For engineers who already know the math—but still lose projects. For the last few years, I’ve been sharing technical guides here on Mechanical Design Handbook —how to size a motor, how to calculate fits, and (as you recently read) how to choose between timing belts and ball screws. But after 25 years in industrial automation, I realized something uncomfortable: Projects rarely fail because the math was wrong. They fail because: The client changed the scope three times in one week. A critical vendor lied about a shipping date (and no one verified it). The installation technician couldn’t fit a wrench into the gap we designed. University taught us the physics. It didn’t teach us the reality. That gap is why I wrote my new book, The Sheet Mechanic . This is not a textbook. It is a field manual for the messy, political, and chaotic space between the CAD model and the factory floor. It captures the systems I’ve used to survive industrial projec...
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.
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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
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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
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