Consider a solid spherical object of uniform density that is rotating on an axis A1. Perpendicular to that axis one can draw another line that passes through the sphere. On this axis, on both sides of the sphere one attaches mass-less springs with spring constants F, and then attaches point masses of mass M to both springs.

If the object is now rotated around the axis A1 and simultaneously both springs are pulled and then let go, what will be observed?

My intuition is as follows:

The rate of rotation of the system will undergo wave-like behavior. It will rotate faster as the point masses attached to springs come closer to the center and it will rotate slower as they move away from the center.

Is this correct?

  • $\begingroup$ I think you need to specify how A1 is located. $\endgroup$
    – Man
    Commented May 27, 2013 at 18:50
  • $\begingroup$ I like the question. I have assumed it to be passing through the centre of the sphere, I am sure thats what you meant $\endgroup$
    – Man
    Commented May 27, 2013 at 18:52
  • 1
    $\begingroup$ Your intuition is correct, are you looking for equations? $\endgroup$
    – Man
    Commented May 27, 2013 at 19:15
  • $\begingroup$ no, at least i would like to take a stab at deriving them myself before I get the answer $\endgroup$ Commented May 27, 2013 at 19:20

2 Answers 2


Let's adopt polar coordinates. Fix the rotating body at the origin. Everything is happening in a plane. I assume that both springs rotate with the rotating body, i.e the follow the movement. Thus, they have the same angular velocity $\theta$.

Let's take the moment of inertia of the rotating body at the origin as J. I also assume a symmetry of the system in the way that both springs are equal and the masses start out at the same distance, noted $r$. The lagrangian of the system is then

$$ \mathcal{L} = \frac{1}{2}M\left(\dot{r}^2 + r^2\dot{\theta}^2\right) + \frac{1}{2}M\left(\dot{r}^2 + r^2\dot{\theta}^2\right) + \frac{1}{2}J\dot{\theta}^2 - \frac{1}{2}kr^2 - \frac{1}{2}kr^2 $$

$\theta$ being a cyclic variable, the total angular momentum is conserved

$$ \frac{\partial \mathcal{L}}{\partial\dot{\theta}} = (2Mr^2+J)\dot{\theta} = c $$

where $c$ is a constant. This is a result you can find also with newtonian mechanics. No force is creating torque on the system since the spring force is orthogonal to the motion. Then, the total moment of inertia with the help of Steiner's theorem is $2Mr^2 + J$ and the angular momentum follows as above. This constant will help us answer your question. To know whether $\dot{\theta}$ will slow down and accelerate etc... we need to know the behaviour of $r(t)$. This is given by the euler-Lagrange equations for $r$, which again you could get with a newtonian yet lengthy force diagram study :

$$ \frac{d}{dt}\frac{\partial{\mathcal{L}}}{\partial \dot{r}} - \frac{\partial \mathcal{L}}{\partial r} = 0 $$ which after carrying it out gives

$$ 2M\ddot{r} - 2Mr\dot{\theta}^2 +2kr = 0 \text{ or } \ddot{r} = r\dot{\theta}^2 - \omega_0^2 r = (\dot{\theta}^2 - \omega_0^2)r $$

where $\omega_0^2 \equiv \frac{k}{M}$. This differential equation is hard to solve, even if we substitute $\dot{\theta}$ from the conserved quantity, but it tells us two things from the last differential equation :

  1. If the centrifugal part is greater than the spring part, the movement obeys approximately diverging exponential $\ddot{r} \sim z r$ with $z > 0$. Intuitively, this means that the mass attached to the spring is under such a strong rotation, that the spring is not enough to hold it on a circular motion, and thus the masses go always further apart.
  2. If both contributions are equal, it means that we're in a steady regime where the spring force is just enough to hold the mass on a circle, and this there is no oscillation and the angular velocity is constant.
  3. If the spring force dominates the centrifugal part, then we have an oscillatory behavior and your intuition is correct.

Note however that for the first point, as $r$ goes away divergently, the conserved quantity tells us that the angular velocity should diminish, thus allowing the mass to come back since the spring force would overtake the centrifugal part. So all in all, the mass would come back, but I don't know how oscillatory this would be. The best thing would be to numerically solve the above differential equation to study all the cases.


As far as I understood, there are two cases:

  1. Rotor: If there is a rotor at the end of A1 axis which rotates with constant rotation rate, then obviously the angular velocity is conserved. However, in this case energy is not conserved at all times(its average over long periods is though); and the rotor will receive some energy at times and it will do work on other times.

  2. Free rotation: In this case, since there is no external source of energy, the total energy is conserved. Also there is nothing to provide torque, so the angular momentum is conserved as well. This, as you stated, means when the two masses get closer to the axis, the system will spin faster(which makes sense with conservation of energy as well).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.