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

Question

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?

Answer 1

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.

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^{*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 $U'_\text{eff}(r) = 0$.}

Answer 2

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 $r_0$.

Answer 3

"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:

$E_{tot} = 0.5mr_a^2\omega_{ta}^2 + 0.5kA^2 \tag6$