# On conservation of angular momentum and energy

The problem is: Two particles with mass $m$ are attached to a spring of negligible mass, with lengh $l_0$ without streching. The spring is streched until it reaches twice it's initial lenght and it's released after a velocity perpendicular to the sprinc of $(v_0, - v_0)$ is transmited to the particles, such that $kl_0^2 = mv_0^2$, where $k$ is the spring constant. Calculate the components $(v_r, v_\theta)$ of the velocity of the particle when the spring passes through it's non-streched position, where $v_r$ is the radial velocity and $v_\theta$ is the tangencial velocity

My attempt:

Since the force that act onthe masses is radial, the angular momentum is conserved such that:

$$\frac{I_i v_i}{R_i} = \frac{I_f v_f}{R_f}$$

Where $I$ is the moment of inertia of the system, $v$ the velocity and $R$ the radius of gyration. The initial moment being the moment of releasing and the final, the moment where the spring is in it's non-streched position. This implies that:

$$v_\theta = 2v_0$$

By conservation of energy, we thus have:

$$U_i + K_i = U_f + K_f$$

Hence:

$$\frac{kl_0^2}{2} + \frac{mv_0^2}{2} = \frac{mv^2}{2}$$ $$kl_0^2 + mv_0^2 = mv^2$$ $$7mv_\theta^2 = mv^2 = m(v_r^2 + v_\theta^2)$$

We thus have:

$$v_r = v_0 \sqrt{3}$$

The books answer says that $v_r = 0$, but I think it's wrong.

• It is true the $v_r=0$, so in one sense the book is right. However, the spring never passes through its unstretched length. The initial conditions are such that the spring force is exactly that required for circular motion. Thus $v_r=0$ and $v_{\theta}=v_0$ always. – Ben51 Jan 16 '18 at 16:20

The book is right. The centripetal force needed to keep the masses spinning at a constant radius is given by $F_{cent}=\frac{mv_0^2}{r}=\frac{kl_0^2}{l_0}=kl_0$. This is exactly the force exerted by the spring.

• How is the book right? The unstretched position is $r=l_0/2$ and as shown below and argued by you as well, the motion is restricted to the circle $r=l_0$. – Borun Chowdhury Jan 16 '18 at 18:22
• The book is right that the radial velocity is zero. It is not right that the spring ever reaches its unstretched length. – Ben51 Jan 16 '18 at 18:28
• @Ben a broken clock shows the right time twice a day. A confined particle in a central potential, when looked at in radial coordinates, will always have at least one place where the velocity is zero. Doesn't make the book right. – Borun Chowdhury Jan 17 '18 at 8:16

The book answer is wrong (or maybe you have made a mistake in understanding). I will treat only one particle below as the problem is symmetric under exchange of particles. This means the spring constant has to be multiplied by 2. Energy conservation gives

$$v_r^2 + v_\theta^2 + \frac{2kl_0^2}{m} (\frac{r}{l_0} - \frac{1}{2})^2 = const$$

Using your relation $k l_0^2 = m v_0^2$ and angular momentum $L=m l_0 v_0$ we can write this as

$$v_r^2 + \frac{v_0^2}{(\frac{r}{l_0})^2} + 2 v_0^2 (\frac{r}{l_0} - \frac{1}{2})^2 = const$$

We can write the scaled radial position $\rho = \frac{r}{l_0}$ and get

$$\frac{3}{2} v_0^2$$

for the constant by putting $v_r=0$ and $\rho=1$. So we get the solution for the problem

$$\frac{v_r}{v_0} = \sqrt{ \frac{3}{2} - \frac{1}{\rho^2} - 2(\rho- \frac{1}{2})^2} \\ = \frac{\rho-1}{\rho} \sqrt{ - (2 \rho^2 +2 \rho +1)}$$

Since the quantity in the square root is always negative, the only solution is $\rho=1$.

An interesting thing to look at is if we say $\frac{kl_0^2}{mv_0^2}=\gamma$. Then we have

$$\frac{v_r}{v_0} = \sqrt{ (1+ \frac{\gamma}{2}) - \frac{1}{\rho^2} - 2 \gamma(\rho-\frac{1}{2})^2} \\$$

and I plot below the allowed regions in solid and disallowed in dashed for three values of $\gamma$.

We see that when $\gamma<1$ (the spring is lax) the allowed region is on the outside of $r=l_0$ while when $\gamma>1$ (spring is tighter) the allowed region is on the inside of $r=l_0$. Of course when $\gamma=1$ (the case in point) we get the allowed region is $r=l_0$.