# Calculating angular frequency for solid disk/spring system

I have a solid disk with radius $R$, mass $M$, and moment of inertia $I = \frac{1}{2}MR^2$ which can freely rotate about a fixed mass in its center. A spring with spring constant $k$ is attached to the top of the disk and to an adjacent wall, as in the figure below.

I'm tasked with calculating the angular frequency $\omega$ of the disk about its central axis. I am given that the angle $\phi$ of initial perturbation is very small, so $sin(\phi) \approx \phi$. Here is what I have so far:

$$\tau = I \alpha = \frac{1}{2}MR^2 \alpha = F_{spring} R = -k x R$$ $$x = R sin( \phi ) \approx R \phi$$ $$\implies -k R^{2} \phi = \frac{1}{2}MR^2 \alpha \implies \alpha = \frac{d^2 \phi}{d t^2} = \frac{-k 2 \phi}{M}$$ $$\implies \frac{d^2 \phi}{d t^2} + \frac{k 2 \phi}{M} = 0$$

At this point, I think we have a fairly simple differential equation: $\phi'' + \frac{k 2}{M}\phi = 0$. But I am not sure how to proceed from here. The solution provided by the instructor gives the differential equation above and then immediately concludes that $\omega = \sqrt{\frac{2k}{M}}$ without any intermediary logic, which suggests that some kinematic formula or something must exist relating the differential equation with the conclusion, but I've poured through the book and am at a complete loss. I also tried solving it as a differential equation and then plugging back in the kinematic equation for $\phi(t)$, but this did not yield anything meaningful. What is the intermediary logic between the step I'm currently at and the conclusion I'm working toward?

Write $$\frac{d^2\phi}{dt^2}=-\frac{2k}{M}\phi$$ and ask yourself: what functions do I know with the property that the second derivative if a negative multiple of the function itself? Surely the answer are the trigonometric functions, so try $$\phi(t)= A\cos(\omega t)+B\sin(\omega t) = C\cos(\omega t+ \beta). \tag{1}$$ where $A,B,C,\beta$ and $\omega$ are constants. Inserting this into your differential equation yields $\ddot{\phi}=-\omega^2 \phi$ so that $$-\omega^2 \phi = -\frac{2k}{M}\phi$$ so that, by inspection, $\omega=\sqrt{\frac{2k}{M}}$. You can then find the remaining constants from the initial conditions. For instance, at $t=0$, $$\phi(0)=A\, ,\qquad \dot{\phi}(0)=\omega B$$ will allow you to determine $A$ and $B$. Alternatively, you can also write $$\phi(0)=C\cos\beta\, ,\qquad \dot{\phi}(0)=-\omega C \sin\beta$$ with $\omega$ as above.

Note that most differential equations are solved by guessing a form to the solution and then specializing this form on the problem. The guess here is the form given by (1) on the basis of the argument presented immediately above that guess.

• I'm accepting this answer because it explains the theory, whereas mine focuses on the mechanics of the proof. In particular, the point of why we elect the trig functions is useful - this step is implicit in my proof, since I use the method of undetermined coefficients, but your explanation explicitly gives motivation for the choice. – Max von Hippel May 9 '18 at 3:15

Your differential equation looks like harmonic oscillator equation (for small oscillations) $\ddot{\theta}+\omega^{2}\theta=0$, where $\omega$ is the frequency of oscillation.
If you want to prove $\omega$ is really the frequency you can try the solution $\theta(t)=A\cos(\omega t+\phi)$, where $A$ and $\phi$ are constant determined by the initial conditions. I don't know if my explanation is good.

• Thanks! Yes, this was helpful - I took this approach in my proof above, although from the opposite direction (I started with the differential equation and worked my way to the form of SHM rather than the other way around). – Max von Hippel May 9 '18 at 3:11

My confusion arose because I was uncertain of the definition of SHM. Simple Harmonic Motion is any motion in which the restoring force $F_r$ is directly proportional to the displacement from equilibrium (and the equilibrium is an attractor with regard to this force). Consequentially, SHM is by definition of the form $F = -kx$, like in Hooke's Law. So:

$F = M_{cm}a_{cm} -kx \implies M\frac{d^2 x}{d t^2} + kx = 0$ $\implies \frac{d^2 x}{d t^2} + \frac{k}{M}x = 0$

The characteristic equation is $\lambda^2 + \frac{k}{M} = 0$, so we get roots at $\lambda = \pm \sqrt{\frac{k}{M}}i$. We therefore guess a homogenous solution of the form $x_h = Acos(\sqrt{\frac{k}{M}}t) + Bsin(\sqrt{\frac{k}{M}}t)$. But we know the motion is simple ipso facto, so I can simplify this to by allowing B = 0. Then I have the very definition of SHM, with a phase shift of 0, amplitude of A, and angular frequency of $\omega = \sqrt{\frac{k}{M}}$.

Applying this to the original problem statement:

$\frac{d^2 \phi}{d t^2} + \frac{2k}{M}\phi = 0$ $= \frac{1}{R}\frac{d^2 x}{d t^2} + \frac{2k}{M}\frac{x}{R} = 0$ (since we are rolling without slipping) $\iff \frac{d^2 x}{d t^2} + \frac{2k}{M}x = 0$

This is the form which I just proved leads directly to our equation for SHM, therefore I can now (like my instructor did) immediately conclude that $\omega = \sqrt{\frac{2k}{M}}$.