Particle slides on incline where incline angle increases with rate $\omega$: why does kinetic energy have a term $(1/2)m(\omega^2 x^2)$?

Question

A particle slides on a smooth inclined plane whose inclination is $\theta$ is increasing at a constant rate $w$. If $\theta = 0$, at time t = 0 at which time the particle start from rest, Find the subsequent motion of the particle.

This problem can be solved with Lagrangian methods. The kinetic and potential energies are

$$T = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m(\omega^2x^2)$$ $$V = mgh = m g x \sin(wt) \, .$$

Where does the $\frac{1}{2}m(\omega^2x^2)$ come from?

Answer 1

Try to think of this problem using a polar coordinate system. $x$ is essentially the radius $r$ or $\rho$, measured from pivotal. $w$ is simply the angular velocity. So the position vector of the object is

$x\hat{\vec r}+\theta\hat{\vec \theta}$

So the velocity vector is

$\dot x\hat{\vec r}+w\hat{\vec \theta}$

The hatted vectors are unit.

So the second term of $T$ is "the rotational kinetic energy".