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

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 $$1/12\cdot m_1 L_1^2$$ and with the central axes theorem I obtain $$J_{1yy} = J_{1zz} = 1/12\cdot m_1 L_1^2 + m_1 l_{c,1}^2$$ So I can express the the kinetic energy as $$1/2 \cdot \dot{\theta}_1^2 (1/12\cdot m_1 L_1^2 + m_1 l_{c,1}^2)$$ Is that right?

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

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\$.