# Lagrangian exercise from Goldstein

Question 14 from the 1st chapter of H.Goldstein's book "Classical Mechanics":

Q: Two points of mass $m$ are joined by a rigid weightless rod of length $l$, the center of which is constrained to move on a circle of radius $a$. Express the kinetic energy in terms of the generalised co-ordinates.

My understanding of the problem tells me that the system has only $1$ degree of freedom, $\theta$, since the center of mass moves in a circle of fixed radius in a fixed plane. So there must be only one generalised co-ordinate. Now if I consider that the position vectors of the $2$ point masses are $\vec r_1$ and $\vec r_2$ respectively,then the constraints of the system are $$\lvert\vec r_2-\vec r_1\rvert=l$$ $$\left\lvert\frac{\vec r_2+\vec r_1}{2}\right\rvert=a$$

But I cannot find out the generalised co-ordinate from here on. Can you tell me if I am right about the degrees of freedom of the system? And am I on the right track? How should I proceed?

Also, one more thing is that there is nothing written if the masses are rotating about their center of mass. Do I have to make a special case for it?

• there are 2 degrees of freedom (as I understand it), I myself would start by parametrising the possible positions of the centre of the rod and then the possible positions of the masses as seen from the centre of the rod and progress from there. – Sanya Aug 11 '16 at 15:22
• The problem doesn't say the center of the rod is constrained to move on a circle at consta – Lewis Miller Aug 11 '16 at 15:26
• I agree with the previous comment. There are two degrees of freedom. First, the circular motion of on a circle of radius a and second, the rotation of the masses around their centre of mass (as you suggested). – physicopath Aug 11 '16 at 15:26
• Continued: velocity, hence two generalized coordinates. The masses are points, so no rotation. – Lewis Miller Aug 11 '16 at 15:28
• The question does not explicitly state that the rod is free to rotate around its center, but I think that that's the intent. I would also assume that all the motion is in one plane. I think this question needs some word-smithing. – garyp Aug 11 '16 at 15:47

This problem, as most problems from Goldstein's "Classical Mechanics", must be carefully analysed. They are not easy.

Having said that, I want to propose a solution to this problem. The original question was about the number of degrees of freedom. We are going to get there, and also write the full expression for the kinetic energy.

We have two particles. In three dimensions, without constraints, the number of degrees of freedom is $n = 2\cdot 3=6$. We may ask now how many constraints are there. Before answering that question, let's remember that any system of two particles may be described by the center of mass vector $\mathbf{R}$, and the positions relative to the center of mass, which we may call $\mathbf{r}'_{1}$ and $\mathbf{r}'_{2}$. In the general case, the total kinetic energy is given by the center of mass kinetic energy plus the kinetic energy of the particles in relation to the center of mass, i.e.

$$T = T_{CM} + T'_{1} + T'_{2} = \frac{M}{2}\dot{\mathbf{R}}^2 + \frac{m_{1}}{2}\dot{\mathbf{r}}_{1}^2+\frac{m_{2}}{2}\dot{\mathbf{r}}_{2}^2$$

The mass of the two particles are equal and the center of the rod is constrained to move in a circle. Therefore, the center of mass (CM) coincides with the rod center and it is also constrained to move in a circle. The CM may be described by one variable only, the angle $\Theta$ in respect to a fixed axis - let's say $X$ - whose origin is fixed in the center of the circular trajectory. If the CM is restricted to a circular motion, then it is equivalent to say that its trajectory is restricted to a plane (-1 degree of freedom) in a circular path (-1 degree of freedom). So we are left with $n = 6-2 = 4$ degrees of freedom.

The $X, Y, Z$ coordinates of the center of mass may be readily written as $$\mathbf{R} = (a\cos\Theta, a\sin\Theta, 0)$$ Deriving it with respect to time gives us the velocity $$\dot{\mathbf{R}} = (-a\dot{\Theta}\sin\Theta, a\dot{\Theta}\cos\Theta, 0)$$ We may now compute now the kinetic energy of the center of mass

$$T_{CM} = \frac{M}{2} (\dot{\mathbf{X}^2}+\dot{\mathbf{Y}^2}+\dot{\mathbf{Z}^2}) = \frac{M}{2} R^{2}\dot\Theta^{2}\,\, .$$

Let's now analyse the movement of the particles with respect to the CM. The statement of the question does not restrict the particles to a plane motion, so we should consider the rod which connects them to move in a 3D space. The constraint eliminates one more degree of freedom, so we are left with $n=4-1=3$. Unless one further restricts the rod motion to a plane, the system has exactly $n=3$ degrees of freedom. That answer your question, but let's now calculate the kinetic energy.

The motion of the particles with respect to the CM may be described by spherical coordinates referred to a frame whose origin is centered at the CM. So let's define a $x',y',z'$ frame whose origin is at the CM and orientation is equal to the previously defined $X,Y,Z$ frame. The positions $\mathbf{r}'_{1}$ and $\mathbf{r}'_{2}$ may be written as
$$\mathbf{r}'_{1,2} = (\frac{l}{2}\sin\theta_{1,2}\cos\phi_{1,2}, \frac{l}{2}\sin\theta_{1,2}\sin\phi_{1,2}, \frac{l}{2}\cos\theta_{1,2}\cos\phi_{1,2})$$
In this expression we have already considered the radius of the trajectory to be fixed and to be given by $l/2$. The angles $\theta_{1}$ and $\theta_{2}$, as well as $\phi_{1}$ and $\phi_{2}$, are related by one phase, and it does not change the value of the kinetic energy. Indeed, if the masses are equal, and they are constrained to move with the same magnitude of angular velocity and speed, then their kinetic energies are equal. So we shall just compute the kinetic energy for one particle and multiply it by two. We will now use the angles $\theta_{1}$ and $\phi_{1}$, to compute $T'_{1}$, dropping the subscripts of them.

$$T'_{1} = \frac{m}{2} (\dot{\mathbf{x}_{1}'^2}+\dot{\mathbf{y}_{1}'^2}+\dot{\mathbf{z}_{1}'^2}) = \frac{m}{2}\frac{l^2}{4}(\dot\theta^2 + \dot\phi^2\sin\theta)$$

We may finally write the total kinetic energy: $$T = T_{CM} + T'_{1} + T'_{2} = \frac{M}{2} R^{2}\dot\Theta^{2} + m\frac{l^2}{4}(\dot\theta^2 + \dot\phi^2\sin\theta)$$

Using the definition of reduced mass, i.e., $\mu=m^2/2m = m/2$, the kinetic energy may be written as $$T = \frac{M}{2} R^{2}\dot\Theta^{2} + \frac{\mu}{2} l^2(\dot\theta^2 + \dot\phi^2\sin\theta)$$ The problem has three degrees of freedoms and the generalized coordinates used to solve this problem are $\Theta, \theta$ and $\phi$, as defined in our solution.