Rotational Motion and RPM
Understanding angular velocity, RPM, rotational kinematics, and gear systems
Rotational Motion and RPM
Understanding rotational motion is essential for motor selection, gear design, velocity control, and predicting robot performance. This section covers the physics and practical applications of rotational motion in robotics.
Revolutions Per Minute (RPM)
RPM is the most common way to specify motor speed. It represents the number of complete 360° rotations an object makes in one minute.
Key Conversions
Converting between different rotational speed units:
| From | To | Formula |
|---|---|---|
| RPM | rad/s | ω (rad/s) = (RPM × 2π) / 60 |
| rad/s | RPM | RPM = (ω × 60) / (2π) |
| RPM | deg/s | deg/s = RPM × 6 |
| RPM | Hz (rev/s) | Hz = RPM / 60 |
| Linear speed | RPM | RPM = (v × 60) / (2πr) |
| RPM | Linear speed | v = (RPM × 2πr) / 60 |
Practical RPM Ranges in Robotics
| Motor Type | Typical RPM | Common Applications |
|---|---|---|
| DC Motor | 5,000-20,000 | Direct drive wheels, small robots |
| Brushless Motor | 3,000-10,000 | Drones, precision systems |
| Stepper Motor | 1,000-2,000 (typical) | 3D printers, CNC |
| Servo Motor | 300-500 (loaded) | Robot arms, controlled motion |
| Gear-reduced DC | 10-300 (typical) | Wheel drive, heavy lifting |
Why Gear Reduction?
Raw motors are too fast and weak for most robotic tasks. Gearboxes reduce speed by a known ratio while multiplying torque by the same ratio.
Example: 3000 RPM motor + 50:1 gearbox = 60 RPM output with 50× more torque
Rotational Kinematics
Just as linear motion has kinematic equations, rotational motion has parallel equations that describe angular position, velocity, and acceleration.
Rotational Motion Equations
Fundamental Rotational Equations
1. Angular Velocity:
ω = ω₀ + αt2. Angular Displacement:
θ = ω₀t + ½αt²3. Velocity-Displacement:
ω² = ω₀² + 2αθWhere:
- ω = Final angular velocity (rad/s)
- ω₀ = Initial angular velocity (rad/s)
- α = Angular acceleration (rad/s²)
- t = Time (seconds)
- θ = Angular displacement (radians)
Converting degrees to radians:
θ (radians) = θ (degrees) × π / 180Linear vs Rotational Analogy
Understanding the direct correspondence helps apply familiar linear concepts to rotation:
| Linear Motion | Rotational Motion | Conversion |
|---|---|---|
| Position: x (m) | Angular Position: θ (rad) | x = θ × r |
| Velocity: v (m/s) | Angular Velocity: ω (rad/s) | v = ω × r |
| Acceleration: a (m/s²) | Angular Acceleration: α (rad/s²) | a = α × r |
| Mass: m (kg) | Moment of Inertia: I (kg·m²) | - |
| Force: F (N) | Torque: τ (N·m) | F = τ / r |
| Linear Equations | Rotational Equations | - |
| v = v₀ + at | ω = ω₀ + αt | |
| x = v₀t + ½at² | θ = ω₀t + ½αt² | |
| v² = v₀² + 2ax | ω² = ω₀² + 2αθ |
Key Insight: The relationship x = θ × r applies to all motion variables when proper units are used (radians for angles, meters for radius).
Deriving the Equations
Starting from angular acceleration:
α = dω/dtIntegrating to get velocity:
ω = ∫α dt = α·t + C
With initial condition ω(0) = ω₀:
ω = ω₀ + αtIntegrating again for displacement:
θ = ∫ω dt = ∫(ω₀ + αt) dt = ω₀t + ½αt² + C
With initial condition θ(0) = 0:
θ = ω₀t + ½αt²For velocity-displacement (eliminating time):
From ω = ω₀ + αt, solve for time: t = (ω - ω₀)/α
Substitute into θ = ω₀t + ½αt²:
θ = ω₀(ω - ω₀)/α + ½α[(ω - ω₀)/α]²
θ = (ω₀ω - ω₀²)/α + (ω² - 2ωω₀ + ω₀²)/(2α)
θ = (ω² - ω₀²)/(2α)
Rearranging:
ω² = ω₀² + 2αθPractical Examples
Example 1: Robot Arm Joint Acceleration
A robot arm joint accelerates from rest (ω₀ = 0) with angular acceleration α = 10 rad/s² for 2 seconds.
Find final angular velocity and displacement:
ω = ω₀ + αt = 0 + 10 × 2 = 20 rad/s
θ = ω₀t + ½αt² = 0 + ½ × 10 × 2² = 20 radians
Converting to degrees: 20 × 180/π = 1146° ≈ 3.2 full rotationsExample 2: Spinning Up a Wheel
A wheel with angular velocity of 50 rad/s must stop in exactly 1 rotation (2π radians) with constant deceleration.
Find the required deceleration:
Using ω² = ω₀² + 2αθ:
0² = 50² + 2 × α × 2π
0 = 2500 + 4πα
α = -2500/(4π) = -198.9 rad/s²
Time to stop:
ω = ω₀ + αt
0 = 50 + (-198.9) × t
t = 0.251 secondsGear Ratios and Speed Reduction
Gears are the primary method to trade motor speed for usable torque and to match motor characteristics to load requirements.
Understanding Gear Ratios
Simple Gear Ratio:
Gear Ratio = Number of Teeth on Driven Gear / Number of Teeth on Drive Gear
Gear Ratio = Output Speed / Input SpeedExample: Two Gears
Driver gear: 20 teeth, spinning at 300 RPM Driven gear: 100 teeth
Gear Ratio = 100 / 20 = 5:1
Output Speed = 300 RPM / 5 = 60 RPM
Output Torque = Input Torque × 5Compound Gear Systems
For larger speed reductions (like 50:1 or 100:1), multiple gear stages are used:
Total Ratio = Ratio₁ × Ratio₂ × Ratio₃ × ...Example: 3-Stage Gearbox
- Stage 1: 10:1 (20 teeth → 200 teeth)
- Stage 2: 8:1 (25 teeth → 200 teeth)
- Stage 3: 5:1 (20 teeth → 100 teeth)
Total Ratio = 10 × 8 × 5 = 400:1
5000 RPM motor → 5000/400 = 12.5 RPM output
1 N·m input → 1 × 400 × 0.95 ≈ 380 N·m output (95% efficiency)Efficiency in Gearboxes
Real gearboxes lose energy to friction:
Output Power = Input Power × Efficiency
Output Torque = Input Torque × Gear Ratio × EfficiencyTypical Gearbox Efficiencies:
| Type | Efficiency |
|---|---|
| Spur Gear (single stage) | 95-98% |
| Helical Gear (single stage) | 96-99% |
| Worm Gear (single stage) | 40-90% |
| Planetary Gear | 90-96% |
| Multi-stage gearbox | 75-95% (cumulative) |
Worm Gears
While worm gears provide very high ratios in compact form, they have poor efficiency (~70-80% per stage) and generate considerable heat. Use only when space is critical and power loss is acceptable.
Motor-Load Matching
Proper gear selection ensures the motor operates in its efficient range:
Example: Mobile Robot Wheel Motor
Requirements:
- Desired wheel speed: 2 m/s
- Wheel radius: 0.1 m
- Required wheel RPM: (2 × 60) / (2π × 0.1) ≈ 191 RPM
Motor selection:
- Use 5000 RPM motor with 26:1 gearbox
- Output: 5000 / 26 ≈ 192 RPM ✓
- Motor runs at ~96% of max (efficient)
- Good acceleration response
Angular Motion in Robot Mechanics
Velocity Conversion: RPM to Linear Speed
For a rotating wheel or drum:
v = (RPM × 2πr) / 60Where:
- v = Linear velocity (m/s)
- RPM = Motor/wheel speed
- r = Wheel/drum radius (meters)
Example: Drive Wheel
3000 RPM motor with:
- 50:1 gearbox: Output = 60 RPM
- 0.1 m wheel radius
v = (60 × 2π × 0.1) / 60 = 0.628 m/s ≈ 0.63 m/s or 2.3 km/hAcceleration Conversion
Angular acceleration translates to linear acceleration at the wheel edge:
a = α × rWhere:
- a = Linear acceleration (m/s²)
- α = Angular acceleration (rad/s²)
- r = Radius (m)
Practical Considerations
Motor Specifications
When selecting a motor, you'll see:
- No-load RPM: Maximum speed with no load (useful for calculation)
- Stall Torque: Maximum torque when motor can't spin (brief, not continuous)
- Continuous Torque/Current: Safe long-term operation point
- Peak Torque/Current: Short-term maximum (seconds only)
Always design for continuous ratings, not peak or stall values.
Selecting RPM for Your Robot
For Speed Priority:
- Use higher RPM motor (3000+ RPM)
- Lower gear ratio (5:1 to 20:1)
- Fast acceleration
- Example: Racing robot, delivery robot
For Torque Priority:
- Use lower RPM motor (500-1500 RPM)
- Higher gear ratio (50:1 to 100:1)
- Strong climbing/lifting
- Slower response
- Example: Heavy loader, climbing robot
For Balance:
- Use medium RPM (1500-3000 RPM)
- Medium gear ratio (20:1 to 50:1)
- Good acceleration and torque
- Example: Most general-purpose robots
Key Takeaways:
✓ RPM describes motor speed, but gearing transforms it to useful speed/torque combinations ✓ Rotational kinematics follows the same principles as linear kinematics with analogous variables ✓ Gear ratios multiply torque and divide speed (or vice versa) through the mechanical advantage ✓ Efficiency matters - multi-stage gearboxes lose significant power ✓ Match motor RPM and gearing to your actual application requirements
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