Free Vibration - Energy Method

ENERGY METHOD

A circular cylinder of mass m and radius r is connected by a spring of modulus of elasticity k as shown in the figure. It is free to roll on the rough horizontal surface without slipping, finding its natural frequency

The kinetic energy KE is stored in the mass by virtue of its velocity, 

$KE =K{E}_{translation} + K{E}_{rotation}$

$KE = \frac{1}{2}m{\dot{x}}^2 + \frac{1}{2}J{\dot{\theta }}^2$

${\mathrm{The\; potential\; energy\; PE\; is\; stored\; in\; the\; form\; of\; strain\; energy\; in\; elastic\; deformation\; or\; work\; done\; in\; a\; force\; field\; such\; as\; gravity}}^{}$

$\mathrm{PE} = \frac{1}{2}k{x}^2$

${\mathrm{KE\; +\; PE\; =\; constant}}^{}$

$KE + \mathrm{KPN}= \frac{1}{2}m{\dot{x}}^2 + \frac{1}{2}J{\dot{\theta }}^2 + \frac{1}{2}k{x}^2$

$\sin ce, J=\frac{1}{2}m{r}^2 x=r\theta \dot{x}= r\dot{\theta }$

$\stackrel{}{The\; total\; energy\; being\; constant.}$

$\stackrel{}{ }$KE + KP = constant 

$KE + KE = \frac{1}{2}m{r}^2{\dot{\theta }}^2 + \frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\dot{\theta }}^2 + \frac{1}{2}k{r}^2{\theta }^2$

ddt(KE + PE) = 0

$\left(\frac{3}{2}m\right)\dot{\theta }\ddot{\theta } +k\theta \dot{\theta }=0$

$\left(\frac{3}{2}m\right)\ddot{\theta } +k\theta =0$

${\omega }_n={\left(\frac{2k}{3m}\right)}^{\frac{1}{2}}$