## Essential University Physics: Volume 1 (3rd Edition)

$T=2\pi \sqrt{\frac{7}{10ga}}$
We first find the kinetic energy: $E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ $E_k = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)\frac{v^2}{r^2}$ $E = \frac{7}{10}mv^2$ $E = \frac{7}{10}m(\frac{dx}{dt})v^2$ Thus, we can find omega. Using the equation for omega, we obtain: $\omega = \frac{x}{\frac{d^2x}{dt^2}}$ $\omega= \sqrt{\frac{7}{10ga}}$ The period is equal to the angular velocity times 2 pi, so we find: $T=2\pi \sqrt{\frac{7}{10ga}}$