A conveyor belt needs 40 N·m at 120 rpm. Is a 0.75 kW motor enough?
Start at the output shaft
First establish the load torque and the required speed at the driven shaft. In a steady-state rotating system, P = T · ω, with ω = 2πn/60. Using P in kW, T in N·m and n in rpm gives P = T · n / 9549.2966. Power describes work per unit time; torque alone doesn't tell you how fast the machine performs that work.
From load to motor
At 40 N·m and 120 rpm, the mechanical output power is about 0.503 kW. With a transmission efficiency of 85%, this requires roughly 0.591 kW at the motor shaft. A 0.75 kW motor looks sufficient in steady state, but that comparison doesn't yet assess starting or acceleration. Check the torque-speed curve and the allowed thermal load of the chosen motor.
Acceleration is a separate load
During acceleration, an inertial torque adds to the process load. For an equivalent inertia J at the shaft under consideration, T = J · α. Convert every inertia consistently to the same shaft and use the desired speed profile. Friction, breakaway torque and cycle time can determine the peak torque, even when the average power stays low.
Assumptions and design check
The calculator describes a steady-state mechanical operating point and uses positive quantities. Electrical input power, motor and drive losses, regeneration, dynamics and safety functions fall outside this model. Check load cases and manufacturer data before selecting components.