Will two bodies initially connected to and revolving around each other, start spinning when disconnected? Two extended bodies are connected with a string and revolve around each other (that is, around the center of mass of this system). No gravity, no external forces.
The string is cut, and they start to depart from each other. Will they spin around their own centers of mass?
For a (representative) particular case when one of the bodies is a dumbbell (two point masses connected by a light rigid rod),

the answer is trivial - each point mass will start its tangential movement with a different velocity, so the pair will be revolving.
On the other hand, for the following configuration,

there seems to be no rotation.
Is there a general approach?
 A: The bodies are spinning before the string is cut, once per revolution. Since neither the string nor the cutting of the string apply torque to the bodies, there will be no change of rotational speed for the bodies, so they will keep spinning.
To think of this another way: at the moment the string is cut, the part of the body connected to the string is moving slower than the part of the body directly opposite due to the circular motion. Cutting the string does not change the speed of any part of the body, so the difference in velocity in different parts of the body will result in an overall rotation.
In your second illustration, the two balls have different velocities, as shown with the arrows below:

These different velocities would result in an overall rotation in the two-ball system after the string is cut.
A: Take a ball of radius $r$ connected to a thin center-post by string of length $R$. The post accelerates up to and stays at a high final rotation speed $\Omega$. Obviously the ball revolves at $\Omega$. Now imagine case 2 where a thin, light ball-bearing is connected to the string and the ball rotates within it.
Lookin down along the axis of rotation, draw an arrow ⬆️ on the top of the ball. In case 2 it always points the same way, and in case 1 it rotates once per revolution.
At the same angular velocity $\Omega$ of the system:
$$ v_1=v_2 =v= \Omega (R+r) $$ $$ \omega_1 = \Omega ~,~ \omega2 = 0$$
By conservation of angular and linear momentum, after the cut:
$$v_{1,f}=v= \Omega (R+r) ~,~ \omega_{1,f}= \Omega$$
$$v_{2,f}=v= \Omega (R+r) ~,~ \omega_{2,f}= 0$$
Yes it spins as it travels away no matter what shape it is.
In your lower pic the tangent velocities are not parallel for the two balls.
By the way, the kinetic energy in case 1 is higher: $$E_1= \frac{1}{2} m v^2 + \frac{1}{2} I_{ball} \omega^2$$ where $I_{ball}$ is the moment of inertia of a sphere around around its axis. The second term is gone for case 2:
$$ E_2= \frac{1}{2} m v^2= \frac{1}{2} I_{mass} \Omega^2$$
where $I_{mass}$ is the moment of inertia of a point-mass in revolution, $m(R+r)^2$ in this case. We also just derived the parallel axis theorem where $E_1= \frac{1}{2} I_{tot} \Omega^2$ and $I_{tot}=I_{mass}+I_{ball}$.
