Calculate ideal unloaded rotor speed
Concerns about the question



The key words here are "ideal" and "unloaded"

I have not regarded the word "nominal" to mean it has a 5% slip at full load because it would be impossible to accurately predict what slip it has when unloaded. I have used this as the definition of the word nominal: -



To my way of thinking this means that the "nominal slip of 5%" means at virtually no-load (barring the windage and friction of the rotor). Having said that, many induction motors will have a full load speed at around 90% to 95% slip; such as is depicted in this picture: -



Nevertheless, I have used the word "nominal" to mean very lightly loaded.

My answer

A nominal slip of 5% (rotor unloaded as per the pasted text in the question) means that it is rotating at 5% lower than synchronous speed. Synchronous speed for a motor with 8 pole pairs operating from a supply of 350 Hz is 5250 rpm (Calculator here) and so 95% of this is 4987.5 rpm.


  But is this what is meant when they ask for the "ideal unloaded rotor
  speed" or have I calculated something different?


I think it is correct.


  But I am not sure WHY induction motors have slip in the first place.


If the rotor was travelling at synchronous speed, it would be travelling at the same rotating speed as the revolving magnetic field hence, to the rotor, there is no \$\frac{d\Phi}{dt}\$ to induce a voltage into the shorted rotor and therefore, there can be no magnetism produced by the rotor hence it can't spin at synchronous speed. 

It has to spin slightly slower than sync speed in order for a voltage to be induced in the rotor so that rotor current can flow in the squirrel cage and, produce a counteracting magnetic field (that allows a torque to develop that produces motion).