# Homework Help: Natural frequency of a simple harmonic system

1. Aug 6, 2015

### TomW17

1. The problem statement, all variables and given/known data
Derive the natural frequency $f_n$ of the system composed of two homogeneous circular cylinders, each of mass $M$, and the connecting link $\text{AB}$ of mass $m$. Assume small oscillations.

2. Relevant equations
The sum of the kinetic + potential energies in an isolated system remains constant.

An undamped, simple harmonic system has an equation of motion of the form $$\ddot{\theta} + \omega^{2}_n \theta = 0$$

3. The attempt at a solution
The sum of the kinetic + potential energy when the system rotates by an angle $\theta$ measured as in the diagram counterclockwise is $$\sum E = V + T = mg r_0 (1 - \cos \theta ) + \tfrac{1}{2} m v_{AB}^{2} + 2 \times \tfrac{1}{2} M v_{O}^2 + 2 \times \tfrac{1}{2} I_O \dot{\theta}^2 = \text{const.}$$

where $v_O$ denotes the velocity of the centre of mass of the cylinders, $\dot{\theta}$ denotes the angular velocity of the system (and by extension the angular velocity of the cylinders) and $v_{AB}$ the velocity of the centre of mass of $\text{AB}$.

$$E = mgr_0 (1 - \cos \theta ) + \tfrac{1}{2} m (r_0 \dot{\theta})^2 + M (r \dot{\theta})^2 + \tfrac{1}{2} M r^2 \dot{\theta}^2 = \text{const.}$$
$$\implies \frac{\text{d}E}{\text{d}t} = mg r_0 \sin \theta \dot{\theta} + m r^{2}_0 \dot{\theta} \ddot{\theta} + 3 Mr^2 \dot{\theta} \ddot{\theta} = 0$$

using small angle approximation and rearranging...

$$\ddot{\theta} + \frac{mgr_0}{3Mr^2 + mr^{2}_0} \theta = 0 \implies \omega_n = \sqrt{\frac{mgr_0}{3Mr^2 + mr^{2}_0}} \implies f_n = \frac{1}{2\pi} \omega_n$$

The answer given in the book is $$f_n = \frac{1}{2\pi} \sqrt{\frac{mgr_0}{3Mr^2 + m(r-r_0)^{2}}}$$

Comparing the answers, my problem appears to be that I've calculated the kinetic energy of the link $\text{AB}$ incorrectly but I can't figure out why. My logic was that the angle $\theta$ is the angle between the vertical through the centre of the cylinder and the line connecting the centre of the cylinder and the end of the link $\text{AB}$, so the velocity of the end of $\text{AB}$ would be equal to $\dot{\theta} r_0$ (angular velocity x arm) and as $\text{AB}$ remains horizontal throughout, the velocity of the centre of mass of $\text{AB}$ would be the same as the velocity of its end.

Help?

Last edited: Aug 6, 2015
2. Aug 6, 2015

### DEvens

You could put in a few more steps to help a tired old duffer understand how you get each thing.

It looks like you have $v_{AB}$ as $r_0 \dot{\theta}$, and $V_O$ as $r \dot{\theta}$. But you need to be sure that these are both in the same frame of reference. Are these both with respect to the ground?

3. Aug 6, 2015

### TomW17

Ahaa, there's my problem. No, I've calculated $v_{AB}$ to be wrt the centre of the cylinder but $v_O$ is wrt the ground.

4. Aug 6, 2015

### TomW17

Got the right answer when I take $v_{AB}$ wrt the ground. Thanks for your help!