# How to use the variational method to solve this problem

I see this mechanical problem here.

I want to solve this problem with the variational method. The Lagrangian of this system is obtained by subtracting potential energy from kinetic energy.

m = 1;
g = 9.8;
R = 1;
EulerEquations[
m*g*R (1 - Cos[θ[t]]) - m*g*R*Cos[θ[t]], θ[
t], t](*L=T-V or L=kinetic energy - potential energy*)


But the result is $$19.6 (\sin (\theta [t]))=0$$, which has no significance to solve this problem.

I know that by listing the Lagrangian of this system, we can get the correct equation of motion:

<< VariationalMethods
L = 1/2 m (R θ'[t])^2 - m g R Cos[θ[t]];
EulerEquations[L, θ[t], t] //
FullSimplify[#, Assumptions -> R > 0 && m > 0] &


But before the object moves along the sphere and does not separate, the kinetic energy of the system $$m g R (1-\cos (\theta (t)))=\frac{1}{2} m \left(R \theta '(t)\right)^2$$.

I want to know what mistakes I made in listing the following Lagrangian of this system(L= m*g*R (1 - Cos[θ[t]]) - m*g*R*Cos[θ[t]]), and how to use the variational method to solve this problem correctly.

• what is it you want to "solve" for? Do you want the angle of separation? if so, I can show you how. I solved this same exact problem before. But all by hand, not using the computer, and the package you are using. – Nasser Jun 6 '20 at 10:39
• @Nasser I think though the question is out of scope of this site (it belongs to physics.SE, I believe), it's relatively clear. OP wants to know what's wrong with the first attempt, and I think I know the answer: "But before the object moves along the sphere and does not separate, the kinetic energy of the system $mgR(1−\cos(θ(t)))=\frac{1}{2}m(Rθ'(t))^2$." This holds only if the initial conditions are $\theta(0)=0$ and $\theta'(0)=0$. If you use these initial conditions for r*θ''[t] == g*Sin[θ[t]], the solution will be $\theta(t)=0$, so the output of the first code sample is correct. – xzczd Jun 6 '20 at 11:35
• You need to add constraint. So your Lagrangian is missing a constraint $\lambda$. As I said, this is standard problem, was a HW problem for a mechanics class I took few years, and I have the latex solution, but I did not solve it using Mathematica, all by hand. So can't help with using these function you want to use. I can post my solution, but I am afraid it has no Mathematica any thing in it, so I am not sure if one is allowed to do so. If this question was in the Physics form I could do this. – Nasser Jun 6 '20 at 11:42
• Ok, I can post it. But if someone gets mad, I can tell them you said it is OK to do. – Nasser Jun 6 '20 at 11:47
• The analysis is just a trivial one based on energy conservation. However, if you ask e.g. "assuming θ[t0]==0, why doesn't L = 1/2 m (R θ'[t0])^2+m*g*R (1 - Cos[θ[t]]) - m*g*R*Cos[θ[t]]` work?", as someone never learned about Lagrangian mechanics, I don't know the answer, either. Once again, I believe this should be asked in physics.SE. – xzczd Jun 8 '20 at 6:17

This is the problem I solved sometime ago. Same problem as you show. The angle $$\theta$$ used is measured from x-axis, positive anti-clockwise, as it was simpler to do so, but it does not affect the solution ofcourse. Since there is no Mathematica stuff in this, I can add a Manipulate later on if needed to show the ball falling of?
There are two coordinates $$r,\theta$$ (polar) which is the position vector of the ball, and one constraint $$$$f\left( r,\theta\right) =r-R=0 \tag{1}$$$$
$$R$$ above is the radius of the hemisphere.
Now we set up the equations of motion for $$m$$ \begin{align*} T & =\frac{1}{2}m\left( \dot{r}^{2}+r^{2}\dot{\theta}^{2}\right) \\ U & =mgr\sin\theta\\ L & =T-U\\ & =\frac{1}{2}m\left( \dot{r}^{2}+r^{2}\dot{\theta}^{2}\right) -mgr\sin\theta \end{align*} Hence the Euler-Lagrangian equations are (we need to add contraint)
\begin{align} \frac{d}{dt}\frac{\partial L}{\partial\dot{r}}-\frac{\partial L}{\partial r}+\lambda\frac{\partial f}{\partial r} & =0\tag{2}\\ \frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}-\frac{\partial L} {\partial\theta}+\lambda\frac{\partial f}{\partial\theta} & =0 \tag{3} \end{align} But \begin{align*} \frac{d}{dt}\frac{\partial L}{\partial\dot{r}} & =m\ddot{r}\\ \frac{\partial L}{\partial\dot{\theta}} & =mr^{2}\dot{\theta}\\ \frac{d}{dt}\left( \frac{\partial L}{\partial\dot{\theta}}\right) & =m\left( 2r\dot{r}\dot{\theta}+r^{2}\ddot{\theta}\right) \\ \frac{\partial L}{\partial r} & =mr\dot{\theta}^{2}-mg\sin\theta\\ \frac{\partial L}{\partial\theta} & =-mgr\cos\theta\\ \frac{\partial f}{\partial r} & =1\\ \frac{\partial f}{\partial\theta} & =0 \end{align*} Hence (2) becomes $$$$m\ddot{r}-mr\dot{\theta}^{2}+mg\sin\theta+\lambda=0 \tag{4}$$$$ And (3) becomes \begin{align} m\left( 2r\dot{r}\dot{\theta}+r^{2}\ddot{\theta}\right) +mgr\cos\theta & =0\nonumber\\ r\ddot{\theta}+2\dot{r}\dot{\theta}+g\cos\theta & =0 \tag{5} \end{align} We now need to solve (1,4,5) for $$\lambda$$. Now we have to apply the constrain that $$r=R$$ in the above to be able to solve (4,5) equations. Therefore, (4,5) becomes \begin{align} -mR\dot{\theta}^{2}+mg\cos\theta+\lambda & =0\tag{4A}\\ R\ddot{\theta}+g\cos\theta & =0 \tag{5A} \end{align} Where (4A,5A) were obtained from (4,5) by replacing $$r=R$$ and $$\dot{r}=0$$ and $$\ddot{r}=0$$ since we are using that $$r=R$$ which is constant (the radius).
From (5A) we see that this can be integrated giving $$$$R\dot{\theta}^{2}+2g\sin\theta+c=0 \tag{6}$$$$ Where $$c$$ is constant. Since if we differentiate the above with time, we obtain \begin{align*} 2R\dot{\theta}\ddot{\theta}+2g\dot{\theta}\cos\theta & =0\\ R\ddot{\theta}+g\cos\theta & =0 \end{align*} Which is the same as (5A). Therefore from (6) we find $$\dot{\theta}^{2}$$ to use in (4A). Hence from (6) $$\dot{\theta}^{2}=-2\frac{g}{R}\sin\theta+c$$ To find $$c$$ we use initial conditions. At $$t=0$$, $$\theta=90^{0}$$ and $$\dot{\theta}\left( 0\right) =0$$ hence $$c=2\frac{g}{R}$$ Therefore \begin{align*} \dot{\theta}^{2} & =-2\frac{g}{R}\sin\theta+2\frac{g}{R}\\ & =2\frac{g}{R}\left( 1-\sin\theta\right) \end{align*} Plugging the above into (4A) in order to find $$\lambda$$ gives \begin{align*} -mR\left( 2\frac{g}{R}\left( 1-\sin\theta\right) \right) +mg\sin \theta+\lambda & =0\\ \lambda & =m\left( 2g\left( 1-\sin\theta\right) \right) -mg\sin\theta\\ \lambda & =2mg-2mg\sin\theta-mg\sin\theta\\ & =mg\left( 2-3\sin\theta\right) \end{align*} Now that we found $$\lambda\,,$$we can find the constraint force in the radial direction \begin{align*} N & =\lambda\frac{\partial f}{\partial r}\\ & =mg\left( 2-3\sin\theta\right) \end{align*} The particle will leave when $$N=0$$ which will happen when \begin{align*} 2-3\sin\theta & =0\\ \theta & =\sin^{-1}\left( \frac{2}{3}\right) \\ & =41.8^{0} \end{align*} Therefore, the angle from the vertical is $$90-41.8=48.2^{0}$$