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I'm studying these lecture notes on dynamic models of robot arm links, slides 33-36, where two examples are given for the kinetic energies for a single link and two link robot arm.

In the single link case, the kinetic energy is posed as:

$$K=\frac{1}{2}I\dot{\theta}^2$$

This appears to account for the rotational kinetic energy of the single link around the joint.

In the two link case, the kinetic energy of the first link is posed as:

$$K=\frac{1}{2}I_1\dot{\theta_1}^2 + \frac{1}{2}m_1a_{c1}^2\dot{\theta_1}^2$$

This appears to mean that the total kinetic energy in the first link of the two link chain contains a contribution from rotation around the first joint ($m_1a_{c1}^2$ term) and from rotation around the center of mass ($I_1$ term).

Comparing the two link and single link cases, shouldn't $I$ from the single link case be decomposed similarly into energies from joint rotation and center of mass rotation as well? Or is the single link case fundamentally different in that one only needs to consider energy from rotation around the joint?

It's my intuition that if the parameters of the single link case are equal to the parameters of the first link (of the two-link case), then we could conceivably write $I=I_1+m_1a_{c1}^2$. Is this correct? If not, why not?

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Yes, your intuition is correct. In the first case, $I$ is computed around the joint point. In the second case, $I_1$ is computed around the center of mass of the link. The Parallel Axis Theorem can be used to yield different inertia values around different axes, yielding to your intuition.

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From my understanding it is not correct as in the case of the single-link $I$ is directly computed around w.r.t. the joint point. However, in the two-link case, each link's inertia is firstly computed w.r.t. the center of mass of the link ($I_1$) and then translated to a reference frame with the origin coinciding to the joint-point via the Huygens-Steiner Theorem.

By the way, this structure is given by the Euler-Lagrange formulation of the robot model and the definition of the inertia matrix $M(q)$. The slides you linked from Prof. Claudio Melchiorri (my former supervisor at the university of Bologna) explains this in the part preceding the example.

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