Motor selection for mechatronic systems

Why it matters

A motor that is too small stalls, overheats or loses position; one that is too large wastes money, space and energy and can make the control loop sluggish. Motor selection is where every earlier topic – motor characteristics, drives, gearing and feedback sensors – comes together into one engineering decision, and it is a standard interview and design-project question.

Key ideas

• Start from the load, not the motor. Define the motion first: the move profile (distance, time, maximum speed, acceleration), the external forces (gravity, friction, process force), the inertia of everything that moves, the duty cycle, and the accuracy required. Only then translate it to the motor shaft.
• Three numbers decide the size.
  • Peak torque – the largest torque in the cycle, usually during acceleration; it must lie inside the motor's intermittent (peak) zone at the speed where it occurs.
  • RMS (thermal) torque – heating depends on I²R, and torque is proportional to current, so the root-mean-square torque over the whole cycle (dwell included) must not exceed the motor's continuous rated torque.
  • Maximum speed – must be below the rated or maximum speed at the required voltage. Plot every (speed, torque) point of the cycle on the motor's torque–speed curve; all must lie inside the envelope with a margin (often 20–50 %).
• Transmission (gears, belts, lead screws). A reduction of ratio n (motor turns n times per load turn) multiplies speed at the motor by n, divides load torque by n (and by the efficiency η), and divides load inertia by n². Gearing lets a small, fast motor drive a slow, heavy load, but adds backlash, compliance, friction and cost.
• Inertia matching. For a purely inertial load, motor torque needed for a given load acceleration is smallest when the reflected load inertia equals the motor inertia, n_opt = √(J_L / J_m). In practice servo makers recommend keeping the load-to-motor inertia ratio below a limit (commonly about 5:1 to 10:1 for good tuning; take the exact figure from the drive manufacturer). High ratios make the loop oscillatory and slow to settle.
• Choosing the motor type.
  • Brushed DC (PMDC) – simple drive, cheap, linear torque–current; brushes wear and limit speed. Small appliances, low-cost robots.
  • Stepper – open-loop positioning, high torque at low speed, no tuning; torque falls steeply with speed and missed steps go undetected. 3D printers, small CNC, valves.
  • AC servo (PMSM) / DC servo – closed loop with encoder or resolver; high peak torque, fast response, accurate at any speed; more expensive and needs tuning. CNC axes, robot joints, pick-and-place.
  • BLDC – brushless, efficient, long life, high speed; needs electronic commutation (Hall sensors or sensorless). Drones, fans, pumps, e-bikes.
  • Induction motor with VFD – rugged, cheap, high power; precise positioning needs vector control and an encoder. Conveyors, pumps, fans, spindles.
  • Linear motors, solenoids, piezo – direct linear motion, short strokes or very fine resolution.
• Other constraints. Supply voltage and current available; environment (temperature, dust and water via the IP rating, explosive atmosphere); mounting and frame size (e.g. NEMA or IEC flange); holding brake for vertical axes; feedback resolution; regenerative energy during braking (braking resistor); noise; efficiency class; cost and availability of spares.
• Gravity and vertical axes. A vertical load needs continuous torque even at standstill, so the RMS torque rises and a brake is needed for power loss.

Formulas

ω = 2π · N / 60, P = T · ω

• ω angular speed (rad/s), N speed (rpm), T torque (N·m), P mechanical power (W).

ω_m = n · ω_L, T_L,m = T_L / (n · η), J_L,m = J_L / n²

• n reduction ratio (motor speed / load speed), η transmission efficiency (0–1), T_L load torque at the load shaft (N·m), J_L load inertia (kg·m²); subscript m means referred to the motor shaft. Inertia is reflected without η in this simple model.

T_m = (J_m + J_L / n²) · α_m + T_L / (n · η), α_m = n · α_L

• J_m motor (plus coupling) inertia (kg·m²), α angular acceleration (rad/s²). Peak motor torque during acceleration.

n_opt = √(J_L / J_m)

• gear ratio that minimises motor torque for a pure inertia load.

T_rms = √( Σ T_i² · t_i / Σ t_i )

• T_i torque in segment i (N·m), t_i its duration (s); the sum of t_i includes dwell time.

T = F · p / (2π · η), J_eq = m · (p / 2π)² (lead screw)

• F axial force (N), p screw lead (m/rev), m translating mass (kg).

T = F · r, J_eq = m · r² (drum, pulley or rack and pinion)

• r drum or pulley radius (m).

Worked examples

Example 1 (standard) – hoist drive. A 20 kg load is lifted by a rope on a drum of radius 0.05 m and inertia 2 × 10⁻³ kg·m². It must accelerate at 2 m/s² to a lifting speed of 0.5 m/s. A 10:1 gearbox (η = 0.9) connects a motor of inertia 1 × 10⁻⁴ kg·m². Find motor speed, peak motor torque and steady lifting power.

1. Rope force while accelerating: F = m·(g + a) = 20 × (9.81 + 2) = 236.2 N.
2. Drum torque: T = F·r = 236.2 × 0.05 = 11.81 N·m. Drum angular acceleration α_L = a/r = 2/0.05 = 40 rad/s²; drum inertia torque = 2 × 10⁻³ × 40 = 0.08 N·m. Total at drum T_L = 11.89 N·m.
3. Drum speed ω_L = v/r = 0.5/0.05 = 10 rad/s; motor speed ω_m = 10 × 10 = 100 rad/s = 955 rpm.
4. Motor acceleration α_m = 10 × 40 = 400 rad/s²; motor's own inertia torque = 1 × 10⁻⁴ × 400 = 0.04 N·m.
5. Peak motor torque T_m = 0.04 + 11.89/(10 × 0.9) = 0.04 + 1.321 = 1.36 N·m.
6. Steady lifting torque = 20 × 9.81 × 0.05/(10 × 0.9) = 1.09 N·m; power = 1.09 × 100 = 109 W (useful power m·g·v = 98.1 W, the rest is gear loss).

Answer: 955 rpm, peak torque ≈ 1.36 N·m, steady power ≈ 109 W. Choose a motor with at least 1.1 N·m continuous and 1.4 N·m peak at about 1000 rpm, plus margin and a holding brake.

Example 2 (GATE level) – servo for a ball-screw axis. A 50 kg table is driven by a ball screw of lead 10 mm (η = 0.9, screw inertia 5 × 10⁻⁵ kg·m²), directly coupled to a servo of inertia 1 × 10⁻⁴ kg·m². Guide friction is 100 N. Cycle: accelerate to 0.25 m/s in 0.1 s, run 0.4 s, decelerate in 0.1 s, dwell 0.4 s. Find the maximum speed, peak torque, RMS torque and inertia ratio. Treat η as acting in both directions (a simplification).

1. ω_max = 2π·v/p = 2π × 0.25/0.01 = 157.1 rad/s = 1500 rpm.
2. a = 0.25/0.1 = 2.5 m/s²; α = 2π·a/p = 1570.8 rad/s²; screw factor p/(2π·η) = 0.01/(2π × 0.9) = 1.768 × 10⁻³ m.
3. Acceleration: T₁ = (J_m + J_s)·α + (m·a + F_f)·p/(2π·η) = 1.5 × 10⁻⁴ × 1570.8 + (125 + 100) × 1.768 × 10⁻³ = 0.236 + 0.398 = 0.634 N·m.
4. Constant speed: T₂ = 100 × 1.768 × 10⁻³ = 0.177 N·m.
5. Deceleration: T₃ = −0.236 + (−125 + 100) × 1.768 × 10⁻³ = −0.236 − 0.044 = −0.280 N·m; dwell T₄ = 0.
6. T_rms = √[(0.634² × 0.1 + 0.177² × 0.4 + 0.280² × 0.1 + 0) / 1.0] = 0.246 N·m.
7. Table inertia at the motor = m·(p/2π)² = 50 × (0.01/2π)² = 1.27 × 10⁻⁴ kg·m²; load inertia = 1.27 × 10⁻⁴ + 0.5 × 10⁻⁴ = 1.77 × 10⁻⁴ kg·m²; ratio to motor = 1.77.

Answer: 1500 rpm, peak ≈ 0.63 N·m, RMS ≈ 0.25 N·m, inertia ratio ≈ 1.8:1. A servo rated about 0.32 N·m continuous (≈100 W at 3000 rpm), with peak torque above 0.7 N·m, fits with margin.

Common mistakes

• Sizing on steady-state power alone and ignoring the acceleration torque, which is usually the peak.
• Leaving dwell time out of the RMS denominator, or averaging torque instead of squaring it.
• Reflecting inertia through a gearbox by n instead of n².
• Dividing by η in the wrong place – load torque reflected to the motor is divided by η when the motor drives the load.
• Forgetting gravity on vertical axes and the need for a brake.
• Mixing rpm and rad/s, or screw lead in mm with forces in N.
• Choosing a stepper for a high-speed, high-dynamics axis where its torque has fallen off, or a servo where a cheap open-loop stepper would do.

For GATE ME

Expect numericals on power from torque and speed, torque and inertia referred through a gear train, the optimum gear ratio for an inertial load, lead-screw torque and RMS torque over a duty cycle. Conceptual MCQs ask which motor suits a given application (positioning, constant speed, high speed, low cost). Practise free-body diagrams of the load, unit conversions and the n² rule for inertia.

Quick check

1. A motor drives a load through a 5:1 reduction. The load inertia is 0.5 kg·m². What inertia does the motor see?
2. Why must dwell time be included when computing RMS torque?
3. A 4 N·m load torque runs at 600 rpm. What power is needed (η = 1)?
4. Find the optimum gear ratio for J_L = 0.05 kg·m² and J_m = 2 × 10⁻⁴ kg·m².
5. Which motor type would you choose for an open-loop, low-cost positioning axis at low speed?

Answers: 1. 0.5/25 = 0.02 kg·m²; 2. The motor cools during dwell, so heating is averaged over the whole cycle time; 3. P = 4 × 62.83 = 251 W; 4. n = √(0.05/2 × 10⁻⁴) = √250 ≈ 15.8; 5. A stepper motor.