How should we interpret a electric rotor in short circuit of a induction machine?
First of all, in an induction motor, the rotor conductors are always shorted by an end-ring. So, there is no case of short circuit of rotor. It is always shorted.

The circuit you have shown is the equivalent circuit for an induction motor. In that circuit, the rotor is already short circuited. The rotor is the secondary of the transformer in the equivalent circuit. 



So, shorting the secondary does not mean that \$R'_2\$ is zero. And \$R'_c\$ is actually \$\frac{R'_2(1 - s)}{s}\$, where \$s\$ is the slip of the induction motor.

If you add \$R'_2\$ and \$R'_c\$, you will get \$\frac{R'_2}{s}\$ (which is the secondary resistance as far as induction motor equivalent circuit is concerned).

We have separated the \$\frac{R'_2}{s}\$ term into \$R'_2\$ and the other term, so we can get back the analogy with the transformer (and consider the other term as load of the transformer) and utilize transformer equations.