Mass inertia tensor for rod non-symmetric regarding coordinate system origin

Question

I want to express the inertia tensor of a rotating rod (total length $L_1$) to use it in Lagrange mechanics for expressing the kinetic energy associated to the rotation with angular velocity $dot{theta}_1$. I know the moment of inertia regarding the center of mass $m_1$ as $1/12cdot m_1 L_1^2$ and with the central axes theorem I obtain
$$
J_{1yy} = J_{1zz} = 1/12cdot m_1 L_1^2 + m_1 l_{c,1}^2
$$
So I can express the the kinetic energy as
$$
1/2 cdot dot{theta}_1^2 (1/12cdot m_1 L_1^2 + m_1 l_{c,1}^2)
$$
Is that right?

The rotation is constrained to the horizontal plane around the origin ($vec{x}_i = vec{0}$)

kamran · Answer

Yes, you are right. Using parallel axis theorem, you add the moment of inertia by $m*r^2$

Sometimes they use$  omega^2  for  dot theta^2  $ as angular velocity. But it is in a context where the need to convert quantities to linear velocity$.

Mass inertia tensor for rod non-symmetric regarding coordinate system origin

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