# Oscillations of a rotating mass on a spring

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A spring of constant $k$ is wrapped around a long rod. One end of the rod and spring are attached to a motor that rotates the rod with constant angular velocity $\Omega$. The other end of the spring is attached to a mass $m$. This mass slides along the rod as the spring expands and contracts. There is no friction or gravity in the problem. The equilibrium length of the spring is $r_{0}.$

What is the oscillation frequency of the mass for small values of $\Omega$? What happens when $\Omega$ is large? Is the motion still simple harmonic motion?

Since there is no gravity, the centrifugal force must equal the spring force so

$$m\Omega^{2}r=k(r-r_{0})$$

which leads to a constant $r$ since $\Omega$ is constant. So I am missing something basic here.

Maybe I should use energy conservation?

asked Oct 21, 2018
edited Oct 22, 2018

The equilibrium length of the spring $r_0$ is not its natural length. $r_0$ is the radius at which the mass orbits the pivot. It increases with $\Omega$.

The question is asking about oscillations in the radial direction.

Suppose we have chosen a value for $\Omega$ and the mass is currently at radius $r$. In the rotating frame of reference there is a centrifugal outward force $mr\Omega^2$ and an elastic inward force $k(r-a)$ where $a$ is the natural length of the spring. When these two forces are exactly balanced and the mass is rotating at a constant radius about the pivot, this is the equilibrium position. so $$mr_0\Omega^2=k(r_0-a)$$ which you can solve to find $r_0$.

However, we could displace the mass a small distance from its equilibrium position. The forces would then no longer be balanced. If the equilibrium position is stable the mass will return to it with a non-zero velocity, overshoot, and oscillate about the equilibrium position. The frequency $\omega$ of this oscillation is what you are being asked to find.

To find $\omega$ you need to write the equation of radial motion of the mass in the usual form for simple harmonic motion $$\ddot x +\omega^2 x=0$$ To obtain the equation of motion suppose that $r$ is increased by a small amount $x$. Then applying $F=m\ddot x$ we have $$m(r_0+x)\Omega^2-k(r_0+x-a)=m\ddot x$$ This can be simplified by substituting for $r_0$ as found above.

Comment

Energy is not conserved because the motor is doing work to keep the mass, spring and axle rotating at constant angular velocity. For the same reason angular momentum is not conserved either. A constant value of $\Omega$ could otherwise be achieved using a flywheel, or making the mass of the axle very much larger than $m$. In this case energy and momentum are conserved because the system, which is isolated, now consists of the axle, mass and spring instead of only the mass and spring.

If the motor is switched off before the mass is displaced then both energy and angular momentum are conserved. In this case the oscillation frequency would be different, because $\Omega$ would vary as $r$ changes, unless the mass of the axle or flywheel are much larger than $m$.

A related question on this site is Spinning connected springs system.

answered Oct 22, 2018 by (26,660 points)
selected Oct 22, 2018
The second  equation above gives a frequency of oscillation

$$\omega^{2}=\frac{k}{m}-\Omega^{2}$$

This is only positive if $\Omega^{2}<\frac{k}{m}$. If $\Omega^{2}=\frac{k}{m}$ the equation becomes a constant acceleration problem (instead of harmonic motion) and presumably the mass flies off the rod. What happens if $\Omega^{2}>\frac{k}{m}$? (what kind of motion is it?)
If $\Omega^2 \gt \frac{k}{m}$ then there is again an outward acceleration, but it is not constant, it increases exponentially. Substitute $x=Ae^{\gamma t}$ instead of $x=Ae^{i\gamma t}$. See eg [Correct way of solving the equation for simple harmonic motion](https://physics.stackexchange.com/q/305262)