# Finding Lagrangian of a Spring Pendulum

I'm trying to understand Morin's example of a spring pendulum. What I don't get is his expression for $T$. I can understand the $\dot x^2$ term in the brackets. But I don't understand the $(l + x)^2\dot \theta^2$.

Also, it seems rather strange to break up Kinetic Energy into tangential and radial components when it is a scalar.

-

$\newcommand{\er}{\hat e_r} \newcommand{\et}{\hat e_\tau} \newcommand{\d}{\dot} \newcommand{\m}{\frac{1}{2}m}$ In radial coordinates, $\d\er=\d\theta \et$, and (useless here) $\d\et= -\d r \er$. $\er,\et$ are unit vectors in radial and tangential directions respectively. Due to this mixing of unit vectors (they move along with the particle), things get a little more complicated than plain 'ol cartesian system, where the unit vectors are constant.

For your particle, writing $x+l\to r$, the position vector is: $$\vec p= r\er$$ $$\therefore \vec v=\d{\vec p}= \d r\er + r\d\er=\d r \er + r\d\theta\et$$ $$\therefore v^2= \vec v\cdot\vec v= \d r^2+r^2\d\theta^2$$

Substituting back the value of $r=x+l,\d r=\d x$ (and mutiplying by $\m$, we get the above expression?

As you can see in my expression for $\vec v$, I had two components of velocity--radial and tangential. Since they are perpendicular, I can just square and add, akin to $T=\m\left(\d x^2 +\d y^2\right)$.

The point is, it may be a scalar, but it contains a vector in its expression:$$T=\m v^2=\m|v|^2=\m \vec v\cdot \vec v=\m(\dot x^2+\dot y^2)$$

-

The Total kinetic energy is the sum of the parts. For this there is kinetic energy in the radial direction $1/2 m \dot{x}^2$ and in the $\hat{\theta}$ direction. The super position of these two energies forms the total. We must take that into account when we write down the total kinetic energy. The first term in the parenthesis is the radial velocity squared, and the second term is the version of $r^2 \dot{\theta}^2$ (The definition of the angular velocity squared) pertaining to this particular problem. Here is a little more on radial and angular velocity. Hope this helps.

-

The Kinetic Energy needs to be $\frac{1}{2}m(v_x^2 + v_y^2)$ where $v_x$ and $v_y$ are the velocity in the respective direction. $x$ is set to be $(l+x)*\sin \theta$ and $y$ is set to be $(l+x)*\cos\theta$ take the time derivative of both, then square them. After doing that, realize that $\cos^2 + \sin^2 = 1$ so do the necessary factoring to use the identity. Then the middle terms with the coefficient 2 cancel. Finally the relation left is kinetic energy equation that is in the book.

-
Welcome to Physics.SE, Cleo. With MathJax active on the site it is possible to write you mathematics in a LaTeX like notation and have it rendered neatly. Clicking the "edit" button will let you see what I've done. – dmckee Apr 27 '12 at 15:02

Don't think of it as components of KE: rather think about it as that the total KE of the body is the summation of it's angular KE and Linear KE(which happens to be in the radial direction in this case)...Hope that helps..

-