Analytical Dynamics
Taught by: Dr. Riccardo Bevilacqua
Taken: FS 2015
Text: Dynamics of Particles and Rigid Bodies: A Systematic Approach
by: Anil Rao
Table of Contents
Kinematics
Kinetics of Particles
Kinematics and Kinetics of Rigid Bodies/Systems of Particles
Analytical Dynamics
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
 Free vectors  & are the same vector if they have the same magnitude and direction
 Sliding vectors  & are the same vector if they have the same magnitude and direction & share the same line of action in
 Ex. force vectors
 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 noncolinear 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
 Have all three vectors {} be mutually orthogonal
 Have all three vectors {} be unit vectors
 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
 Pick an origin
 Define the three righthanded 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 rewrite 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 rewrite 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
 Start with a few important defintions of angles
 There is no vector such that its derivative is equal to the angular acceleration
 For there will never be a vector such that , ,
 Rotations do NOT commute
 Order is extremely important
 There is no vector such that its derivative is equal to the angular acceleration
 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 zaxis
 The second rotation is around the new yaxis
 Last rotation is around the new xaxis
 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
 The instantaneous point of contact between the two rigid bodies which doesn't belong to either body
 The point C belonging to rigid body R
 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)
 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
 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)
 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
 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
 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
 For a system of particles
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
 Where
 Lagrange's Equation for a rigid body is given by the following expression
 where
 for i = 1, ..., 6
Examples
Examples Content
Kinematics
Kinetics
Kinetics for a System of Particles
Kinetics for Rigid Bodies
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
 Start with the tangential vector
 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
Friction Force Models
Incomplete
Linear Spring Force Models
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
Angular Momentum
Basic Rigid Body Problem
Gyroscope Example
Pendulum Example