Meaning of 2 kinetic energy terms in the equations

Question

I have this problem (The two rods will be called links. Link 1 has length $a_1$ while link 2 has length $a_2$. The distance of the center of mass of each link to their respective joint is $l_i$):

And I want to find the kinetic energy in order to find the dynamics of the system using the euler-lagrange method. The system has two degrees of freedom ($θ_1,θ_2$). I started by assuming one degree of freedom as a variable and all the others equal to 0.

So the kinetic energy when $θ_2=0, θ_1 \neq 0$ is:

$$T_1 = \frac{1}{2} (I_1+m_1 l_1^2)\dot{θ}_1^2 + \frac{1}{2} \left(I_2+m_2(a_1^2 + l_2^2+2a_1l_2cos(θ_2) \right)\dot{θ}_1^2 $$

Each of this terms has a physical meaning:

1. Rotational energy of $link1$ about joint1: $\frac{1}{2} (I_1+m_1 l_1^2)\dot{θ}_1^2$

2. Rotational energy of $link2$ about joint1: $\frac{1}{2} \left(I_2+m_2(a_1^2 + l_2^2+2a_1l_2cos(θ_2) \right)\dot{θ}_1^2$

The kinetic energy when $θ_1=0, θ_2 \neq 0$ is:

$$ T_2 = \frac{1}{2} (I_2+m_2 l_2^2)\dot{θ}_2^2 $$

which is the rotational energy of $link2$ when rotating about joint 2.

The total kinetic energy is : $T = T_1 + T_2$. However, this is wrong.

If we write the kinetic energy as:

$$ T = \frac{1}{2}m_1|u_1|^2 + \frac{1}{2}I_1ω_1^2 + \frac{1}{2}m_2|u_2|^2 + \frac{1}{2}Ι_2ω_2^2 $$

and then we write: $$ x_1 = l_1 cos(θ_1), \ y_1 = l_2 sin(θ_1) $$ $$ x_2 = a_1 cos(θ_1) + l_2cos(θ_1+θ_2), \ y_2 = a_2 sin(θ_1)+ l_2sin(θ_1+θ_2) $$

and differentiate this expression, the kinetic energy ends up like that:

$$T = \frac{1}{2} (I_1+m_1 l_1^2)\dot{θ}_1^2 + \frac{1}{2} \left(I_2+m_2(a_1^2 + l_2^2+2a_1l_2cos(θ_2) \right)\dot{θ}_1^2 + \frac{1}{2} (I_2+m_2 l_2^2)\dot{θ}_2^2 + \frac{1}{2} I_2 (2\dot{θ}_1\dot{θ}_2) + \frac{1}{2} m_2(l_2^2+a_1l_2cos(θ_2) )(2\dot{θ}_1\dot{θ}_2) $$ There are two extra terms in that expression:

1. $ \frac{1}{2} I_2 (2\dot{θ}_1\dot{θ}_2) $
2. $\frac{1}{2} m_2(l_2^2+a_1l_2cos(θ_2) )(2\dot{θ}_1\dot{θ}_2) $

The question: What is the physical meaning of these two terms? And could I predict the kinetic energy without writing the explicit expressions of the velocity of each center of mass?

Answer 1

Start always with the position vectors to the center of mass given in inertial system

$\def \b {\mathbf}$

Position vectors to the center of mass \begin{align*} &\b R_1=l_1\,\begin{bmatrix} \cos(\vartheta_1) \\ \sin(\vartheta_1) \\ \end{bmatrix}\quad, \b R_2=\frac{a_1}{l_1}\,\b R_1+ \underbrace{l_2\,\begin{bmatrix} \cos(\vartheta_1+\vartheta_2) \\ \sin(\vartheta_1+\vartheta_2) \\ \end{bmatrix}}_{\b R_{12}}\\ \end{align*} velocities \begin{align*} &\b v_1= \frac{d}{dt}\b R_1=\frac{\partial\b R_1}{\partial \vartheta_1}\dot{\vartheta}_1+ \frac{\partial\b R_1}{\partial \vartheta_2}\dot{\vartheta}_2\\ &\b v_2=\frac{a_1}{l_1}\b v_1+\frac{\partial\b R_{12}}{\partial \vartheta_1}\dot{\vartheta}_1+ \frac{\partial\b R_{12}}{\partial \vartheta_2}\dot{\vartheta}_2 \end{align*} kinetic energy \begin{align*} &T=\frac 12\left(m_1\b v_1\cdot\b v_1+m_2\b v_2\cdot\b v_2+I_1\dot{\vartheta}_1^2+I_2\dot{\vartheta}_2^2\right) \end{align*}

potential energy \begin{align*} &U=-m_1\,g\,\left(\b{R}_1\right)_y-m_2\,g\,\left(\b{R}_2\right)_y \end{align*}