Maybe I'm misunderstanding something. Rotor Brake Start: Nr=0. Np>0. Explain.
Rotor brake start:
Rotor brake = on, start engines
Ng=otherwise normal idle
Nr=0 (because the rotor brake is holding it)
Np=0* (because the freewheeling unit makes Nr => Np <-- two ways of saying the same thing --> Np =< Nr )
Rotor brake = off/disengage
Ng=still otherwise normal
Np accelerates* and takes Nr with it
Some helicopters (H-34, R-22, others) also feature an additional on/off clutch to completely engage and disengage the engine from the drivetrain, but that works more like the clutch pedal in a manual transmission car (or shifting an automatic between D and N). Note that without the load of the rotors, when Ng is at a steady idle speed then Np will normally seek some equilibrium speed of slightly less than 100%. Also note that this note does not apply to the R-22 and piston-engined variants of the H-34. Also also note that the H-34 and R-22 also have a freewheeling unit (aka overrunning clutch) for the same reason as any helicopter. Also also also note that there are a few unlucky helicopters do not feature a freewheeling unit.
*The main problem with a rotor brake start is it puts a lot of stress on the drivetrain because the engines make a lot of torque when Np is at or near 0. That is a basic characteristic of free power turbine turboshaft engines. (Random related fact, the experimental Chrysler turbine cars in the 1960s had a lot of snap off the line for this reason.) The torque indications in the 60 don't really show this well but only because they are designed to be accurate when the Np is rotating at normal speeds.