# Analytical Dynamics

Taught by: Dr. Riccardo Bevilacqua
Taken: FS 2015
Text: Dynamics of Particles and Rigid Bodies: A Systematic Approach
by: Anil Rao

## Scalars

• Scalars do not need reference frames or coordinate systems
• Derivates of scalars follow the formal definition of the derivative
• Ex. Let y(t) be a scalar function of time t,

## Vectors

• Vectors are quantities in (3D Euclidian Space) that have both magnitude and direction
• Magnitude is defined as
• Direction of the vector is defined by its unit vector
• Vectors are reference frame independent
• Vectors hold their true magnitude and direction no matter what reference frame you're observing from
• There are three types of vectors
1. Free vectors - & are the same vector if they have the same magnitude and direction
2. Sliding vectors - & are the same vector if they have the same magnitude and direction & share the same line of action in
• Ex. force vectors
3. Bound vectors - & are the same bound vector if they are the same sliding vector & share the same point of application
• Ex. position vectors

## Reference Frames

• A reference frame is any set of at least 3 non-colinear points whose mutual distances are constant
• A reference frame is a rigid body and vice versa
• Deciding reference frames allows for the formulation of problems
• They define a point of view put not a basis of measurement

## Coordinate System

• Coordinate systems are a set of three linearly independent vectors {} in (3D Euclidian Space)
• There are three main conviences that are made when creating a coordinate system
1. Have all three vectors {} be mutually orthogonal
2. Have all three vectors {} be unit vectors
3. Create a right handed system
• Coordinate systems are built on reference frames to allow measurement within the frame or allow a problem to be implemented
• The coordinate system moves with its reference frame
• Coordinate systems must be fixed with only one reference frame whereas each reference frame can have infinite coordinate systems
• Bigger problems come from mistakes in the formulation of the problem (deciding reference frames) than mistakes in the implementation of the problem (deciding and using the coordinate systems)
• Two steps to defining a coordinate system
1. Pick an origin
2. Define the three right-handed orthogonal unit vectors, {}

## Vector Derivatives

• This class will be using Newtonian Mechanics rather than Relativistic Mechanics
• Time is the independent variable ( Galilean Invariance)
• Although vectors are reference frame independent, the geometric meaning of their derivative vectors are not reference frame independent (though the derivative vectors themselves being vectors are reference frame independent)
• Development of the general form of a derivative of a vector as viewed from reference frame A
• The derivatives of the scalars are reference frame independent but the derivatives of the vectors are not
• Utilizing the multiplication rule of derivatives
• The first three terms represent the rate of change of vector in the coordinate system defined by the vectors
• The last three terms represent the rate of change of the coordinate system defined by the vectors with respect to reference frame A
• If the vectors {} are fixed in reference frame A then the last three terms in the derivative equal zero

## Transport Theorem

• The transport theorem is a simplification of the vector derivative and can be stated as where {} are fixed in reference frame B and is the angular velocity of frame B as seen by an observer in reference frame A
• Some useful properties of the angular velocity are:

### Transport Theorem Derivation

• Taking the resultant formula for the derivative of a vector from above where {} are fixed in reference frame B we have    or
• Next we need to use some of the mathematical definitions of the {} vectors
• The vectors are all unit vectors therefore for i = 1, 2, 3
• The vectors are all mutually orthogonal therefore for i j
• Taking the derivative of the unit vector rule we obtain
• A result of this is that it shows that a unit length vector, , will always be perpendicular to its derivative
• Taking the derivative of the orthogonal rule we obtain    or
• The next step is to write out the derivate of each unit vector in terms of a component of , the angular velocity, multiplied by each of the unit vectors
• We know from the derivative of the unit vector rule that the derivative of a unit vector will not contain any component along the direction of that unit vector ()
• We also know from the result of the derivative of the orthogonal rule that the component of the derivative of first vector in the second vectors direction is the negative of the second vectors derivative in the first vector's direction (, , )
• With these two rules we can re-write the three equations
• We now define the following three 's: , and
• With the redefinitions we can introduce , the angular velocity of frame B as seen by an observer in frame A as
• This definition of angular velocity allows the equation to produce the three above equations
• Substituting the equation for into the very first equation results in
• To simplify further separate the angular velocity out of the cross products:
• Now taking the definition of as we can re-write the above expression as which is the transport theorem

## Cylindrical Coordinate Systems

• In the given diagram a cylindrical coordinate system can be established in the reference frame defined by the plane containing points O, P and Q (or vectors r and )
• The coordinate system in this reference frame will be:
• Origin at O
• = along
• =
• =
• This coordinate system rotates with and as it rotates the line forms the outside of a cylinder which gives the coordinate system its name

## Spherical Coordinate Systems

• For the given diagram we will create a new reference frame that conatains the vector and is perpendicular to the plane created by and
• Note that this new reference plane differs from the reference frame containing the cylindrical coordinate system by an angle
• This reference frame rotates with and
• The spherical coordinate system in this new reference frame will be:
• Origin at O
• = along
• = (from cylindrical coordinate system)
• =
• For a constant , the full rotation of point P with respect to both and produces a sphere which is where the name for this coordinate system originates

## Euler Angles

1. There is no vector such that its derivative is equal to the angular acceleration
• For there will never be a vector such that , ,
2. Rotations do NOT commute
• Order is extremely important
• There are many different rotation orders (12 combinations for three axes)
• Can repeat numbers so long as the repeated numbers are not sequential
• The most commonly used rotation sequence is the 321 sequence (z, y, x)

• First rotation in the 321 sequence is around the z-axis

• The second rotation is around the new y-axis
• Last rotation is around the new x-axis
• Careful with the second and third rotations as they are around the NEW axis formed from the previous rotations
• We now have 4 different coordinate systems fixed in four different reference frames
• We can now define the angular velocity of the final, body frame as seen by the initial frame as:
• Due to the fact that the rotations can often be measured in reference frame R by gyroscopes, it is often desired to have the angular velocity defined in that coordinate system

### Limitations of Euler Angles

• For each sequence of rotations there is a rotation that will result in a singularity where there is an infinite combination of values for the first two rotations to arrive at that same orientation
• For a 321 sequence for example there is a singularity for a 90 degree rotation as the second rotation causing an infinite number of combinations of and to arrive at the same final orientation
• Singularities basically occur when the same axis in space is reused

## Intrinsic Coordinates

• An intrinsic coordinate system is one that is based within the motion and trajectory of a particle
• If there's no motion there can not be an intrinsic coordinate system
• The three unit vectors of an intrinsic coordinate system are the tangential vector (), the principal normal to the trajectory () and the principal binormal to the trajectory ()
• When defined for a using the motion of point P in the general reference frame A the three vectors are defined as:
• Vector points to the center of curvature of the trajectory of point P in reference frame A and so difficulties can be encountered when the trajectory is a straight line

## Motion Between Two Points in the Same Reference Frame

• For two points Q and P in the same rigid body, a relation can be derived for their velocities and accelerations
• For general reference frame A the position vector between points Q and P fixed in reference frame R can be given as:
• Now taking the time derivative of in the A reference frame and including the transport theorem between the A reference frame and the R reference frame leads to the following expression
• Noting that the distance between points Q and P do not change by definition of a rigid body the first term on the right hand side of the above expression equals zero
• By taking another time derivative of the above expression the difference in accelerations of the two points can be obtained
• Now keeping in mind that is the angular acceleration and that is the velocity expression that has already been found, the acceleration expression can be simplified to the following expression
• It is important to remember that cross products are not commutative and that the order in which they are performed is important

## Rolling and Slipping

• The condition for rolling between two rigid bodies R and S in a general reference frame is the following expression
• where point C is the point of contact between rigid bodies R and S
• Sliding between two rigid bodies occurs anytime the rolling condition is not met
• Note that when considering rolling motion there are three point C's to consider
1. The instantaneous point of contact between the two rigid bodies which doesn't belong to either body
2. The point C belonging to rigid body R
3. The point C belonging to rigid body S
• Therefore in the above condition for rolling the term stands for the velocity of point C belonging to rigid body R in reference frame A

## Inertial Reference Frames

• An inertial reference frame is one for which we will assume exists as a set of absolutely stationary points
• Inertial reference frames are an imaginary concept that is an approximation of reality since all motion is relative the question arises to whom are the set of points stationary
• We can pick reference frames to be our inertial reference frame when the degree of motion of the points is negligible compared to the motion that we are studying

## Three Laws of Mechanics

• When studying kinetics there the particles are no longer massless and we are considering the effect of action on the particles (forces/vectors)
1. The first law of mechanics or the inertia law states that an object at rest tends to remain at rest and an object in motion tends to remain in motion
2. The second law of mechanics relates forces to momentum as it's most common form shows ( where is the acceleration in an inertial reference frame)
3. The last of the three laws of mechanics states that for every action there is an equal and opposite reaction
• If the resultant force lies on the same line of action as the initial force (sliding vectors) then it is considered the strong form of the 3rd law
• The weak form of the third law is when the resultant force and the initial force are not on the same line of action

## Angular Momentum

• Some problems are easier to solve by using the concept of angular momentum where the definition of angular momentum for a particle P about point Q is given as
• where and are measured relative to the origin of N and m is the mass of particle P
• Another useful relation from the angular momentum is the time rate of change of the angular momentum
• Note that the first term involves taking the cross product of a vector with the same vector and is therefore zero and so the expression can be rewritten as
• The above expression can as the cross product of the distance between the two points with each acceleration individually
• Since m is the mass of particle P the term is the sum of forces acting on point P by Newton's third law of mechanics
• The term is the moment of all of the forces acting on particle P with respect to point Q and so the expression can be rewritten one more time as the following

## Relative Velocity

• Relative velocity is the vector that represents the motion of one point compared to another
• For two points P and Q the relative velocity can be expressed as follows
• Note that the relative velocity is reference frame independent

## Friction Force Models

• Irregardless of the model the friction force will oppose the relative motion between two surfaces
• Caution as it may not oppose the direction of overall motion of the object

Coulomb Friction

• is the resultant force normal to the point of contact
• coefficient of friction

Viscous Friction

• coefficient of vicsous friction

## Linear Spring Force Model

• For particle with mass and point being connected by a linear spring the force model can be written as follows
• the spring length when it is unstreched/uncompressed
• the spring constant

## Center of Mass

• Whether working with a system of particles or a rigid body the center of mass is an important point to know the position of while trying to solve problems
• For a system of particles the center of mass is found by taking the sum of the mass of individual particles multiplied by their position vectors and dividing by the overall mass of the system
• Finding the center of mass of a rigid body is practically the same process but due to the rigid body having potentially infinite points with mass the sum in the above expression naturally turns into an integral
• Additional quantities that are useful are the velocity and acceleration of the center of mass and these quantities are easy to find when considering that the mass of the system/rigid body is not time dependent
• For a system of particles
• For a rigid body

## Tensors

• Tensors are geometrical objects similar to vectors
• Tensors are reference frame independent
• Letting be a vector in , we define a tensor as a linear operator which takes vectors as inputs and returns vectors
• An example of a tensor that has been used often in this class already is the cross product
• If a tensor is represented by a matrix it must include a basis or it is not a tensor just a matrix
• A tensor is represented as operating on a vector by a large dot
• Tensor operating on vector

Tensor Product

• One common operation that will need to be defined is that of the tensor product
• is the tensor
• The dot on the right hand side is a dot product not an "operates on" operator therefore the result is a scalar multiplied by the vector
• Example of the tensor created by

## Angular Momentum of Rigid Bodies/Systems of Particles

• For a system of particles the angular momentum about point Q is given by the following expression. The first time derivative of the angluar momentum for a system of particles is also presented
• Where is the real moment given by
• For a rigid body the angular momentum about point Q is given by the following expression. The first time derivative of the angular momentum expression is also provided
• for generic point
• where is on rigid body
• if is the center of mass
• can be found in tables in text books for general shapes/configurations
• The generic expression of angular momentum of a rigid body about generic point Q can be found from the angular momentum about its center of mass from the following expression
• The parallel axis theorem can be found from this expression but this is a stronger expression and as such should stick to this one instead

## Pure Torque

• Pure torque is a moment on a rigid body that is independent of the point on the body for which the moment is being determined
• Can be represented by a force couple which creates a moment with zero net force
• Symbolically represented as
• Including pure torque in the expression for the sum of moments for a system of particles about generic point Q yields the following expression

## Euler's Laws

First Law

• Where N is inertial

Second Law

• Where O is a point that is inertially fixed
• When moved to a generic point that does not have to be inertially fixed the second law becomes the following

## Parameterize a Problem

• When looking at how to parameterize a problem we first must consider the degrees of freedom (M) the system has and how many of those degrees are constrained (P)

## Lagrange Equations

• Lagrange equations provide a means of formulating problems without forces caused by constraints on the system
• Lagrange equations are derived from Newton's Second Law ()
• We will need the different expressions for kinetic energy when working with Lagrange Equations

Kinetic Energy of a Particle

Kinetic Energy for a System of Particles

Kinetic Energy for a Rigid Body (Koenig Decomposition)

• The Fundamental Form of Lagrange's Equations is the following expression
• Where

• Lagrange's Equation for a rigid body is given by the following expression
• where
• for i = 1, ..., 6

## Examples

### Transport Theorem Examples

• First Transport Theorem Example:
• Given a disk (D) rotating about its centerpoint (O), find the velocity and acceleration of point P as seen by the disk (D) and the ground (G)
• Start by defining reference frames and then coordinate systems within them
• For this problem two reference frames will be used: the disk (D) and the ground (G)
• Coordinate system fixed in the disk reference frame (D)
• Origin at point O
• = along the line
• = out of the page (positive with theta)
• =
• Coordinate system fixed in the ground reference frame (G)
• Origin at point O
• =
• = along the line @
• =
• Now to find the velocity and acceleration of point P in the disk reference frame we take the time derivative of the vector from the origin O to point P ()
• Neither r nor are changing in the disk reference frame therefore
• Next the acceleration of point P in the disk reference frame is to be found by taking a time derivative of the velocity of point P in the disk reference frame
• Now the velocity needs to be found for point P in the ground reference frame and two different approaches will be shown
• First approach involves writing all of the equations with respect to the coordinate system fixed in reference frame G
• The second approach involves leaving the equations with terms referenced in the disk reference frame and uses the transport theorem
• The second approach leaves a much cleaner equation for further derivatives and so this is the form we'll use when finding the acceleration
• Second Transport Theorem Example:
• This example is the same as the previous one execept it has an additional rotation around its base
• It's good practice to have an additional reference frame for each rotation and so in this example we'll use three reference frames: disk (D), disk at (C) and the ground (G)
• Always start by defining the reference frames and then the coordinate systems within them
• Coordinate system fixed in the disk reference frame (D)
• Origin at point O
• = along
• = perpendicular to the disk, positive with
• =
• Coordinate system fixed in the ground reference frame (G)
• Origin at point O
• = Out of the page
• = along the line @ ,
• =
• Coordinate system fixed in the disk at reference frame (C) (rotates with )
• Origin at point O
• = along at
• =
• Just as in the previous example the velocity and acceleration in the disk reference frame (D) will equal zero
• Now to find the velocity of point P in the ground reference frame we are going to take advantage of the transport theorem again

### Cylindrical Coordinate System Example

• Find the equation for the velocity and acceleration of point P in the A reference frame ()
• For this problem we will use two different approaches:
• The first approach will be to derive the velocity and acceleration using a coordinate system fixed in reference frame A
• The second approach will be to develop and use a cylindrical coordinate system fixed in a reference frame C
• For the first approach the reference frame is already defined in the following figure
• The vector will be defined in these coordinates as
• We can now find
• The same process can be used to find
• For the second approach we need more formal clarifications of our reference frames and coordinate systems
• We will be using two reference frames: reference frame A in which the {} are already fixed and reference frame B which is the plane containing points O, P and Q (or vectors and )
• Coordinate system fixed in reference frame B (Cylindrical Coordinates)
• Origin at point O
• = along line
• =
• =
• Now we're going to define in the cylindrical coordinates (in reference frame B)
• We can now find using the cylindrical coordinates
• Now that we have the velocity we can find the acceleration

### Spherical Coordinate System Example

• The problem is to find the velocity and acceleration of point P in the A reference frame ( and )
• Since the example for cylindrical coordinate systems already found the expression using the coordinate system in reference frame A and using the cylindrical coordinate system, this example will only be using the spherical coordinate system approach
• In order to define the spherical coordinate system three reference frames will be used:
• The given reference frame (A)
• The reference frame fixed in the plane containing points O, P and Q (or vectors and ) (B)
• The reference frame fixed in the plane containing vector that is perpendicular to the plane containing vectors and (C)
• Coordinate system in reference frame A is already given and origin at point O
• Coordinate system fixed in reference frame B (Cylindrical Coordinates)
• Origin at point O
• = along line
• =
• =
• Coordinate system fixed in referenc frame C (Spherical Coordinates)
• Origin at point O
• First note that the angular accelertation between reference frames A and C can be expressed as
• Now we can find the velocity of point P in the A reference frame by making use of the transport theorem and the following definition of :
• The same process can be used to find the acceleration of point P in reference frame A, however, the expression becomes very large and so it will be omitted here for space

### Intrinsic Coordinate System Example

• Given the above set up find the intrinsic coordinate system for point P
• To begin the velocity and acceleration of point P will be found in reference frame A
• Two different reference frames will be used: the given reference frame (A) and the reference frame using the plane containing vectors and
• Coordinate system fixed in reference frame A
• Origin at O
• to the right
• out of the page
• Coordinate system fixed in reference frame B
• Origin at O
• along
• Now we can find the velocity of point P in reference frame A using
• Now the that the velocity has been found the acceleration of point P in reference frame A can be found
• Let us now find the vectors of the intrinsic coordinate system attached to point P
• Now that the tangential vector has been found the principle normal to the trajectory can be found
• Finally the princible binormal to the trajectory can be determined

### Rolling Motion

• This example also includes examples on motion between two points fixed in the same rigid body
• The diagram for the problem is given as follows
• It is desired to find the velocity and acceleration of point O and point P given that there is roll without slip between the disk and the wedge
• To begin we will define two different reference frames, the disk (D) and the ground (G)
• Coordinate system fixed in the ground reference frame (G)
• Origin at point O at time = 0
• = into the page
• = along
• =
• Coordinate system fixed in the disk reference frame (D)
• Origin at point O
• = along
• =
• =
• We will begin by stating the condition for rolling without slip
• In this case we will use the condition in the ground reference frame (A = G)
• Now we can note that point C in the ground reference frame is fixed and therefore its velocity is zero and as a consequence the point C in the disk reference frame also has zero velocity
• Now we will make use of the velocity relations for two points fixed in the same rigid body to relate the velocity between points C and O
• From
• We can use the same relation between two points fixed in a rigid body to now relate the velocities of point O and point P
• With both of the desired velocities found we can now find the desired accelerations. We will be begin with the acceleration of point O
• In order to find the acceleration of point P we are going to use the relation between the accelerations of point P and point O
• As a bonus the acceleration of point C on the disk will be found. It should be noted that the rolling condition is strictly for velocities and not the velocities ()

### Basic Equation of Motion

• The problem is to determine the equations of motion for mass located at point P which is attached by a rod to point O which is fixed in the ground and the ground reference frame can be considered intertial
• Only one equation of motion is expected due to the mass being constrained to a trajectory that lies on a single line
• We will begin by defining our reference frames. For this problem two frames will be used due to the presence of a rotation
• Ground reference frame (G)
• Plane perpendicular to the page containing (A)
• Coordinate system fixed in the ground reference frame (G)
• Origin fixed at O
• = out of the page
• = along at = 0
• =
• Coordinate system fixed in reference frame A
• Origin fixed at point O
• = along
• =
• =
• Now we will begin by finding the position velocity and acceleration of point P in the ground reference frame
• With the acceleration of point P in the ground reference frame found we can now move on to the kinetics portion of the problem
• We will begin this portion with a free body diagram to determine what forces are present
• The sum of forces is therefore as follows
• With the sum of forces found we can now express all fo the terms in Newton's Second Law
• In the above expression the only unknown value is , therefore we can find the desired equation of motion by taking the dot product of each term by

### Angular Momentum

• The problem in this example is to find the equations of motion for the point P in the folling figure where point Q slides around the circle at given speed
• For this problem two different reference frames will be use
• Ground (G), treated as inertial
• Frame attached to rod along (A)
• Coordinate system fixed in Ground reference frame (G)
• Origin at point O
• given in the figure
• Coordinate system fixed in reference frame (A)
• Origin at point Q
• = along
• =
• =
• The first thing that will need to be done will be to solve for the kinematics of point P in the inertial reference frame (G)
• Next a free body diagram will be used so that the equations of motion can be found
• The equations of motion will be found using two different methods. The first will use Newton's second law for reference when the angular momentum approach is used.
• Using Newton's second law will lead to the following result

Incomplete

Incomplete

Incomplete

### Basic System of Particles

• For the above diagram it is desired to obtain the equations of motion for both the wedge (W) and the block (B)
• This should lead to two equations of motion as both objects are constrained to a single degree of freedom
• To begin solving the problem 2 reference frames will be used and then coordinate systems will be defined in each reference frame
• Ground (G)
• Wedge (W)
• Coordinate system in Ground reference frame (G)
• Origin at O
• = to the right
• =
• =
• Coordinate system in Wedge reference frame (W)
• Origin at O'
• = along the slant away from O'
• =
• =
• Next the kinematics need to be solved for the Wedge
• Now the kinematics for the block will be determined
• h is the height of the wedge
• The relative velocity between the block and the wedge will also be needed when determining the friction force
• Last bit of kinematics will be for the wedge block system
• The next step towards obtaining the equations of motion will be to determine the forces acting on each part of the system and the system as a whole
• Now we can start to apply Newton's Second Law and its derived result on the wedge, block and system
• Newton's Second Law for the wedge results in the following expression
• Newton's Second Law for the block results in the following expression
• The derived result of Newton's Second Law will now be used on the system of particles
• The first equation of motion can be found by using the equation for the system and taking the dot product of each individual term with
• Where therefore
• For the second and final equation of motion take the result of Newtons Second Law from the block and take the dot product of each term with
• The two equations of motion for the system are therefore as follows