Spring pendulum, Kinetic Energy

[KEVmathematics]

In the included picture, I don't get how they get to the kinetic energy part. I would say, that the traveled distance is equal to (l + x(t))*θ. Then I would take the time derivative, resulting in dx(t)/dt * θ + (l + x(t))* dθ/dt. Then I would square this result and multiply that with 1/2 m. But then I would get a totally different kinetic energy.

 

[mfb]

I would say, that the traveled distance is equal to (l + x(t))*θ

What does that mean? The traveled distance depends on the trajectory. You'll need a position (as vector) to get a meaningful derivative.

 

[Vishwaas]

You must also consider the radial component. The velocity component in the radial direction is the time derivative of x(t) (since it is a spring, it can compress or expand). Hence the total kinetic energy is the sum of energies in both radial and angular directions.

 

[vanhees71]

In polar coordinates, with the polar angle relative to the direction poiinting downwards, you have
$$\vec{x}=r \begin{pmatrix}
\sin \varphi \\ \cos \varphi
\end{pmatrix}.$$
Then, after straight-forward algebra, you have
$$T=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2).$$
Now in the textbook, they set
$$r=l+x,$$
where $l$ is the length of the relaxed spring.

For the potential energy you have the part from the gravitational field of the Earth and the spring:
$$V=-m g x_2+\frac{k}{2} x^2 = -m g (x+l) \cos \varphi+\frac{k}{2} x^2.$$