When building a CNC router or upgrading a 3D printer, the first question is usually: "Is NEMA 17 enough, or do I need NEMA 23?" Most beginners look at the Holding Torque and stop there. This is a mistake. A NEMA 23 motor isn't just "stronger"—it is physically different in ways that affect your speed, your driver choice, and your machine's ability to avoid missed steps. If you choose a NEMA 17 for a heavy gantry, it is far more likely to overheat or lose steps under cutting load. If you choose NEMA 23 for a fast 3D printer, it might actually run slower than the smaller motor. This guide explains the engineering limits of each frame size. Table of Contents 1. Physical Difference (The Frame Size) 2. Torque & Speed (The Inductance Trap) 3. Driver Compatibility 4. Selection Summary Advertisement 1. Physical Difference (The Frame Size) "NEMA" is just a standard for ...
In practice, most machines involve rotary motion as well as linear motion. Typical examples include electric motors, gears, pulleys, flywheels, and internal combustion engines. If we wish to calculate how quickly a machine reaches its full operating speed—determining the acceleration of its components—we must consider rotary acceleration and the associated torques.
Advertisement
Fortunately, Newton’s second law of motion applies equally well to rotary motion, provided that the correct rotational form of the equation is used.
The Challenge of Non-Uniform Motion
Consider a solid disc mounted on a shaft and rotated by a pull cord wrapped around its rim. We cannot apply the standard linear form of Newton’s second law, F = ma, because the resulting motion is rotational. Furthermore, material close to the axle travels very little distance, while material at the rim moves at a much higher speed. This non-uniform motion means that mass distribution plays a critical role.
Adapting Linear Kinematics to Rotation
Equations of uniform linear motion are adapted for rotary motion by substituting equivalent angular quantities:
- Linear displacement (s) → Angular displacement (θ)
- Linear velocity (v) → Angular velocity (ω)
- Linear acceleration (a) → Angular acceleration (α), where α = a / r
- Force (F) → Torque (Ï„), defined as Ï„ = F × r
1. Mass Moment of Inertia (I)
Mass represents an object’s resistance to linear acceleration. In rotational motion, resistance to angular acceleration depends not only on mass but also on how that mass is distributed. The quantity describing this resistance is the mass moment of inertia (I), measured in kg·m².
For a uniform solid disc (pulleys, flywheels):
I = (m · r²) / 2
If a manufacturer specifies a radius of gyration (k), the moment of inertia is calculated as:
I = m · k²
Advertisement
2. Newton’s Second Law for Rotation
We can now write the rotational equivalent of Newton’s second law. This equation is the foundation for sizing electric motors and analyzing dynamic loads:
Ï„ = I · α
The Engineering Role of Flywheels
A flywheel is a massive wheel mounted on a shaft specifically to provide rotational inertia. Its purpose is to smooth out speed fluctuations in machines with pulsating inputs, such as internal combustion engines.
Example Calculation
Problem: An electric motor accelerates a flywheel from rest to 3000 rev/min in 40 seconds. The flywheel is a uniform disc (m = 96 kg, r = 0.75 m). Calculate the required torque.
Figure 1: Flywheel acceleration parameters.
1. Find Final Angular Velocity (ω):
ω = (3000 rev/min × 2Ï€) / 60 = 314.2 rad/s
2. Calculate Angular Acceleration (α):
α = (ω2 - ω1) / t = (314.2 - 0) / 40 = 7.854 rad/s²
3. Calculate Moment of Inertia (I):
I = (m · r²) / 2 = (96 × 0.75²) / 2 = 27 kg·m²
4. Determine Required Torque (Ï„):
Ï„ = I · α = 27 × 7.854 = 212 N·m
Comments