since $\dot{\theta}$ is a constant – Oswald Nov 13 '15 at 18:36

## 1 Answer

Here we have a holonomic constraint: $\theta-\omega t=\theta_0$ (or $\dot{\theta}=\omega$). The question is where we should use it in solving the Lagrange's equations. At first you obtained equations from the Lagrangian for the free particle, solved them, and then used the constraint $\dot{\theta}=\omega$. But let us remember how to solve the classical problem of a simple pendulum. That is a problem of motion of a point particle in the gravity field of the Earth with the condition (a holonomic constraint): $r=l$ — the length of the pendulum. And it (the constraint) was used in constructing the Lagrangian of the system. That is instead of the Lagrangian $$L =\frac m 2 (\dot r^2+r^2\dot\phi^2)+mgr\cos\phi$$ we was used $$L =\frac m 2 l^2\dot\phi^2+mgl\cos\phi.$$ Thus in your problem after a correct using of the condition $\dot{\theta}=\omega$, the function $L$ is $$\frac m 2 (\dot r^2+r^2\omega^2)$$ and not depend from $\theta, \dot\theta.$ Therefore the equations of motion will be: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right)=\frac{\partial L}{\partial r} \implies \ddot{r} = \omega^2r \implies r=r_o e^{\omega t},\\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)=\frac{\partial L}{\partial \theta} \implies 0=0.$$