# Mass fixed to a wheel - mechanics problem [closed]

I am struggling with a problem from classical mechanics. Imagine a massless wheel (to make it simpler) with a mass $m$ fixed to it rolling without slipping on a horizontal ground. If we now try to find the equations of motion of the wheel (for instance the angle $\alpha$ it turns) we will find that all the forces are independent of velocity, so $\alpha ''=f(\alpha)$.

After doing that I decided to solve this problem using Euler-Langrange equations (since friction does no work). I came up with $L=\frac{1}{2} m R^{2} (\frac{d\alpha}{dt})^{2} (1+\cos \alpha)-mgR(1+\cos \alpha)$ which, upon solving, gives $\alpha ''$ as a function of both $\alpha$ and $\alpha '$. What is my problem?

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## closed as off-topic by tpg2114, Qmechanic♦Nov 3 '13 at 2:25

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Maybe I'm just being stupid, but I don't see where the $1+\cos\alpha$ in the first term of the Lagrangian comes from. The kinetic energy term in the Lagrangian is just ${1\over 2}mR^2\dot\alpha^2$. –  Ted Bunn Mar 27 '11 at 14:25
Ted's correction will also eliminate the $\alpha'$-dependent terms from the equation for $\alpha''$. –  Luboš Motl Mar 27 '11 at 15:22
Sorry, forgot the most important part - the wheel is rolling without slipping on the horizontal ground. I'll edit this asap –  malina Mar 27 '11 at 16:33
Sorry, Malina, we were obviously assuming that the wheel wasn't slipping. The comments above fully apply and your last comment makes no difference. –  Luboš Motl Mar 27 '11 at 16:36
The kinetic energy of the mass is due to moving around the center of the wheel and due to the linear motion of the wheel. There is no slipping, so the linear displacement of the wheel equals $R \alpha$. Now the displacement in x-direcetion is $R\alpha + R \sin \alpha$, where the first term is due to displacement of the wheel, and the second one is the relative position of the mass with respect to the centre of the wheel. In y-direction its $R+R \cos \alpha$. What's given in the post is $\frac{m}{2}((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)$ –  malina Mar 27 '11 at 16:48

(Updated to reflect my better understanding of the problem, based on our discussion in the comments.)

In my comments above, I didn't understand the question. Now that I do, it seems to me that your Lagrangian is correct, except that the factor $1\over 2$ in the kinetic energy term doesn't belong. The equation of motion does involve $\dot\alpha$. I don't think I understand why that's a problem. I can't think of a general argument that proves that there shouldn't be any first derivatives in the Euler-Lagrange equations in a system like this.

For instance, consider a force-free particle moving in two dimensions. If we express things in polar coordinates, the Lagrangian is $$L={1\over 2}m(\dot r^2+r^2\dot\phi^2).$$ The Euler-Lagrange equation for $\phi$ reduces to $$2\dot r\dot\phi+r\ddot\phi=0.$$ This involves first derivatives, even though there's no force at all (hence a fortiori no velocity-dependent force).

In subsequent comments (below) you explained your reason for not expecting there to be $\dot\alpha$ terms: calculating the torque about the instantaneous point of contact and setting that equal to $I\ddot\alpha$ yields an equation with no $\dot\alpha$'s. The reason that doesn't work is that the torque is equal to $d{\bf l}/dt=d(I\dot\alpha)/dt$. Since $I$ is a function of time, this is not equal to $I\ddot\alpha$. There's a bit more detail in the comments below. Anyway, the final equation of motion, derived either from the Euler-Lagrange equation or from torque considerations, comes out to $$2(1+\cos\alpha)\ddot\alpha=(\dot\theta^2+g/R)\sin\alpha.$$

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Thank you for your answer. Could you then help me detect the fault in the following reasoning to the problem I posted: the point of contant isn't moving and we can consider it to be an instantaneous center of rotation. The only external force that has a moment with respect to that point is the gravity on the mass. Then the equation of motion is $I\alpha''=mgd \sin \frac{\alpha}{2}$ where $d$ is the distance from the point of contact to the mass. Now $d=2R\cos \frac{\alpha}{2}$ and $I=md^2$, hence we have an equation of motion independent of the angular velocity –  malina Mar 27 '11 at 18:02
I think the answer is that you want to set the torque equal to the rate of change of angular momentum, which is ${d\over dt}(I\dot\alpha)$. The moment of inertia $I$ is time-dependent, so this is not equal to $I\ddot\alpha$. –  Ted Bunn Mar 27 '11 at 21:10
I managed to work through the problem both ways (Euler-Lagrange and torque=rate of change of angular momentum) and got the same equation of motion. I hadn't noticed before, but there is a factor of 2 off in your Lagrangian: the kinetic energy term shouldn't have that $1\over 2$. Sketch of the torque calculation in the next comment. –  Ted Bunn Mar 28 '11 at 17:45
Hah, I just realised that was the mistake and ther reason different methods didn't agree. Anyway, thanks for the second comment, it was very useful! –  malina Mar 28 '11 at 17:49
Say we're evaluating the torque at time $t_0$, when the angle is $\alpha_0$. Our axis is the point of contact at time $t_0$. For arbitrary time $t$, the position of the mass, relative to our chosen origin, has Cartesian components $${\bf r}=R(\alpha-\alpha_0+\sin\alpha,1+\cos\alpha)$$. The speed is $${\bf v}=\dot{\bf r}=R\dot\alpha(1+\cos\alpha,-\sin\alpha)$$. Find the angular momentum $l={\bf r}\times m{\bf v}$. Calculate $dl/dt$ at $t=t_0$ and set this equal to the torque. The result is $$2(1+\cos\alpha)\ddot\alpha=(\dot\alpha^2+g/R)\sin\alpha,$$ which matches the Euler-Lagrange equation. –  Ted Bunn Mar 28 '11 at 17:52