# Formula for a double pendulum

In Bence, Hobson, Riley Mathematical Methods for Physics pg 317, a double pendulum consisting of a rod attached to a string is pictured as follows:

I'm trying to understand the formula for the rotational kinetic energy(T or eq.1 below). The motion of the center of the rod is described as follows:

Considering the term on the right for the rotational kinetic energy, I would have guessed the moment of inertia without two components would be $I=M(\frac{3}{2}l)^2$ while in this problem the angle $\theta_2$ folds the length by $l\frac{1}{2}cos(\theta)$ so the radius of $I$ would be $l +l\frac{1}{2}cos(\theta)$ however above $l$ appears to be constant in the right term. Also in the formula (1) why is the angular velocity $\omega^2$ independent of $\dot{\theta_1}$: ($\dot{\theta_2}^2$) and no indication of a changing radius. In short the entire term $\frac{1}{24}Ml^2\dot{\theta_2}^2$ confuses me. I see the tangential velocity of the translational kinetic energy on the left term, and I'm also confused about why there is no translational motion in the y direction as highlighted in the text above if the double pendulum supposedly has motion in the x-direction.

Ok let us take a step back and derive this kinetic energy from first principles.

The kinetic energy can be split into two parts:

1. Motion of the center of mass.
2. Motion about the center of mass.

# Motion of the center of mass

(I am going to do this via a very neat quick method you can do it via finding $\vec r$ for the center of mass and differentiating but this is tedious.)


# Motion about the center of mass

Now onto motion about the center of mass. The moment of inertia of a rod about the center is given by: $$I=\f{1}{12}Ml^2$$ The rod rotates about the center of mass at an angular velocity of $\dot \theta_2$ and as such this produces a rotational kinetic energy of: $$T_2=\f{1}{2}I \dot \theta_2^2$$ $$=\f{1}{24}M\dot \theta_2^2$$

• In equations 9.4 and 9.5 above, do you know where the approximation $(1-\cos(\theta))=\theta^2/2$ comes from? Jun 9, 2017 at 2:56
• @user135711 You are simply using the Taylor expansion of $\cos(\theta)=1-\frac{\theta^2}{2}+...$ keeping only terms to second order. Jun 9, 2017 at 7:16