DOCSLIB.ORG

Moments of Inertia of Rigid Bodies∗

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Moments of Inertia of Rigid Bodies∗ Connexions module: m14292 1 Moments of inertia of rigid bodies∗ Sunil Kumar Singh This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License y Abstract Moment of inertia of rigid body depends on the distribution of mass about the axis of rotation. In the module titled Rotation of rigid body 1, we derived expressions of moments of inertia (MI) for dierent object forms as : 1: For a particle : I = mr2 P 2 2: For a system of particles : I = miri 3: For a rigid body : I = R r2m In this module, we shall evalaute MI of dierent regularly shaped rigid bodies. 1 Evaluation strategy We evaluate right hand integral of the expression of moment of inertia for regularly shaped geometric bodies. The evaluation is basically an integration process, well suited to an axis of rotation for which mass distribution is symmetric. In other words, evaluation of the integral is easy in cases where mass of the body is evenly distributed about the axis. This axis of symmetry passes through "center of mass" of the regular body. Calculation of moment of inertia with respect to other axes is also possible, but then integration process becomes tedious. There are two very useful theorems that enable us to calculate moment of inertia about certain other relevant axes as well. These theorems pertaining to calculation of moment of inertia with respect to other relevant axes are basically "short cuts" to avoid lengthy integration. We must, however, be aware that these theorems are valid for certain relevant axes only. If we are required to calculate moment of inertia about an axis which can not be addressed by these theorems, then we are left with no choice other than evaluating the integral or determining the same experimentally. As such, we limit ourselves in using integral method to cases, where moment of inertia is required to be calculated about the axis of symmetry. In this module, we will discuss calculation of moment of inertia using basic integral method only, involving bodies having (i) regular geometric shape (ii) uniform mass distribution i.e uniform density and (iii) axis of rotation passing through center of mass (COM). Application of the theorems shall be discussed in a separate module titled " Theorems on moment of inertia 2". ∗Version 1.10: May 23, 2011 2:48 am GMT-5 yhttp://creativecommons.org/licenses/by/3.0/ 1"Rotation of rigid body" <http://cnx.org/content/m14278/latest/> 2"Theorems on moment of inertia" <http://cnx.org/content/m14294/latest/> http://cnx.org/content/m14292/1.10/ Connexions module: m14292 2 As far as integration method is concerned, it is always useful to have a well planned methodology to complete the evaluation. In general, we complete the integration in following steps : 1. Identify an innitesimally small element of the body. 2. Identify applicable density type (linear, surface or volumetric). Calculate elemental mass "dm" in terms of appropriate density. 3. Write down the expression of moment of inertia (I) for elemental mass. 4. Evaluate the integral of moment of inertia for an appropriate pair of limits and determine moment of inertia of the rigid body. Identication of small element is crucial in the evaluation of the integral. We consider linear element in evaluating integral for a linear mass distribution as for a rod or a plate. On the other hand, we consider thin concentric ring as the element for a circular plate, because we can think circular plate being composed of innite numbers of thin concentric rings. Similarly, we consider a spherical body, being composed of closely packed thin spherical shells. Calculation of elemental mass "m" makes use of appropriate density on the basis of the nature of mass distribution in the rigid body : mass = appropriate density x geometric dimension The choice of density depends on the nature of body under consideration. In case where element is considered along length like in the case of a rod, rectangular plate or ring, linear density (λ) is the appropriate choice. In cases, where surface area is involved like in the case of circular plate, hollow cylinder and hollow sphere, areal density (σ) is the appropriate choice. Finally, volumetric density (ρ) is suitable for solid three dimensional bodies like cylinder and sphere. The elemental mass for dierent cases are : m = λ x m = σ A (1) m = ρ V Elemental mass (a) (b) Figure 1: (a) Elemental mass of a rod for MI about perpendicular bisector (b) Elemental mass of a circular disc for MI about perpendicular axis through the center http://cnx.org/content/m14292/1.10/ Connexions module: m14292 3 Elemental mass Figure 2: Elemental mass of a sphere for MI about a diameter The MI integral is then expressed by suitably replacing "dm" term by density term in the integral expression. This approach to integration using elemental mass assumes that mass distribution is uniform. 2 Evaluation of moment of inertia In this section, we shall determine MI of known geometric bodies about the axis of its symmetry. 2.1 MI of a uniform rod about its perpendicular bisector The gure here shows the small element with repect to the axis of rotation. Here, the steps for calculation are : http://cnx.org/content/m14292/1.10/ Connexions module: m14292 4 Moment of inertia Figure 3: MI of a rod about perpendicular bisector (i) Innetesimally small element of the body : Let us consider an small element "dx" along the length, which is situated at a linear distance "x" from the axis. (ii) Elemental mass : Linear density, λ, is the appropriate density type in this case. M λ = L where "M" and "L" are the mass and length of the rod respectively. Elemental mass (m) is, thus, given as : M m = λ x = L x (iii) Moment of inertia for elemental mass : Moment of inertia of elemental mass is : 2 2 M I = r m = x L x (iv) Moment of inertia of rigid body : R 2 R 2 M I = r m = x L x http://cnx.org/content/m14292/1.10/ Connexions module: m14292 5 While setting limits we should cover the total length of the rod. The appropriate limits of integral in this case are -L/2 and L/2. Hence, L R 2 M 2 I = − L L x x 2 Taking the constants out of the integral sign, we have : L M R 2 2 ) I = L − L x x 2 h 3 i L 2 M x 2 ML (2) ) I = L 3 − L = 12 2 2.2 MI of a rectangular plate about a line parallel to one of the sides and passing through the center The gure here shows the small element with repect to the axis of rotation i.e. y-axis, which is parallel to the breadth of the rectangle. Note that axis of rotation is in the place of plate. Here, the steps for calculation are : Moment of inertia Figure 4: MI of rectangular plate about a line parallel to its length and passing through the center (i) Innetesimally small element of the body : Let us consider an small element "dx" along the length, which is situated at a linear distance "x" from the axis. http://cnx.org/content/m14292/1.10/ Connexions module: m14292 6 (ii) Elemental mass : Linear density, λ, is the appropriate density type in this case. M λ = a where "M" and "a" are the mass and length of the rectangular plate respectively. Elemental mass (dm) is, thus, given as : M m = λ x = a x (iii) Moment of inertia for elemental mass : Moment of inertia of elemental mass is : 2 2 M I = r m = x a x (iv) Moment of inertia of rigid body : Proceeding in the same manner as for the case of an uniform rod, the MI of the plate about the axis is given by : Ma2 (3) ) I = = 12 Similarly, we can also calculate MI of the rectangular plate about a line parallel to its length and through the center, Mb2 (4) ) I = = 12 2.3 MI of a circular ring about a perpendicular line passing through the center The gure here shows the small element with respect to the axis of rotation. Here, we can avoid the steps for calculation as all elemental masses are at a xed (constant) distance "R" from the axis. This enables us to take "R" directly out of the basic integral : http://cnx.org/content/m14292/1.10/ Connexions module: m14292 7 Moment of inertia Figure 5: MI of a circular ring about a perpendicular line passing through the center I = r2 m = R2 m (5) ) I = R I = R2 R m = MR2 We must note here that it was not possible so in the case of a rod or a rectangular plate as elemental mass is at a variable distance from the axis of rotation. We must realize that simplication of evaluation in this case results from the fact that masses are at the same distance from the axis of rotation. This means that the expression of MI will be valid only when thickness of the ring is relatively very small in comparison with its radius. 2.4 MI of a thin circular plate about a perpendicular line passing through the center The gure here shows the small ring element with repect to the axis of rotation. Here, the steps for calculation are : http://cnx.org/content/m14292/1.10/ Connexions module: m14292 8 Moment of inertia Figure 6: MI of a thin circular plate about a perpendicular line passing through the center (i) Innetesimally small element of the body : Let us consider an small circular ring element of width "dr", which is situated at a linear distance "r" from the axis.
Load more

Recommended publications
  • Rotational Motion (The Dynamics of a Rigid Body)
    University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body) Henry Semat City College of New York Robert Katz University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/physicskatz Part of the Physics Commons Semat, Henry and Katz, Robert, "Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body)" (1958). Robert Katz Publications. 141. https://digitalcommons.unl.edu/physicskatz/141 This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 11 Rotational Motion (The Dynamics of a Rigid Body) 11-1 Motion about a Fixed Axis The motion of the flywheel of an engine and of a pulley on its axle are examples of an important type of motion of a rigid body, that of the motion of rotation about a fixed axis. Consider the motion of a uniform disk rotat­ ing about a fixed axis passing through its center of gravity C perpendicular to the face of the disk, as shown in Figure 11-1. The motion of this disk may be de­ scribed in terms of the motions of each of its individual particles, but a better way to describe the motion is in terms of the angle through which the disk rotates.
    [Show full text]
  • Basic Galactic Dynamics
    5 Basic galactic dynamics 5.1 Basic laws govern galaxy structure The basic force on galaxy scale is gravitational • Why? The other long-range force is electro-magnetic force due to electric • charges: positive and negative charges are difficult to separate on galaxy scales. 5.1.1 Gravitational forces Gravitational force on a point mass M1 located at x1 from a point mass M2 located at x2 is GM M (x x ) F = 1 2 1 − 2 . 1 − x x 3 | 1 − 2| Note that force F1 and positions, x1 and x2 are vectors. Note also F2 = F1 consistent with Newton’s Third Law. Consider a stellar− system of N stars. The gravitational force on ith star is the sum of the force due to all other stars N GM M (x x ) F i j i j i = −3 . − xi xj Xj=6 i | − | The equation of motion of the ith star under mutual gravitational force is given by Newton’s second law, dv F = m a = m i i i i i dt and so N dvi GMj(xi xj) = − 3 . dt xi xj Xj=6 i | − | 5.1.2 Energy conservation Assuming M2 is fixed at the origin, and M1 on the x-axis at a distance x1 from the origin, we have m1dv1 GM1M2 = 2 dt − x1 1 2 Using 2 dv1 dx1 dv1 dv1 1 dv1 = = v1 = dt dt dx1 dx1 2 dx1 we have 2 1 dv1 GM1M2 m1 = 2 , 2 dx1 − x1 which can be integrated to give 1 2 GM1M2 m1v1 = E , 2 − x1 The first term on the lhs is the kinetic energy • The second term on the lhs is the potential energy • E is the total energy and is conserved • 5.1.3 Gravitational potential The potential energy of M1 in the potential of M2 is M Φ , − 1 × 2 where GM2 Φ2 = − x1 is the potential of M2 at a distance x1 from it.
    [Show full text]
  • Rigid Body Dynamics 2
    Rigid Body Dynamics 2 CSE169: Computer Animation Instructor: Steve Rotenberg UCSD, Winter 2017 Cross Product & Hat Operator Derivative of a Rotating Vector Let’s say that vector r is rotating around the origin, maintaining a fixed distance At any instant, it has an angular velocity of ω ω dr r ω r dt ω r Product Rule The product rule of differential calculus can be extended to vector and matrix products as well da b da db b a dt dt dt dab da db b a dt dt dt dA B dA dB B A dt dt dt Rigid Bodies We treat a rigid body as a system of particles, where the distance between any two particles is fixed We will assume that internal forces are generated to hold the relative positions fixed. These internal forces are all balanced out with Newton’s third law, so that they all cancel out and have no effect on the total momentum or angular momentum The rigid body can actually have an infinite number of particles, spread out over a finite volume Instead of mass being concentrated at discrete points, we will consider the density as being variable over the volume Rigid Body Mass With a system of particles, we defined the total mass as: n m m i i1 For a rigid body, we will define it as the integral of the density ρ over some volumetric domain Ω m d Angular Momentum The linear momentum of a particle is 퐩 = 푚퐯 We define the moment of momentum (or angular momentum) of a particle at some offset r as the vector 퐋 = 퐫 × 퐩 Like linear momentum, angular momentum is conserved in a mechanical system If the particle is constrained only
    [Show full text]
  • Rigid Body Dynamics II
    An Introduction to Physically Based Modeling: Rigid Body Simulation IIÐNonpenetration Constraints David Baraff Robotics Institute Carnegie Mellon University Please note: This document is 1997 by David Baraff. This chapter may be freely duplicated and distributed so long as no consideration is received in return, and this copyright notice remains intact. Part II. Nonpenetration Constraints 6 Problems of Nonpenetration Constraints Now that we know how to write and implement the equations of motion for a rigid body, let's consider the problem of preventing bodies from inter-penetrating as they move about an environment. For simplicity, suppose we simulate dropping a point mass (i.e. a single particle) onto a ®xed ¯oor. There are several issues involved here. Because we are dealing with rigid bodies, that are totally non-¯exible, we don't want to allow any inter-penetration at all when the particle strikes the ¯oor. (If we considered our ¯oor to be ¯exible, we might allow the particle to inter-penetrate some small distance, and view that as the ¯oor actually deforming near where the particle impacted. But we don't consider the ¯oor to be ¯exible, so we don't want any inter-penetration at all.) This means that at the instant that the particle actually comes into contact with the ¯oor, what we would like is to abruptly change the velocity of the particle. This is quite different from the approach taken for ¯exible bodies. For a ¯exible body, say a rubber ball, we might consider the collision as occurring gradually. That is, over some fairly small, but non-zero span of time, a force would act between the ball and the ¯oor and change the ball's velocity.
    [Show full text]
  • Center of Mass, Torque and Rotational Inertia
    Center of Mass, Torque and Rotational Inertia Objectives 1. Define center of mass, center of gravity, torque, rotational inertia 2. Give examples of center of gravity 3. Explain how center of mass affects rotation and toppling 4. Write the equations for torque and rotational inertia 5. Explain how to change torque and rotational inertia Center of Mass/Gravity • COM – point which is at the center of the objects mass – Middle of meter stick – Center of a ball – Middle of a donut • Donuts are yummy • COG – center of object’s weight distribution – Usually the same point as center of mass Rotation and Center of Mass • Objects rotate around their center of mass – Wobbling baseball bat tossed through the air • They also balance there – Balancing a broom on your finger • Things topple over when their center of gravity is not above the base – Leaning Tower of Pisa Torque • Force that causes rotation • Equation τ = Fr τ = torque (in Nm) F = force (in N) r = arm radius (in m) • How do you increase torque – More force – Longer radius Rotational Inertia • Resistance of an object to a change in its rotational motion • Controlled by mass distribution and location of axis • Equation I = mr2 I = rotational inertia (in kgm2) m = mass (in kg) r = radius (in m) • How do you increase rotational inertia? – Even mass distribution Angular Momentum • Measure how difficult it is to start or stop a rotating object • Equations L = Iω = mvtr • It is easier to balance on a moving bicycle • Precession – Torque applied to a spinning wheel changes the direction of its angular momentum – The wheel rotates about the axis instead of toppling over Satellite Motion John Glenn 6 min • Satellites “fall around” the object they orbit • The tangential speed must be exact to match the curvature of the surface of the planet – On earth things fall 4.9 m in one second – Earth “falls” 4.9 m every 8000 m horizontally – Therefore a satellite must have a tangential speed of 8000 m/s to keep from hitting the ground – Rockets are launched vertically and rolled to horizontal during their flights Sputnik-30 sec .
    [Show full text]
  • Rotation: Moment of Inertia and Torque
    Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. This tutorial discusses the dynamics of an object rotating about a fixed axis and introduces the concepts of torque and moment of inertia. These concepts allows us to get a better understanding of why pushing a door towards its hinges is not very a very effective way to make it open, why using a longer wrench makes it easier to loosen a tight bolt, etc. This module begins by looking at the kinetic energy of rotation and by defining a quantity known as the moment of inertia which is the rotational analog of mass. Then it proceeds to discuss the quantity called torque which is the rotational analog of force and is the physical quantity that is required to changed an object's state of rotational motion. Moment of Inertia Kinetic Energy of Rotation Consider a rigid object rotating about a fixed axis at a certain angular velocity. Since every particle in the object is moving, every particle has kinetic energy. To find the total kinetic energy related to the rotation of the body, the sum of the kinetic energy of every particle due to the rotational motion is taken. The total kinetic energy can be expressed as ..
    [Show full text]
  • Lecture 18, Structure of Spiral Galaxies
    GALAXIES 626 Upcoming Schedule: Today: Structure of Disk Galaxies Thursday:Structure of Ellipticals Tuesday 10th: dark matter (Emily©s paper due) Thursday 12th :stability of disks/spiral arms Tuesday 17th: Emily©s presentation (intergalactic metals) GALAXIES 626 Lecture 17: The structure of spiral galaxies NGC 2997 - a typical spiral galaxy NGC 4622 yet another spiral note how different the spiral structure can be from galaxy to galaxy Elementary properties of spiral galaxies Milky Way is a ªtypicalº spiral radius of disk = 15 Kpc thickness of disk = 300 pc Regions of a Spiral Galaxy · Disk · younger generation of stars · contains gas and dust · location of the open clusters · Where spiral arms are located · Bulge · mixture of both young and old stars · Halo · older generation of stars · contains little gas and dust · location of the globular clusters Spiral Galaxies The disk is the defining stellar component of spiral galaxies. It is the end product of the dissipation of most of the baryons, and contains almost all of the baryonic angular momentum Understanding its formation is one of the most important goals of galaxy formation theory. Out of the galaxy formation process come galactic disks with a high level of regularity in their structure and scaling laws Galaxy formation models need to understand the reasons for this regularity Spiral galaxy components 1. Stars · 200 billion stars · Age: from >10 billion years to just formed · Many stars are located in star clusters 2. Interstellar Medium · Gas between stars · Nebulae, molecular clouds, and diffuse hot and cool gas in between 3. Galactic Center ± supermassive Black Hole 4.
    [Show full text]
  • Lecture Notes for PHY 405 Classical Mechanics
    Lecture Notes for PHY 405 Classical Mechanics From Thorton & Marion's Classical Mechanics Prepared by Dr. Joseph M. Hahn Saint Mary's University Department of Astronomy & Physics November 21, 2004 Problem Set #6 due Thursday December 1 at the start of class late homework will not be accepted text problems 10{8, 10{18, 11{2, 11{7, 11-13, 11{20, 11{31 Final Exam Tuesday December 6 10am-1pm in MM310 (note the room change!) Chapter 11: Dynamics of Rigid Bodies rigid body = a collection of particles whose relative positions are fixed. We will ignore the microscopic thermal vibrations occurring among a `rigid' body's atoms. 1 The inertia tensor Consider a rigid body that is composed of N particles. Give each particle an index α = 1 : : : N. The total mass is M = α mα. This body can be translating as well as rotating. P Place your moving/rotating origin on the body's CoM: Fig. 10-1. 1 the CoM is at R = m r0 M α α α 0 X where r α = R + rα = α's position wrt' fixed origin note that this implies mαrα = 0 α X 0 Particle α has velocity vf,α = dr α=dt relative to the fixed origin, and velocity vr,α = drα=dt in the reference frame that rotates about axis !~ . Then according to Eq. 10.17 (or page 6 of Chap. 10 notes): vf,α = V + vr,α + !~ × rα = α's velocity measured wrt' fixed origin and V = dR=dt = velocity of the moving origin relative to the fixed origin.
    [Show full text]
  • Chapter 8: Rotational Motion
    TODAY: Start Chapter 8 on Rotation Chapter 8: Rotational Motion Linear speed: distance traveled per unit of time. In rotational motion we have linear speed: depends where we (or an object) is located in the circle. If you ride near the outside of a merry-go-round, do you go faster or slower than if you ride near the middle? It depends on whether “faster” means -a faster linear speed (= speed), ie more distance covered per second, Or - a faster rotational speed (=angular speed, ω), i.e. more rotations or revolutions per second. • Linear speed of a rotating object is greater on the outside, further from the axis (center) Perimeter of a circle=2r •Rotational speed is the same for any point on the object – all parts make the same # of rotations in the same time interval. More on rotational vs tangential speed For motion in a circle, linear speed is often called tangential speed – The faster the ω, the faster the v in the same way v ~ ω. directly proportional to − ω doesn’t depend on where you are on the circle, but v does: v ~ r He’s got twice the linear speed than this guy. Same RPM (ω) for all these people, but different tangential speeds. Clicker Question A carnival has a Ferris wheel where the seats are located halfway between the center and outside rim. Compared with a Ferris wheel with seats on the outside rim, your angular speed while riding on this Ferris wheel would be A) more and your tangential speed less. B) the same and your tangential speed less.
    [Show full text]
  • Galactic Mass Distribution Without Dark Matter Or Modified Newtonian
    1 astro-ph/0309762 v2, revised 1 Mar 2007 Galactic mass distribution without dark matter or modified Newtonian mechanics Kenneth F Nicholson, retired engineer key words: galaxies, rotation, mass distribution, dark matter Abstract Given the dimensions (including thickness) of a galaxy and its rotation profile, a method is shown that finds the mass and density distribution in the defined envelope that will cause that rotation profile with near-exact speed matches. Newton's law is unchanged. Surface-light intensity and dark matter are not needed. Results are presented in dimensionless plots allowing easy comparisons of galaxies. As compared with the previous version of this paper the methods are the same, but some data are presented in better dimensionless parameters. Also the part on thickness representation is simplified and extended, the contents are rearranged to have a logical buildup in the problem development, and more examples are added. 1. Introduction Methods used for finding mass distribution of a galaxy from rotation speeds have been mostly those using dark-matter spherical shells to make up the mass needed beyond the assumed loading used (as done by van Albada et al, 1985), or those that modify Newton's law in a way to approximately match measured data (for example MOND, Milgrom, 1983). However the use of the dark-matter spheres is not a correct application of Newton's law, and (excluding relativistic effects) modifications to Newton's law have not been justified. In the applications of dark-matter spherical shells it is assumed that all galaxies are in two parts, a thin disk with an exponential SMD loading (a correct solution for rotation speeds by Freeman, 1970) and a series of spherical shells centered on the galaxy center.
    [Show full text]
  • Multiple Integrals
    Chapter 11 Multiple Integrals 11.1 Double Riemann Sums and Double Integrals over Rectangles Motivating Questions In this section, we strive to understand the ideas generated by the following important questions: What is a double Riemann sum? • How is the double integral of a continuous function f = f(x; y) defined? • What are two things the double integral of a function can tell us? • Introduction In single-variable calculus, recall that we approximated the area under the graph of a positive function f on an interval [a; b] by adding areas of rectangles whose heights are determined by the curve. The general process involved subdividing the interval [a; b] into smaller subintervals, constructing rectangles on each of these smaller intervals to approximate the region under the curve on that subinterval, then summing the areas of these rectangles to approximate the area under the curve. We will extend this process in this section to its three-dimensional analogs, double Riemann sums and double integrals over rectangles. Preview Activity 11.1. In this activity we introduce the concept of a double Riemann sum. (a) Review the concept of the Riemann sum from single-variable calculus. Then, explain how R b we define the definite integral a f(x) dx of a continuous function of a single variable x on an interval [a; b]. Include a sketch of a continuous function on an interval [a; b] with appropriate labeling in order to illustrate your definition. (b) In our upcoming study of integral calculus for multivariable functions, we will first extend 181 182 11.1.
    [Show full text]
  • 8.09(F14) Chapter 2: Rigid Body Dynamics
    Chapter 2 Rigid Body Dynamics 2.1 Coordinates of a Rigid Body A set of N particles forms a rigid body if the distance between any 2 particles is fixed: rij ≡ jri − rjj = cij = constant: (2.1) Given these constraints, how many generalized coordinates are there? If we know 3 non-collinear points in the body, the remaing points are fully determined by triangulation. The first point has 3 coordinates for translation in 3 dimensions. The second point has 2 coordinates for spherical rotation about the first point, as r12 is fixed. The third point has one coordinate for circular rotation about the axis of r12, as r13 and r23 are fixed. Hence, there are 6 independent coordinates, as represented in Fig. 2.1. This result is independent of N, so this also applies to a continuous body (in the limit of N ! 1). Figure 2.1: 3 non-collinear points can be fully determined by using only 6 coordinates. Since the distances between any two other points are fixed in the rigid body, any other point of the body is fully determined by the distance to these 3 points. 29 CHAPTER 2. RIGID BODY DYNAMICS The translations of the body require three spatial coordinates. These translations can be taken from any fixed point in the body. Typically the fixed point is the center of mass (CM), defined as: 1 X R = m r ; (2.2) M i i i where mi is the mass of the i-th particle and ri the position of that particle with respect to a fixed origin and set of axes (which will notationally be unprimed) as in Fig.
    [Show full text]