- #1
TomW17
- 10
- 2
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?
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