What's the relation between the tensions on a rope around a non-ideal pulley? I know that on one side of the pulley the tension must be larger to let the torque do its job and rotate the pulley. But how does the relation look like?
$$T_1 = k \cdot T_1'$$
What's $k$?
 A: By an ideal pulley we mean that it is massless, and there is no friction between pulley and string. We also assume the string to be massless irrespective of pulley(considering mass of string would make the situation extremely complex as tension would vary with height). So for an ideal pulley, the pulley actually does not rotate. The tension throughout a string is same. The pulley does not experience a net force or torque(m=0) and the string slips over the pulley(the pulley does not rotate). 
In case of a non-ideal pulley, for calculations we neglect friction at the axle. We assume pulley to have a mass m. And consider STATIC friction between pulley and string. That means pulley would now rotate on its axis. So we need the moment of inertia of pulley(radius and mass). Since the pulley rotates, the tension varies in the string wrapped over the pulley. The pulley rotates in sense of net torque and the relation between the tensions on two ends of wrap is given by $T_1=T_2 e^{\mu \theta}$ where $\mu $ is coefficient of static friction between string and pulley and $\theta$ is the angle of wrap of string around the pulley. You can apply $\sum F=ma $for the pulley in this case in case it has an acceleration. Note that the tension at two ends is different but it is same throught the string when it is not in contact with the pulley. So in this case the pulley rotates and if there is enough friction, we assume there is no slipping between the string and pulley(static friction).
Hope this clarified your doubts!
