yes of course because the electromagnetic torque which is proportional to the stator current will be equal to the acceleration torque + friction torque + load torque so it will be greater than the case where the motor starts without load

Using the simplified per phase equivalent circuit of the induction motor, one can get the following equation of electromagnetic torque:

Te= 3 (P/2). (Rr/sWs). (Ir)2

the equation clearly indicates that the torque is proportional to the square of rotor current. Thus more load implies more rotor current. At no-load the Small current flows in the rotor that corresponds to the windage and friction torque.

the stator current is nothing but the sum of magnetizing current and rotor current (in complex quantities). In general as the stator voltage is constant, the magnetizing current is constant and independent of load. Therefore, stator current depends mainly on the rotor current.

The starting current corresponding to starting torque is the sum of inertia plus load torque, that is :

Te= J(dw/dt)+Tl, where Tl is the load torque and J(dw/dt) is the required torque to bring rotor from standstill to the speed that makes Te=Tl.

You can notice that J(dw/dt) is proportional to J, moment of inertia which reflects the size of the rotor or the motor. The bigger the motor the stronger starting torque must develop to start.

Following the previous analysis, starting torque depends also on Tl. Maximum starting torque corresponds to full load and hence maximum starting current.

Hope my answer is useful for you.

Assuming the machine under study is in speed control mode; it must have sufficient torque for acceleration from a stall condition to overcome inertia, friction and load. The torque is typically a function of the current and the pole pairs on the machine. The rate change of the speed reference (acceleration) and the rate of change of the acceleration reference (jerk) will affect the torque required.