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 $l_i$):

enter image description here

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?

  • $\begingroup$ Are there “+” signs missing? You have separated your kinetic energy in a sum of translational and rotational term, but this is only possible if the origin of the coordinate system is at the center of mass, which doesn’t seem to be the case from the way you have defined your coordinates. Otherwise there are cross terms linking the translational and rotational degrees of freedom. $\endgroup$ Jan 7 at 16:11

1 Answer 1


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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*}

  • $\begingroup$ I understand mathematically how we get the kinetic energy. But i do not understand the physical meaning of the terms i pointed out at my post. Are they just a byproduct of the relative motion of the second link? And as the systems we study become more complicated, and we have relative motions, we have kinetic energy in forms that are not easily interpretable? $\endgroup$ Jan 8 at 17:54
  • $\begingroup$ these are centrifugal forces and other velocities depending forces , but if your motion is "slow" they are negligible , so you take out these forces from the equation of motion . $\endgroup$
    – Eli
    Jan 8 at 18:39

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