Chapter 21: Three Dimensional Kinetics

Chapter 21: Three Dimensional Kinetics
of a Rigid Body

Chapter Objectives • To introduce the methods for finding the moments of inertia and products of inertia of a body about various axes. • To show how to apply the principles of work and energy and linear and angular momentum to a rigid body having three-dimensional motion. • To develop and apply the equations of motion in three dimensions. • To study the motion of a gyroscope and torque-free motion.

Chapter Outline Moments and Products of Inertia Angular Momentum
Kinetic Energy Equations of Motion Gyroscopic Motion Torque-Free Motion

21.1 Moments and Products of Inertia
Moment of Inertia Consider the rigid body The moment of inertia for a differential element dm of the body about any one of the three coordinate axes is defined as the product of the mass of the element and the square of the shortest distance from the axis to the element From the figure,

Moment of Inertia For mass moment of inertia of dm about the x axis, For each of the axes,

Moment of Inertia Moment of inertia is always a positive quantity Hence, summation of the product of mass dm is always positive

Product of Inertia Defined with respect to a set of orthogonal planes as the product of the mass of the element and the perpendicular (or shortest) distances from the plane to the element For product of inertia dIxy for element dm, dIxy = xy dm Note dIyx = dIxy

Product of Inertia For each combination of planes,

Product of Inertia Unlike the moment of inertia, which is always positive, the product of inertia may be positive, negative or zero Result depends on the sign of the two defining coordinates, which vary independently from one another If either one or both of the orthogonal planes are planes of symmetry for the mass, the product of inertia with respect to these planes will be zero

Product of Inertia In such cases, the elements of mass will occur in pairs on each side of the plane of symmetry On one side of the plane, the product of inertia will be positive while on the other side, the product of inertia will be negative, the sum therefore yielding zero Consider the y-z plane of symmetry, Ixy = Ixz = 0

Product of Inertia Calculation of Iyz will yield a positive result since all elements of mass are located using only positive y and z coordinates Consider the x-z and y-z being planes of symmetry, Ixy = Iyz = Ixz = 0

Parallel Axis and Parallel Plane Theorems If G has coordinates xG, yG, zG defined from the z, y, z axes, the parallel axis equations used to calculate the moments of inertia about the x, y, z axes are Products of inertia of a composite body are computed in the same manners as the body’s moments of inertia

Parallel Axis and Parallel Plane Theorems Parallel axis theorem is used to transfer products of inertia of the body from a set of three orthogonal planes passing through the body’s mass center to a corresponding set of parallel planes passing through some other points O For parallel axis equations,

Inertia Tensor Inertial properties of a body are completely characterized by nine terms, six of which are independent of one another For an inertia tensor, An inertia tensor has a unique set of values for a body when it is computed for each location of the origin O and orientation of the coordinates origin

Inertia Tensor For point O, specify a unique axes inclination for which the products of inertia for the body are zero when computed with respect to these axes For a diagonalized inertia tensor, For principal moments of inertia for the body

Inertia Tensor Of these three principal moments of inertia, one will be a maximum and one will be a minimum If the coordinates axes are orientated such that two of the three orthogonal planes containing the axes are planes of symmetry for the body, then all the products of inertia for the body are zero with respect to the coordinate planes, hence the coordinate axes are principle axes of inertia

Moment of Inertia About an Arbitrary Axis Consider the body where nine elements of the inertia tensor have been computed for the x, y, z axes having an origin at O Determine the moment of inertia of the body about the Oa axis, for which the direction is defined by vector ua IOa = ∫b2 dm where b is the perpendicular distance from dm to Oa

Moment of Inertia About an Arbitrary Axis Position of dm is located using r, b = rsinθ, which represents the magnitude of the cross-product ua x r For moment of inertia, Provided

Moment of Inertia About an Arbitrary Axis Hence, For moment of inertia,

Moment of Inertia About an Arbitrary Axis If the inertia tensor is specified for the x, y, z axes, the moment of inertia of the body about the inclined Oa axis can be found Direction cosines ux, uy and uz can be determined Direction angles α, β and γ made between the positive Oa axis and the positive x, y, z axes can be determined from the direction cosines ux, uy and uz, respectively

Example 21.1 Determine the moment of inertia of the bent rod about the Aa axis. The mass of each of the three segments are shown in the figure.

View Free Body Diagram Solution For moment of inertia of a slender rod, I = 1/12 ml2 For each segment of the rod,

Solution For unit vector of the Aa axis, Thus, Hence,

21.2 Angular Momentum Consider the rigid body having a mass m and center of mass at G X, Y, Z coordinate system represent an inertial frame of reference and its axes are fixed or translating with a constant velocity Angular momentum as measured from this reference will be computed relative to the arbitrary point A

21.2 Angular Momentum Position vectors rA and ρA are drawn from the coordinates to point A and from A to the ith particle of the body If the particle’s mass is mi, for angular momentum about point A, (HA)i = ρA x mivi where vi represent the particle’s velocity measured from the X, Y, Z coordinate system If the body has an angular velocity ω, vi = vA + ω x ρA

21.2 Angular Momentum Hence, (HA)i = ρA x mi(vA + ω x ρA )
= (ρAmi) x vA + ρA x (ω x ρA)mi Summing all the particles of the body, HA = (∫m ρA dm) x vA + ∫m ρA x (ω x ρA) dm

21.2 Angular Momentum Fixed Point O
If A becomes fixed point O in the body, vA = 0 HO = ∫m ρO x (ω x ρO) dm

21.2 Angular Momentum Center of Mass G
If A is located at the center of mass G, ∫m ρA dm = 0 HG = ∫m ρG x (ω x ρG) dm

21.2 Angular Momentum Arbitrary Point A
In general, A may be some point other than O or G HA = ρG/A x mvG + HG Angular momentum consists of two parts – the moment of the linear momentum mvG of the body about point A added (vectorially) the angular momentum HG

21.2 Angular Momentum Rectangular Components of H
Choosing a second set of x, y, z axes having an arbitrary orientation relative to the X, Y, Z axes, H = ∫m ρ x (ω x ρ) dm Expressing in terms of x, y, z components, Hxi + Hyj + Hzk = ∫m (xi + yj + zk ) x [(ωxi + ωyj + ωzk ) x (xi + yj + zk )]dm Expanding and equating the i, j and k components, Hx = Ixxωx - Ixyωy - Ixzωz Hy = Iyyωy - Iyxωx - Iyzωz Hz = Izzωz - Izxωx - Izxωx

21.2 Angular Momentum Rectangular Components of H
If the x, y, z coordinate axes are oriented such as they become the principal axes of inertia for the body at that point If these axes are used, for products of inertia, Ixy = Iyz = Izx = 0 If the principal moments of inertia about the x, y, z axes are represented as Ix = Ixx, Iy = Iyy, Iz = Izz, for components of angular momentum, Hx = Ixωx, Hy = Iyωy, Hz = Izωz

21.2 Angular Momentum Principle of Impulse and Momentum
For principle of impulse and momentum, In 3D, each vector term can be represented by 3 scalar components and 6 scalar equations 3 equations relate the linear impulse and momentum in the x, y, z directions and the other 3 equations relate the body’s angular impulse and momentum about the x, y, z axes

21.3 Kinetic Energy Consider the rigid body which has a mass m and center of mass at G For kinetic energy of the ith particle of the body having a mass mi and velocity vi measured relative to the inertial X, Y, Z frame of reference, Provided the velocity of an arbitrary point A of the body is known,

21.3 Kinetic Energy For kinetic energy of the particle,
The kinetic energy for the entire body is obtained by summing the kinetic energies of the body The last term on the right can be re-written as

21.3 Kinetic Energy Hence,

21.3 Kinetic Energy Fixed Point O If A is a fixed point O in the body,

21.3 Kinetic Energy Center of Mass G
If A is located at the center of mass G of the body, Kinetic energy consists of the translational kinetic energy of the mass center and the body’s rotational kinetic energy

21.3 Kinetic Energy Principle of Work and Energy
Used to solve problems involving force, velocity and displacement Only one scalar equation can be written for each body T1 + ∑U1-2 = T2

21.4 Equations of Motion Equations of Translation Motion
Defined in terms of acceleration of the body’s mass center, which is measured from an inertial X, Y, Z reference For equations of translation motion in vector form, For scalar equations of translation motion, For sum of all external forces acting on the body,

21.4 Equations of Motion Equations of Rotational Motion
which states that sum of moments about a fixed point O of all the external forces acting on a system of particles (contained in a rigid body) is equal to the time rate of change of the total angular momentum of the body about point O When moments of external forces acting on the particles are summed about the system’s mass center G, one again obtain summation ∑MG to the angular momentum HG

Consider a system of particles where X, Y, Z represents an inertial frame of reference and the x, y, z axes with origin at G, translate with respect to this frame In general, G is accelerating, so by definition, the translating frame is not an inertial reference

For the angular momentum of the ith particle with respect to this frame, Taking the time derivatives, By definition, Thus, the first term on the right side is zero since the cross-product of equal vectors equals zero.

21.4 Equations of Motion Equations of Rotational Motion Since Hence,
When the results are summed, for the time change of the total angular momentum of the body computed relative to point G,

For relative acceleration for the ith particle, where ai and aG represent accelerations of the ith particle and point G measured with respect to the inertial frame of reference By vector cross-product, By definition of the mass center, since position vector r relative to G is zero,

Using the equation of motion, For the rotational equation of motion for the body, If the scalar components of the angular momentum HO or HG are computed about x, y, z axes that are rotating with an angular velocity Ω, which may be different from the body’s angular velocity ω, then the time derivative

must be used to account for the rotation of the x, y, z axes as measured from the inertial X, Y, Z axes For the time derivative of H, There are 3 ways to define the motion of the x, y, z axes

21.4 Equations of Motion x, y, z Axes having motion Ω = 0
If the body has general motion, the x, y, z axes may be chosen with origin at G, such that the axes only translate to the inertial X, Y, Z frame of reference However, the body may have a rotation ω about these axes, and therefore the moments and products of inertia of the body would have to be expressed as functions of time

21.4 Equations of Motion x, y, z Axes having motion Ω = ω
The x, y, z axes may be chosen with origin at G, such that they are fixed in and move with the body The moments and the products of inertia of the body relative to these axes will be constant during the motion

For a rigid body symmetric with respect to the x-y reference plane, and undergoing general plane motion,

21.4 Equations of Motion x, y, z Axes having motion Ω = ω Hence,
If the x, y and z axes are chosen as principal axes of inertia, the products of inertia are zero, Ixx = Iz For the Euler equations of motion,

Since the x, y, z components are rotating at Ω = ω, Since Hence,

21.4 Equations of Motion x, y, z Axes having motion Ω ≠ ω
Choose the x, y, z axes having an angular velocity Ω which is different from the angular velocity ω of the body This is particular suitable for the analysis of spinning tops and gyroscopes, which are symmetrical about their spinning axes When this is the case, the moments and products of inertia remain constant during the motion

21.4 Equations of Motion x, y, z Axes having motion Ω ≠ ω
Any one of these sets of moment equations represents a series of three first order nonlinear differential equations These equations are coupled since the angular velocity components are present in all the terms

21.4 Equations of Motion Procedure for Analysis FBD
Draw a FBD of the body at the instant considered and specify the x, y, z coordinate system Origin of this reference must be located either at the body’s mass center G or at point O, considered fixed in an inertial reference frame and located either in the body or a massless extension of the body Unknown reactive forces can be shown having a positive sense of direction

21.4 Equations of Motion Procedure for Analysis FBD
Depending on the nature of the problem, decide which type of rotational motion Ω = 0, Ω = ω, or Ω ≠ ω When choosing, one should keep in mind that the moment equations are simplified in such a manner that they represent principal axes of inertia for the body at all times Compute the necessary moments and products of inertia for the body relative to the x, y, z axes

21.4 Equations of Motion Procedure for Analysis Kinematics
Determine the x, y, z components of the body’s angular velocity and compute the time derivatives of ω If Ω = ω then, ώ = (ώ)xyz and we can either find the components of ω along the x, y, z axes when the axes are oriented in a general position and take the time derivatives of these components (ώ)xyz or we can find the time derivatives of ω with respect to the X, Y, Z axes, ώ, and then determine the components of ώx, ώy, ώz

21.4 Equations of Motion Procedure for Analysis Equations of Motion
Apply either the two vector equations or the six scalar component equation appropriate for the x, y, z coordinate axes chosen for the problem

21.5 Gyroscopic Motion Starting the X, Y, Z and the x, y, z axes in coincidence, the final position of the top id determined by 1. Rotate the top about the Z (or z) axis through an angle Φ (0 ≤ Φ ≤ 2π) 2. Rotate the top about the x axis through an angle θ (0 ≤ θ ≤ π) 3. Rotate the top about the z axis through an angle ψ (0 ≤ ψ ≤ 2π) to obtain the final position Sequence must be maintained

21.5 Gyroscopic Motion Since these finite rotation are not vectors and in this case, the differential rotation are vectors and thus, the angular velocity ω of the top can be expressed in terms of the time derivatives of the Euler angles The angular components are known as precession, nutation and spin

21.5 Gyroscopic Motion

21.5 Gyroscopic Motion It can be seen that these vectors are not all perpendicular to one another However, ω of the top can still be expressed in terms of these three components The body (top) is symmetric with respect to the z or spin axis If the top is orientated so that at the instant, the spin angle equals 0 and the x, y, z axes follow the motion of the body only in nutation and precession, i.e. Ω = ωp + ωn, the nutation and spin are always directed along the x and z axes respectively

21.5 Gyroscopic Motion Hence, for the angular velocity of the body specified only in terms of the Euler’s angle θ, Since the motion of the axes is not affected by the spin component, The x, y, z axes represent the principal axes of inertia of the body for any spin of the body about these axes

21.5 Gyroscopic Motion Hence, the moments of inertia are constant and will be represented by Ixx = Iyy = I and Izz = Iz Since Ω ≠ ω, Each moment summation applies only at the fixed point O or at the center of mass G of the body

21.5 Gyroscopic Motion Since the equations represent a coupled set of nonlinear second order differential equations, in general, a closed-form solution may not be obtained Instead, the Euler’s angles may be obtained graphically as functions of time using numerical analysis and computer techniques As special case, steady precession, occurs when the nutation angle, precession and spin all remains constant

21.5 Gyroscopic Motion The equations reduce to Furthering simplifying,
or Note that the effects the spin has on the moment about the x axis

21.5 Gyroscopic Motion When θ = 90°, consider the spinning rotor, equations reduce to or It can be seen that the vectors ∑Mx, Ωy and ωz all act along their respective positive axes and therefore are mutually perpendicular

21.5 Gyroscopic Motion One would expect the rotor to fall under the influence of gravity, however, provided that product of IzΩyωz is correctly chosen to counter moment ∑Mx = WrG of the rotor’s weight about O This unusual moment is referred as the gyroscopic effect Studying the action of the gyroscope, or known as gyro, a gyro is a rotor that spins at a very high rate about its axis of symmetry This rate of spin is greater than its precessional rate of rotation about the vertical axis

21.5 Gyroscopic Motion Hence, the angular momentum of the gyro can be assumed directed among its axis of spin Thus, for the gyro rotor, ωz >> Ωy, and the magnitude of the angular momentum about point O reduces to the form HO = Izωz Since both the magnitude and directions of HO are constant as observed from the x, y, z, direct application yields ∑Mx = Ωy x HO

21.5 Gyroscopic Motion Ωy always swings HO (or ωz) towards the sense of ∑Mx In effect, the change in direction of the gyro’s angular momentum, dHO is equivalent to the angualr impulse caused by the gyro’s weight about O dHO = ∑Mx dt When a gyro is mounted in gimbal rings, it become free of external momentums applied to its base

21.5 Gyroscopic Motion In theory, its angular momentum H will never precess but, instead maintain its same fixed orientation along the axis of spin when the base is rotated This type of gyroscope is called a free gyro and is useful as a gyrocompass when the spin axis of the gyro is directed north In reality, the gimbal mechanism is never completely free of friction, so such a device is useful only for the local navigation of ships and aircraft

21.6 Torque-Free Motion When the only external force acting on a body is caused by gravitation, the general motion of the body is referred to as torque-free motion In order to describe the characteristics of this motion, the distribution of the body’s mass will be assumed axisymmetric, as shown in figure, where the z axis represents an axis of symmetry

21.6 Torque-Free Motion The origin of the x, y, z coordinates is located at the mass center G, such that Izz = Iz and Ixx = Iyy = I for the body Since gravitation is the only external force present, the summation of moments about the mass center is zero This required the angular momentum of the body to be constant,

21.6 Torque-Free Motion At the instant considered, it will assumed that the inertial frame of reference is oriented so that the positive Z axis is directed along HG and the y axis lies in the plane formed by the z and Z axes. The Euler angle formed between Z and z is θ, and therefore, with this choice of axes the angular momentum may be expressed as

21.6 Torque-Free Motion Since
Equating the respective i, j, k components of the above two equations yields or

21.6 Torque-Free Motion Equating the respective i, j, k components of
into We obtain

21.6 Torque-Free Motion Solving, we get

21.6 Torque-Free Motion Thus for torque-free motion of an axisymmetrical body, the angle θ formed between the angular-momentum vector and the spin of the body remains constant Furthermore, the angular momentum HG, precession Φ, and spin Ψ for the body remain constant at all times during the motion . .

21.6 Torque-Free Motion As shown, the body precesses about the Z axis, which is fixed in direction, spinning about z axis. These two components of angular motion may be studied by using a simple cone model The space cone defining the precession is fixed from rotating, since the precession has a fixed direction, while the body cone rotates around the space cone’s outer surface without slipping

21.6 Torque-Free Motion On the basis, an attempt should be made to imagine the motion The interior angle of each cone is chosen such that the resultant angular velocity of the body is directed along the line of contact of the two cones This line of contact represents the instantaneous axis of rotation for the body cone and hence the angular velocity of both the body cone and the body must be directed along this line

21.6 Torque-Free Motion Since the spin is a function of the moments of inertia I and Iz of the body, the cone model is satisfactory for describing the motion, provided I > Iz Torque-free motion which meets these requirements is called regular precession If I > Iz, the spin is negative and precession is positive

21.6 Torque-Free Motion This motion is represented by the satellite motion shown below (I < Iz) The cone model may again be used to represent; however, to preserve the correct vector addition of spin and precession to obtain the angular velocity ω, the inside surface of the body cone must roll on the outside surface of the (fixed) space cone This motion is called retrograde precession

Chapter Review Moments and Products of Inertia
A body has 6 components of inertia for any specified x, y, z axes 3 are moments of inertia about each of the axes, Ix, Iy, Iz, and 3 are products of inertia , each defined from two orthogonal planes Ixy, Iyz and Ixz If either one or both of the planes are planes of symmetry, then the product of inertia with respect to these planes will be zero Moments and products of integrations can be determined by direct integration

Chapter Review Moments and Products of Inertia

Chapter Review Moments and Products of Inertia
If these quantities are to be determined with respect to axes or planes that do not pass through the mass center, parallel axes theorems must be used Provided the 6 components of inertia are known, the moment of inertia about any axis may be determined using the transformation equation

Chapter Review Principle Moments of Inertia
At any point on or off the body, the x, y, z axes can be oriented so that the products of inertia will be zero. The resulting moments of inertia are called the principal moments of inertia, one of which will be a maximum and the other a minimum moment of inertia for the body.

Chapter Review Principle of Impulse and Momentum
The angular momentum for a body can be determined about any arbitrary point A. Once the linear and angular momentum for the body have been formulated, then the principle of impulse and momentum can be used to solve problems that involve force, velocity, and time.

Chapter Review Principle of Impulse and Momentum

Chapter Review Principle of Work and Energy
Kinetic energy of a body is usually determined relative to a fixed point or the body’s mass center Provided the axes are principal axes of inertia, for a fixed point, And relative to the mass center,

Chapter Review Equations of Motion
There are 3 scalar equations of translational motion for a rigid body moving in 3D There are 3 scalar equations of rotational motion depending upon the location of the x, y, z reference Most often, these axes are oriented so that the axes are principal axes of inertia

Chapter Review Equations of Motion
If the axes are fixed in and move with the rotation ω of the body, the equations are referred to as the Euler’s equations of motion (Ω = ω), If the axes have a rotation (Ω ≠ ω),

Chapter Review Gyroscopic Motion
The angular motion is best described using the changes in motion of the three Euler angles These angular velocity components are the precession, nutation and the spin If spin equals zero and the precession and nutation are constant, the motion is referred as steady precession For rotational equations of motion,

Chapter Review Gyroscopic Motion
It is the spin of the gyro motor that is responsible for holding the rotor from falling downward, and instead causing it to precess about a vertical axis This phenomenon is called the gyroscopic effect

Chapter Review Torque-Free Motion
A body that is subjected to a gravitational force that will have no moments on it about its mass center, and so the motion is described as torque-free motion The angular momentum for the body will remain constant and this causes the body to have both a spin and a precession

Chapter Review Torque-Free Motion
The behavior depends on the size of the moment of inertia of a symmetric body about the spin axis Iz versus that about a perpendicular axis I If I > Iz, regular precession occur If I < Iz, retrograde precession occur

Chapter 21: Three Dimensional Kinetics

Similar presentations

Presentation on theme: "Chapter 21: Three Dimensional Kinetics"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chapter 21: Three Dimensional Kinetics

Similar presentations

Presentation on theme: "Chapter 21: Three Dimensional Kinetics"— Presentation transcript:

Similar presentations

About project

Feedback