# Angular frequency of inverted pendulum

A thin vertical rod of steel is clamped at its lower end. When you push the upper end to one side, bending the rod, the upper end moves (approximately) along an arc of a circle of radius R, and the rod opposes your push with a restoring force F = — k$$\theta$$, where $$\theta$$ is the angular displacement and K is a constant. If you attach a mass m to the upper end, what will be the frequency of small oscillations? $$\color {red} {Treat \space the\space rod\space as\space massless\space in\space your\space calculations.}$$ (Hint: Think of the rod as an inverted pendulum of length R, with an extra restoring force $$—k\theta$$.)

let's find the angular frequency from SHO:

$$\tau = I\alpha = Rk\theta-Rmg\theta$$ $$\alpha = \frac{(Rk-Rmg)\theta}{mR^2}$$ $$\omega = \sqrt{\frac{(Rk-Rmg)}{mR^2}}$$

however, this answer is wrong the true answer is $$\omega = \sqrt{\frac{(k-Rmg)}{mR^2}}$$

what did I get wrong?

• In the differential equation I think we should use negative sign for the restoring force.
– user
Commented Jun 1, 2023 at 13:51

It seems that the restoring action is considered as a torque, indeed the differential equation becomes

$$I\alpha = -k_{\theta}\theta+Rmg\theta \iff I \ddot\theta+(k_{\theta}-Rmg)\theta=0$$

with $$\omega = \sqrt{\frac{k_{\theta}-Rmg}{mR^2}}$$

As an alternative, assuming the restoring action as a horizontal force we should have $$F=-kR\theta$$ the differential equation becomes

$$I\alpha = -kR^2\theta+Rmg\theta \iff I \ddot\theta+(kR^2-Rmg)\theta=0$$

$$\omega = \sqrt{\frac{kR^2-Rmg}{mR^2}}$$
• By definition of torsional spring $\tau = -k\theta$ Commented Jun 1, 2023 at 14:50