Natural frequency of a simple harmonic system

[TomW17]

Homework Statement

Derive the natural frequency [itex]f_n[/itex] of the system composed of two homogeneous circular cylinders, each of mass [itex]M[/itex], and the connecting link [itex]\text{AB}[/itex] of mass [itex]m[/itex]. Assume small oscillations.

Homework 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 [tex]\ddot{\theta} + \omega^{2}_n \theta = 0[/tex]

The Attempt at a Solution

The sum of the kinetic + potential energy when the system rotates by an angle [itex]\theta[/itex] measured as in the diagram counterclockwise is [tex]\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.}[/tex]

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

[tex]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.}[/tex]
[tex]\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[/tex]

using small angle approximation and rearranging...

[tex]\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[/tex]

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

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

Help?

 

[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?

 

[TomW17]

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?

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

 

[TomW17]

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?

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