Figure 1: Visual comparison . Steppers (Left) are dense and simple. Servos (Right) are longer and include a visible feedback encoder housing on the rear. The Million Dollar Question: "Which Motor Do I Need?" If you are designing a CNC machine, a packaging robot, or a conveyor system, you face the same dilemma every time: Stepper or Servo? Make the wrong choice, and you face two disasters: The Stepper Trap: Your machine "loses steps" (positional error) without knowing it, scrapping parts. The Servo Trap: You spend $5,000 on a system that could have been done for $500, blowing your budget. This guide bridges the gap between mechanical requirements and electrical reality. 1. The Stepper Motor: The "Digital Ratchet" Think of a Stepper Motor like a very strong, magnetic ratchet. It divides a full rotation into equal steps (typically 200 steps per revolution, or 1.8°). Pros: Incredible Holding Torque: Ste...
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
—in other words, determine the acceleration of its components—
we must consider rotary acceleration and the associated torques,
in addition to linear acceleration.
Fortunately, Newton’s second law of motion applies equally well to rotary motion, provided that the correct rotational form of the equation is used.
Consider a solid disc mounted on a shaft and rotated by a pull cord wrapped around its rim. In this case, we cannot apply the standard linear form of Newton’s second law, F = ma, because although a linear force is applied through the tension in the cord, the resulting motion is rotational rather than linear. There is no single straight-line acceleration that can be expressed in m/s2.
Furthermore, not all parts of the disc move in the same way. Material close to the axle does not travel any significant distance, while material at the rim moves at a much higher speed. This non-uniform motion means that mass distribution plays a critical role.
From kinematics, we know that equations of uniform linear motion can be adapted for rotary motion by substituting equivalent angular quantities. Linear displacement s is replaced by angular displacement q, linear velocity by angular velocity w, and linear acceleration a by angular acceleration a, where a = a / r.
Since we are dealing with rotation, force must also be replaced by torque. Torque t is defined as the product of force and radius:
t = Fr
The remaining task is to identify the rotational equivalent of mass.
Mass represents an object’s resistance to linear acceleration and is often referred to as inertia. An object with a larger mass accelerates more slowly than one with a smaller mass under the same applied force.
In rotational motion, resistance to angular acceleration depends not only on mass but also on how that mass is distributed. For example, a flywheel with most of its mass concentrated near the rim offers much greater resistance to rotation than a uniform wheel of the same mass.
The quantity that describes this resistance is called the mass moment of inertia, or simply the moment of inertia, denoted by I. Its units are kilograms metres squared (kg·m2).
For a uniform solid disc, which is a common approximation for engineering components such as pulleys and flywheels, the moment of inertia is given by:
I = mr2 / 2
We are now in a position to write Newton’s second law in a form suitable for rotary motion:
t = Ia
This equation allows engineers to analyze real systems that involve rotating components, and it forms the foundation for calculating motor torque requirements, acceleration times, and dynamic loads in machinery.
One of the most common applications of rotary inertia is the flywheel. A flywheel is a massive rotating wheel mounted on a shaft specifically to provide resistance to angular acceleration. Its purpose is to smooth out speed fluctuations, particularly in machines driven by pulsating inputs, such as internal combustion engines.
This analytical method can be extended to more complex shapes. Many rotating components, such as stepped pulleys, can be approximated as a series of uniform discs on a common axis. Because all parts rotate about the same axis, their individual moments of inertia may be summed to obtain the total moment of inertia.
Manufacturers sometimes provide moment of inertia data directly. Alternatively, it may be specified using the radius of gyration, denoted by k. In this case, the moment of inertia is calculated using:
I = mk2
Example: An electric motor accelerates a flywheel from rest to a speed of 3000 rev/min in 40 seconds. The flywheel is a uniform disc with a mass of 96 kg and a radius of 0.75 m. Calculate the angular acceleration and the torque required from the motor.
The first part of this problem uses rotary kinematics. The relevant equation is:
w2 = w1 + at
The initial angular velocity is zero since the flywheel starts from rest. The final angular velocity is:
3000 rev/min = 3000 × 2p / 60 = 314.2 rad/s
Substituting into the equation:
314.2 = 0 + 40a
a = 7.854 rad/s2
The moment of inertia of the flywheel is:
I = mr2 / 2 = 96 × 0.752 / 2 = 27 kg·m2
The required driving torque is therefore:
t = Ia = 27 × 7.854 = 212 N·m
Fortunately, Newton’s second law of motion applies equally well to rotary motion, provided that the correct rotational form of the equation is used.
Consider a solid disc mounted on a shaft and rotated by a pull cord wrapped around its rim. In this case, we cannot apply the standard linear form of Newton’s second law, F = ma, because although a linear force is applied through the tension in the cord, the resulting motion is rotational rather than linear. There is no single straight-line acceleration that can be expressed in m/s2.
Furthermore, not all parts of the disc move in the same way. Material close to the axle does not travel any significant distance, while material at the rim moves at a much higher speed. This non-uniform motion means that mass distribution plays a critical role.
From kinematics, we know that equations of uniform linear motion can be adapted for rotary motion by substituting equivalent angular quantities. Linear displacement s is replaced by angular displacement q, linear velocity by angular velocity w, and linear acceleration a by angular acceleration a, where a = a / r.
Since we are dealing with rotation, force must also be replaced by torque. Torque t is defined as the product of force and radius:
t = Fr
The remaining task is to identify the rotational equivalent of mass.
Mass represents an object’s resistance to linear acceleration and is often referred to as inertia. An object with a larger mass accelerates more slowly than one with a smaller mass under the same applied force.
In rotational motion, resistance to angular acceleration depends not only on mass but also on how that mass is distributed. For example, a flywheel with most of its mass concentrated near the rim offers much greater resistance to rotation than a uniform wheel of the same mass.
The quantity that describes this resistance is called the mass moment of inertia, or simply the moment of inertia, denoted by I. Its units are kilograms metres squared (kg·m2).
For a uniform solid disc, which is a common approximation for engineering components such as pulleys and flywheels, the moment of inertia is given by:
I = mr2 / 2
We are now in a position to write Newton’s second law in a form suitable for rotary motion:
t = Ia
This equation allows engineers to analyze real systems that involve rotating components, and it forms the foundation for calculating motor torque requirements, acceleration times, and dynamic loads in machinery.
One of the most common applications of rotary inertia is the flywheel. A flywheel is a massive rotating wheel mounted on a shaft specifically to provide resistance to angular acceleration. Its purpose is to smooth out speed fluctuations, particularly in machines driven by pulsating inputs, such as internal combustion engines.
This analytical method can be extended to more complex shapes. Many rotating components, such as stepped pulleys, can be approximated as a series of uniform discs on a common axis. Because all parts rotate about the same axis, their individual moments of inertia may be summed to obtain the total moment of inertia.
Manufacturers sometimes provide moment of inertia data directly. Alternatively, it may be specified using the radius of gyration, denoted by k. In this case, the moment of inertia is calculated using:
I = mk2
Example: An electric motor accelerates a flywheel from rest to a speed of 3000 rev/min in 40 seconds. The flywheel is a uniform disc with a mass of 96 kg and a radius of 0.75 m. Calculate the angular acceleration and the torque required from the motor.
The first part of this problem uses rotary kinematics. The relevant equation is:
w2 = w1 + at
The initial angular velocity is zero since the flywheel starts from rest. The final angular velocity is:
3000 rev/min = 3000 × 2p / 60 = 314.2 rad/s
Substituting into the equation:
314.2 = 0 + 40a
a = 7.854 rad/s2
The moment of inertia of the flywheel is:
I = mr2 / 2 = 96 × 0.752 / 2 = 27 kg·m2
The required driving torque is therefore:
t = Ia = 27 × 7.854 = 212 N·m
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