Consider a body some orbiting at some distance around a fixed point and spinning about its center of mass (such that both rotations are in the same plane).

The orbital angular momentum will be that of a point particle positioned at the center of mass: $$L_{orbit} = mr_{CM}^2\omega_{orbit}$$

The spin angular momentum will be: $$L_{spin} = I_{CM}\omega_{spin}$$

So the total angular momentum is: $$L = mr_{CM}^2\omega_{orbit} + I_{CM}\omega_{spin}$$

Now try to picture that a body rotating at angular frequency $\omega$ about a point a distance $d$ from its center of mass is nothing but the previously considered system, with that point as the center of orbit and where the orbit is in "tidal lock", i.e. $r_{CM} = d$ and $\omega_{orbit} = \omega_{spin} = \omega$. The angular momentum is thus

$$L = md^2\omega + I_{CM}\omega = (md^2+I_{CM})\omega$$

And hence we arrive at the parallel axis theorem:

$$I = md^2 + I_{CM}$$

Consider a body orbiting at some distance around a fixed point and spinning about its center of mass (such that both rotations are in the same plane).

The orbital angular momentum will be that of a point particle positioned at the center of mass: $$L_{orbit} = mr_{CM}^2\omega_{orbit}$$

The spin angular momentum will be: $$L_{spin} = I_{CM}\omega_{spin}$$

So the total angular momentum is: $$L = mr_{CM}^2\omega_{orbit} + I_{CM}\omega_{spin}$$

Now try to picture that a body rotating at angular frequency $\omega$ about a point a distance $d$ from its center of mass is nothing but the previously considered system, with that point as the center of orbit and where the orbit is in "tidal lock", i.e. $r_{CM} = d$ and $\omega_{orbit} = \omega_{spin} = \omega$. The angular momentum is thus

$$L = md^2\omega + I_{CM}\omega = (md^2+I_{CM})\omega$$

And hence we arrive at the parallel axis theorem:

$$I = md^2 + I_{CM}$$

Consider a body some orbiting at some distance around a fixed point and spinning about its center of mass (such that both rotations are in the same plane).

The orbital angular momentum will be that of a point particle positioned at the center of mass: $$L_{orbit} = mr_{CM}^2\omega_{orbit}$$

The spin angular momentum will be: $$L_{spin} = I_{CM}\omega_{spin}$$

So the total angular momentum is: $$L = mr_{CM}^2\omega_{orbit} + I_{CM}\omega_{spin}$$

Now try to picture that a body rotating about a point other than its center of mass is nothing but the previously considered system, with that point as the center of orbit and where the orbit is in "tidal lock", i.e. $r_{CM} = d$ and $\omega_{orbit} = \omega_{spin}$. The angular momentum is thus

$$L = md^2\omega + I_{CM}\omega = (md^2+I_{CM})\omega$$

And hence we arrive at the parallel axis theorem:

$$I = md^2 + I_{CM}$$

Consider a body some orbiting at some distance around a fixed point and spinning about its center of mass (such that both rotations are in the same plane).

The orbital angular momentum will be that of a point particle positioned at the center of mass: $$L_{orbit} = mr_{CM}^2\omega_{orbit}$$

The spin angular momentum will be: $$L_{spin} = I_{CM}\omega_{spin}$$

So the total angular momentum is: $$L = mr_{CM}^2\omega_{orbit} + I_{CM}\omega_{spin}$$

Now try to picture that a body rotating at angular frequency $\omega$ about a point a distance $d$ from its center of mass is nothing but the previously considered system, with that point as the center of orbit and where the orbit is in "tidal lock", i.e. $r_{CM} = d$ and $\omega_{orbit} = \omega_{spin} = \omega$. The angular momentum is thus

$$L = md^2\omega + I_{CM}\omega = (md^2+I_{CM})\omega$$

And hence we arrive at the parallel axis theorem:

$$I = md^2 + I_{CM}$$