# Rotating Mechanical Systems; Printable Collection

This document is a compilation of all of the pages regarding rotating mechanical systems and is useful for printing.

Contents

# Rotating Mechanical Systems Background

Animation by Ames Bielenberg

## Introduction

This page describes the development of mathematical models of rotating mechanical systems, though some translate (move back and forth) as well.  It is suggested that you read about translating mechanical systems first.  Since many of the concepts involved with rotating systems are analogous to those in translating systems, many of the discussions in these pages will be brief.

A list of the fundamental units of interest is listed below.  The next tab (above: "System Elements") gives a description of the building blocks of these system (inertia, spring and friction elements). This is followed by a description of  methods to go from a drawing of a system to a mathematical model of a system in the form of differential equations ("Mathematical Model").  Methods for solving the equation are given elsewhere.  The last section discusses topics relevant to energy storage and dissipation in these systems ("Energy Power").

This page does not discuss the solution of these equations, only the development of the equations through a physical model of the system.

## Table of units

The table below lists commonly used units of measure for Rotating   mechanical systems in SI units.  More complete tables are available.

 Fundamental Quantities SI unit Time - t second (s) Moment of Inertia - J kilogram (kg-m²) Angle - θ radians (rad) Torque - τ Newton (N-m) Energy - w Joule (J) [W-s, N-m] Power - p Watt (W) [J/s] Spring Constant - Kr (N-m/rad) Friction Coefficient - Br (N-m-s/rad)

Erik Cheever       Department of Engineering         Swarthmore College

# Elements of Rotating Mechanical Systems

Contents

As with translating mechanical systems, there are three fundamental physical elements that comprise rotating mechanical system: inertia elements, springs and friction elements.

## Analogy to Translating Systems

The quantities used in rotating systems are very similar to those in translating systems.  The table below lists the analogous quantities for rotating and translating systems.  The constitutive equations for inertial elements, springs and friction elements follow.

 Rotating Systems Translating Systems Quantity Unit Quantity Unit Moment of Inertia - J kg-m² Mass - m kg Torque - τ N-m Force - f N Angle - θ rad Length - l m Angular velocity - rad/sec Velocity - m/s Angular acceleration - rad/sec² Acceleration - m/s² Spring Constant - Kr N-m/rad Spring Constant - k N/m Friction Coefficient - Br N-m-s/rad Friction Coefficient - g N-s/m

Note that the spring and friction elements for the rotating systems will use capital letters with a subscript r (Kr, Br), while the translating systems will use a lowercase letter.

One of the difficulties in working with rotating systems (as opposed to those that translate) is that there are often multiple ways to make diagrams of the systems.   In examples below, a few alternatives are given; you should try to become comfortable with the multiple representations.

## Intertia Elements

In rotating mechanical systems, the inertia elements are masses that rotate and are characterized by moment of inertia.  The moment of inertia for some common shapes are given below.

 Shape Image Moment of Inertia, J Cylinder, radius=r, mass=mRotating about center axis Solid Sphere, radius=r, mass=mRotating about center Uniform Rod, length=ℓ, mass=mRotating about end Uniform Rod, length=ℓ, mass=mRotating about center Mass at end of massless rod,length=ℓ, mass=mRotating about end

The relationship between torque, moment of inertia and angular acceleration is given by:

(Translating system equivalent: )

## Springs

A rotational spring is an element that is deformed (wound or unwound) in direct proportion to the amount of torque applied.  Ideal springs have no inertia.  The relationship between torque, spring constant and angle is given by:

(Translating system equivalent: )

A photo of typical rotational springs is shown.  You should be able to visualize how an applied torque to wind (or unwind) the spring would be opposed.

A shaft connected between two elements can also act as a rotational spring.  Click the start button to see an animation of a flexible shaft (i.e., rotational spring) that is fixed on the left but free to rotate on the right.  Note how the shaft twists as the flywheel (with moment of inertia, J), turns (however the rotation doesn't slow because there is no friction).

In practice a rotational spring is drawn as a shaft (with an associated spring constant) or with the same symbol as a translating spring, but with the spring constant as a capital letter with a subscript.  Both images below represent the system from the animation, with theta defined positive in the counterclockwise direction.

 Flexible Shaft Spring

## Friction Elements

As with the translating systems, friction is the most difficult of the three elements to model accurately and we will generally only consider viscous friction.  The constitutive equation relating angular velocity, torque and friction coefficient is

(Translating system equivalent: )

A rotation friction element often consists of an object moving in a fluid, very similar to the translating dashpot, but with a rotary motion.  An example of such a device is used in some exercise equipment.  The photo below shows an exercise trainer with a paddlewheel in liquid (the blue circular object).  As you exercise, the vanes of the paddlewheel move against the fluid, creating a torque in resistance to your motion.

There are several ways of drawing these friction elements.  Consider the animation shown below with one flywheel (J1) attached by a flexible shaft (Kr) to a wall (as before), but now there is a second flywheel (J2) that is driven by friction between the two flywheels (Br1), and friction to the stationary ground (Br2).  J1 starts with an initial value, so it starts oscillating back and forth as the simulation begins.  This drives J2, through Br1, but the energy in the system decays over time because energy is lost to the friction.

This system can be shown schematically in a few ways.  In the first diagram below, the shaft is shown schematically as a spring, the friction Br1 is drawn as a dashpot, while the friction Br2 is shown as hash marks against ground.  In the second diagram the shaft is not drawn as a spring but is shown to be a spring with the label Kr; both friction elements are shown as dashpots.  It is important to be comfortable with both drawings (and others that could be drawn).

 Kr=SpringBr1=dashpotBr2=hashmarks Kr=Flexible ShaftBr1=dashpotBr2=dashpot

## Key points

Three elements were introduced, springs, friction elements and inertial elements (masses).  An ideal linear spring has no mass and a linear relationship between force and elongation.  For viscous friction there is a linear relationship between force and velocity.  Friction may either be between two surfaces (depicted as hash marks) or between two objects (depicted as a dashpot).  An ideal dashpot is also massless.  Masses have a linear relationship between force and acceleration.

##### Key Concept: Constitutive Equations for Rotating Mechanical Elements
 Spring: Friction: Inertia:

Erik Cheever       Department of Engineering         Swarthmore College

# Developing Mathematical Models of Rotating Mechanical Systems

Contents

The discussions on this page follow closely, and draw on, the techniques used to model translating mechanical systems.  It is assumed that you are familiar with those techniques.  We can also combine translating and rotating systems as discussed here.

## D'Alembert's Law

For rotating mechanical systems

(translating system equivalent: )

If we consider the J·α term to be a torque, we are left with D'Alembert's law

(translating system equivalent)

We will call the J·α term D'Alembert's torque, or inertial torque.  The inertial torque is always in a direction opposite to the defined positive direction.

##### Key Concept: D'Alembert's Law

Equations of motion for translating mechanical system depend on the application of D'Alembert's law.  Using this principle we say that the sum of all torques on an object is equal to zero,

but we must also take the inertial torque (J·α) as one of these torques.  This inertial torque is in the opposite direction from the defined positive direction.

## Free Body Diagrams

We can apply D'Alembert's law to develop equations of motion for rotating mechanical systems through the use of free body diagrams.  To do this we draw a free body diagram for each unknown position in a system.   This is very similar to the way this was done for translating mechanical systems

##### Example: Equations of Motion for a Rotating System

In the system shown one flywheel (J1) is attached by a flexible shaft (Kr) to ground (the unmoving wall) and has an applied torque, τa.  A second flywheel (J2) is driven by friction between the two flywheels (Br1).  The second flywheel also has friction to the ground (Br2).  Derive equations of motion for the system shown.

Solution:  First we must define our variables of motion.  In this case there are two - the angles of the flywheel.  It is generally a good idea to define the variables in the same direction, so we arbitrarily define them as positive in the counterclockwise direction.

We now create our free body diagrams

 Free body diagram at θ1 Free body diagram at θ2 There are 4 torques acting: The external torque, τa, clockwise. The torque due to Kr. If θ1 increases (counterclockwise), Kr causes a clockwise torque on J1. The resulting torque is Kr·θ1, clockwise. The torque due to Br1. If θ1 increases, the resulting torque on J1 is Br1·ω1 in the clockwise direction. If θ2 increases, the resulting torque on J1 is Br1·ω2 in the counterclockwise direction. The torque due to Br1 is thus Br1·(ω1-ω2), clockwise. The torque due to J1 (don't forget this - the inertial torque!). The resulting torque is J1·α1 clockwise (the inertial torque is always in the opposite direction from the define positive direction). There are 3 torques acting: The torque due to Br2. If θ2 increases (counterclockwise), the resulting torque is Br2·ω2, clockwise. The torque due to Br1. If θ1 increases, the resulting torque on J2 is Br1·ω1 in the counterclockwise direction. If θ2 increases, the resulting torque on J2 is Br1·ω2 in the clockwise direction. The torque due to Br1 is thus Br1·(ω2-ω1), clockwise (or  Br1·(ω1-ω2) counterclockwise).(Note: this result is trivial because the torque on one end of Br1 must be equal and opposite to the torque on the other end (calculated for the free body diagram for θ1).) The torque due to J2 (the inertial torque). The resulting torque is J2·α2 clockwise.

## From System Diagram to State Space Model

It is possible to go directly from the system diagram to a state space model.  Details are here system diagram to a state space model.

## Solving the model

Thus far we have only developed the differential equations that represent a system.  To solve the system, the model must be put into a more useful mathematical representation such as transfer function or state space.   Details about developing the mathematical representation are here.

Erik Cheever       Department of Engineering         Swarthmore College

# Gears and Systems with both Rotation and Translation

Contents

## Gears

Gears perform many functions, in this section we look at gears that increase or reduce angular velocity (while simultaneously decreasing or increasing torque, such that energy is conserved).  In many ways gears act in rotating systems as do levers in translating systems.   A picture of gears is shown below, along with a schematic representation.

photograph from: http://www.geardesign.co.uk/spur-gears.htm

We will specify gears by their radii.  When you buy gears you normally specify the number of teeth and the number of teeth per inch (or centimeter).  These two measures are equivalent: the number of teeth and teeth per inch give the circumference, which in turn determines the radius.   If the number of teeth are doubled, the radius is doubled.

If we consider two gears in equilibrium and in contact with each other, we can two very useful relationships.

First we note the geometric relationship that results from the path that the arc lengths along their circumference must be equal as the gears turn.

Since the arc lengths (shown with a heavy blue line) must be equal

r1θ1 = r2θ2 = arc length

If we had defined θ2 in the opposite direction, this expression would have a negative sign (r1θ1 = -r2θ2).

We derive a second relationship from a torque balance.  Before we can do so we must define a force between the gears termed a "contact force."  This force must be equal and opposite across the interface between the two gears, but its direction is arbitrary.  We start by drawing free body diagrams with a contact force where the gears meet.

The contact force is tangent to both gears and so produces a torque that is equal to the radius times the force.

We can do a torque balance on each of the two gears

 Gear 1 Gear 2

We are not usually interested in fc, so we remove from the equations and we get

It is easy to show that the same relationship exists even if the contact forces are defined in the opposite direction (up on gear 1, down on gear 2).

##### Example: Equations of motion for a system with gears

In the system below, a torque, τa, is applied to gear 1 (with moment of inertia J1).  It, in turn, is connected to gear 2 (with moment of inertia J2) and a rotational friction Br.  The angle θ1 is defined positive clockwise, θ2 is defined positive clockwise.  The torque acts in the direction of θ1

Derive a differential equation for the system below in terms of θ1 (i.e., θ2 and fc should not be in the final result).

We start by drawing free body diagrams, including a contact force that we will arbitrarily choose to be down on J1 and up on J2.  The directions of the reaction forces due to inertia and friction are chosen, as always, opposite to the defined positive direction.

We convert the force to a torque (note: we could have skipped the previous step and done this directly)

This yields the two equations of motion

We can easily solve for fc and eliminate it from the equations, but we also need to eliminate θ2.  To do this we use the relationship between θ1 and θ2 (from equal arc lengths).

r1θ1 = -r2θ2

Note that we have a negative sign here because of the way θ1 and θ2 were defined (if θ1 moves in the positive direction, then θ2 is negative).  When you use the arc length expression you must be careful of signs.

We can put this into the equation for J1 and solve (in standard form with the output (θ1) on the left, and the input (τa) on the right.

Note: the same result is obtained independent of the chosen direction for θ2 and fc.

The final result from the example is important deserves some notice.  The multiplication of J2 and Br by the square of the ratio of the radii is a well known, and general phenomenon.  If a load is attached to a system through gears, the perceived effect is multiplied by the square of the ratio of the radii.  We can use this to our advantage to make a load appear smaller (if we have a weak driving torque - for example, a small motor).  The reason bicycles have gears is so that the load perceived by the bicycle rider can be kept relatively constant even as s/he goes up or down hills.

##### Aside: Conservation of power in gear trains.

Note that if we combine the geometric relationship

with the torque relationship,

the product of torque and velocity becomes:

You will see on the next web page that the product of angular velocity and torque is power.  So the last equation simply states that power is preserved across a set of gears.  The gears allow one to trade torque for angular velocity (i.e., if the gears are of unequal size, the larger gear will experience higher torque  but lower angular velocity than the smaller gear).

## Pulleys

While this document doesn't discuss pulleys directly, the concepts are closely related (i.e., you must perform a torque balance around the pivot).  Pulleys are discussed here.

## Rotation and Translation

Systems that have parts that rotate and translate are handled in much the same way as with gears:

• Define relationships between rotation and position (i.e., relate arc length of translating elements to angle of rotating elements)
• Define a contact force between moving objects.
• After those steps are accomplished, all that is needed is to draw free body diagrams.  The only difference here is that if an element both translates and rotates, that two free body diagrams are needed for that element: one that rotates, and one that translates, but these motions can be considered independently.  (Ref)

The process will be illustrated with an example.

##### Example: Elements that rotate and translate

Derive equations of motion for the system shown along with an expression relating horizontal position to angle of rotation.  There is a force, fa, applied to a mass m.  A uniform cylinder sits on top of the mass and is free to roll and translate, but it rolls without slipping.

First we need to define positions, in this case three of them.  We need to define one for the position of the mass, m, and one for the angle through which the cylinder has turned about its center.   We also need to define one for the lateral position of the cylinder.

Now we can draw free body diagrams (again, we need three), and write the equations of motion.  Here we have taken fc to be to the right on m, and to the left on the cylinder.  Torques are taken about the center of mass.

 Freebody diagram at x1 Freebodyaround θ Freebodydiagram at x2 (note: this uses mass of cylinder)

The last thing we need is a relationship between position and arc length for the cylinder.  In this case it is not so obvious, but is easily found with a little thought.

• If x1 changes while x2 remains constant, then r·θ=x1
• If x2 changes while x1 remains constant, then r·θ=-x2

Therefore, by superposition

r·θ=x1-x2

##### Key Concept: Solving problems involving gears and/or systems that rotate and translate

When solving problems with gears, you always need to include two facts:

1. Relate arc lengths (for gears r1θ1 = r2θ2; for rotation and translation use rθ=x).
2. There is a contact force (equal and opposite) across the interface.

If you are not using both of these relationships, you probably won't be able to solve the problem.

The motion of an body can be separated into two independent component, translation and rotation.  If an element both rotates and translates a separate freebody diagram is needed for each component.

Erik Cheever       Department of Engineering         Swarthmore College

# Energy and Power in Rotating Mechanical Systems

This section discusses energy and power in rotating mechanical systems.  It is assumed you already know about energy and power in translating mechanical systems.  Accordingly, this page is rather brief.

## ENERGY

### Potential Energy

The potential energy, U, stored in a rotational spring with spring constant Kr that is wound to an angle θ is

(Translating system equivalent: )

### Kinetic energy

The kinetic energy, T,  stored in a rotating mass with moment of inertia J and angular velocity ω is

(Translating system equivalent: )

### Dissipated Energy

Dissipated energy in a friction element, Br, with angular velocity, ω,  is given by

(Translating system equivalent: )

## POWER

Power in rotating systems is given by

(Translating system equivalent: )

where τ is the torque and ω is the angular velocity.

Note: the previous page showed that power is conserved when there are gears in the system, but there is a tradeoff between angular velocity and torque.

Erik Cheever       Department of Engineering         Swarthmore College