# What is the motion of a mass 'circling' a spring?

Consider a situation whereby there is a mass connected to a spring which is initially unstretched and the system lies on a horizontal plane. An impulse is then applied to the mass giving it an initial velocity. What would the resulting motion of the mass look like?

I have considered forming a differential equation using polar coordinates for the acceleration of the mass and force in the spring however this yields a nonlinear DE.

I understand that the particle essentially oscillates between two circles of different radii (call $r_b$ and $r_a$) and that these can be found by consideration of angular momentum and conservation of energy.

However, I do not know why this occurs, why does the mass oscillate between two circles?

• If you write the equations of motion in Cartesian coordinates, wouldn't they form a decoupled system of two one dimensional equations of a mass on a spring? So basically, you would have a 1D mass on a spring for the x coordinate and a 1D mass on a spring for the y coordinate. So the solutions would be linear combinations of sines and c0sines for both coordinates. And then by taking the sum of the squares of the x and the y coordinates you get why the mass oscillates between two circles. – Futurologist May 16 '18 at 0:34
• @Futurologist: That's true if the equilibrium position is $r = 0$, but that wasn't explicitly stated in the question. – Michael Seifert May 16 '18 at 1:27
• @MichaelSeifert Ah, ok I see. – Futurologist May 16 '18 at 16:03

## 3 Answers

The total energy of the system is $$E = \frac{1}{2} m \vec{v}^2 + U(r),$$ where $U(r)$ is the potential energy. The velocity of the mass is $$\vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta},$$ and so $\vec{v}^2 = \dot{r}^2 + r^2 \dot{\theta}^2$: $$E = \frac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2) + U(r),$$ But we also know that the angular momentum of the mass is constant: $\ell = m r^2 \dot{\theta}$ doesn't change with time. This allows us to get rid of $\dot{\theta}$ altogether: $$E = \frac{1}{2} m \dot{r}^2 + \underbrace{\frac{\ell^2}{2 m r^2} + U(r)}_{{} \equiv U_\text{eff}(r)}$$

If we look at this problem, this looks exactly like a particle moving in one dimension ($r$) with a potential energy given by the last two terms. We can therefore define $U_\text{eff}(r)$ to be the effective potential for the radial motion; and we can use the "shape" of this potential to describe the radial motion of the mass.

In the particular case of a mass on a spring, we have $$U_\text{eff}(r) = \frac{\ell}{2 m r^2} + \frac{1}{2} k (r - r_0)^2,$$ where $r_0$ is equilibrium length of the spring. It is not too hard to see that $U_\text{eff}$ diverges as $r \to 0$ and $r \to \infty$; and a bit of algebra shows that $U_\text{eff}(r)$ has a unique minimum at a single value of $r$.* Thus, the effective potential is generally U-shaped, and all radial motion must involve oscillations between some maximum and minimum value of $r$. (In fact, this type of motion is always present if $U(r)$ is unbounded as $r \to \infty$, even if the potential energy is due to something other than a spring; the argument is pretty much the same.)

On a physical level, why does the $\ell^2/2 m r^2$ (sometimes called the "centrifugal barrier") term arise? The best way to think about it is to note that angular momentum is always a constant, and that $\ell = r v_\perp$, where $v_\perp$ is the tangential component of the velocity. This means that for the mass to get close to the origin (i.e., $r$ is small), we must have $v_\perp$ large to keep $\ell$ constant. But $v_\perp$ also contributes to the kinetic energy, and the total energy of the mass is fixed, so $v_\perp$ is (usually) bounded above. This therefore implies that there's a minimum radius that is consistent with energy and angular momentum conservation.

*Actually finding this value of $r$ requires solving a quartic polynomial, but this polynomial will always have one positive root. A simple argument is to show that $U'_\text{eff}(r)$ is monotonically increasing on $r \in (0, \infty)$, is negative as $r \to 0$, and is positive as $r \to \infty$. This allows us to conclude that there is a single value of $r$ for which $U'_\text{eff}(r) = 0$.

• This is just a generic argument on a polar decomposition of a mechanical problem, at no point the specific potential of the problem was used to show the particularities of using a spring (harmonic potential). – ohneVal May 15 '18 at 15:52
• @ohneVal: I would argue that it's important for the OP to realize that the oscillations between maximum and minimum $r$ are present in many systems. But I've edited my answer to include how this applies specifically to the mass on a spring. – Michael Seifert May 15 '18 at 16:25

You can think of this system first in just one dimension, in which case it is clear that the motion is harmonic, it being periodic and with some fixed amplitude, $A$. So if the relaxed position of the spring is taken to be the origin, the mass will oscillate between $-A$ and $A$. Nothing changes if there is some extra component in the velocity (namely going to 2 dimensions). The spring-mass system begins to rotate (assuming the other end of the spring is fixed but allowed to rotate).

If it serves, think of it as a planetary system. However, instead of gravity which is always attractive, you have a $\sim (r-r_0)^2$ potential which confines the mass to certain amplitude (fixed by the energy given in its initial impulse) because it will pull back if the mass goes above $r_0$ but will push away if the position of the mass falls below $r_0$.

• Doesn't your argument imply that gravitational orbits shouldn't oscillate between a minimum and maximum value of $r$? Even though we know that elliptical orbits do precisely that? – Michael Seifert May 15 '18 at 15:34
• No, it just says that for a harmonic potential is more directly to conclude such confinement. While in a $1/r$ potential the existence of stable orbits is more delicate. – ohneVal May 15 '18 at 15:49
• Another critique: This argument only applies for small oscillations. Suppose that the potential energy of the spring is $\frac{1}{2} k (r - r_0)^2$, and I give the mass enough of an initial impulse that it has an energy greater that $\frac{1}{2} k r_0^2$. If we just take the spring's energy into account, we might conclude that it can reach $r \leq 0$. – Michael Seifert May 15 '18 at 16:28

"An impulse is then applied to the mass giving it an initial velocity."

If the impulse gets mass moving in a circular motion, then at least some of the impulse is in the tangential direction. You may soon realise that the type of circular motion is not simple, and that it depends on the initial magnitude of tangential velocity. Illustration at the bottom accompanies explanation.

A spring-less scenario (grey trajectory) would mean that the mass finds its equilibrium rotational velocity and keeps rotating at that velocity if there are no energy losses. In the case of a spring, circular motion will cause radial motion as the centrifugal force experienced by the mass stretches the spring (see diagram).

With spring (blue trajectory in figure): It is convenient to divide the motion into circular and radial for analysis. The total energy of the system $E_{tot}$ is:

$E_{tot} = E_{tan} + E_{rad}\tag1$

where $E_{tan}$ is the energy due to circular motion only (if there were no spring) and $E_{rad}$ is the radial motion energy (associated with the spring oscillations). In turn, the radial energy has two components, kinetic ($K_{rad}$) and potential ($U_{rad}$), so that:

$E_{tot} = E_{tan} + (K_{rad} + U_{rad})\tag{1.1}$

For the circular motion, the applicable energy equation is the usual:

$E_{tan} = 0.5mv^2 = 0.5mr_a^2\omega_{ta}^2\tag2$

where for the mass, $\omega_{ta}$ is the average rotational angular velocity, and $r_a$ is the average distance from the centre of rotation.

For the radial oscillations, we have simple harmonic motion with a spring, position (ref) is expressed by: $r_i(t) = Acos(\omega_r t + \varphi) \tag3$

where $A$ is the maximum amplitude of the radial oscillation, $\omega_r$ is the radial (spring induced) oscillation 'angular' velocity, and $\varphi$ is the phase. The radial direction instantaneous potential energy is:

$E_{rad} = K_{rad}+U_{rad} = 0.5m\dot r_i^2 + 0.5kr_i^2= 0.5m\omega_r^2 A^2sin^2(\omega_r t + \varphi) + 0.5kA^2cos^2(\omega_r t + \varphi) \tag4$

As $\omega_r t$ varies, the kinetic energy $K_{rad}$ oscillates, achieving a max when $U_{rad}$ is at a minimum, and vice-versa. We can express instantaneous values of $E_{tan}$ in equation $2$ by replacing average rotational radius $r_a$ by the instantaneous rotational radius $r_i$, and $\omega_{ta}$ with $\omega_t$. $\omega_t$ in $E_{tan}$ must also oscillate to conserve angular momentum as the rotational radius of the mass $r_i$ changes. See this related question.

Equations $2$ and $4$ are therefore the total energy expression. Note that the (tangential) rotational angular velocity $\omega_{t}$ is different from $\omega_{r}$, the radial (spring induced) oscillation 'angular' velocity. There is an important caveat:

If the tangential velocity is high enough, there will be no radial direction oscillations (orange trajectory in figure). Why is this? The tangential velocity results in an outward centrifugal force: $F_{out} = v^2/r_i$, and at this max radius $r_i$ the spring is pulling the mass towards the centre with maximum force possible for the given spring = $F_{inward} = kr_i$. However:

$F_{out} > F_{inward} \implies v^2/r_i > kr_i$

at all times, so there are no radial oscillations and:

$E_{rad} = K_{rad}+U_{rad} = 0 + 0.5kA^2 \tag5$

and by substituting into equation 1.1:

$E_{tot} = 0.5mr_a^2\omega_{ta}^2 + 0.5kA^2 \tag6$ • I see a couple of problems with this answer. (1) I don't think you can assume that $E_{tan}$ and $E_{rad}$ are separately conserved. For example, suppose that you have a free particle on a trajectory that passes a distance $a > 0$ from the origin. Its energy is largely "radial" when it is far away from the origin, and largely "tangential" when it is closest to the origin. But it has no forces on it at all. (2) What do you mean by "the largest force possible for a given spring"? Are you assuming that the spring simply can't be extended beyond a certain distance? If not, what do you mean? – Michael Seifert Jun 1 '18 at 13:13
• @MichaelSeifert (1) Let's look at it this way: is there simultaneously tangential and radial direction motion (in the general case)? Yes. Is it accepted practice, for example in projectile motion, to account for these separately. Yes. (2) The 'largest possible force' means, the pulling force of the spring, at maximum extension without damaging the spring. – Dlamini Jun 1 '18 at 22:57
• The fact that you can separate the various components of the motion in the case of projectile motion is a special property of Cartesian components; when expressed in these coordinates, the component of the acceleration vector in one coordinate direction is simply the second derivative of that coordinate. This is not the case for any other coordinate system; see, for example, the derivation of the radial acceleration vector (eq. 4) in these MIT course notes. ... – Michael Seifert Jun 3 '18 at 15:50
• ... Since the acceleration in the $r$-direction depends on the $\theta$-velocity (and vice versa), the radial motion and the tangential motion do not decouple in the way they do for projectile motion in Cartesian components. This means that you can't assume that $E_{tan}$ and $E_{rad}$ are both constant; only their sum is constant. – Michael Seifert Jun 3 '18 at 15:52
• @MichaelSeifert I fail to find a place in my answer where it says that $E_{tan}$ and $E_{rad}$ are constant. I do not see a problem in considering radial and tangential components separately, even if they have a connection. Think again. The spring makes the connection between the two non-trivial. If you spin the mass fast enough, and its tether has a spring, then no oscillations occur, but energy is stored in the spring. Slow the rotation down enough, and the spring starts to contract, releasing some of its energy to radial kinetic energy. – Dlamini Jun 4 '18 at 0:26