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Application of Rigid Bodies Dynamics to Snooker - Assignment Example

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The assignment "Application of Rigid Bodies Dynamics to Snooker" focuses on the critical analysis of the main laws and principles of rigid body dynamics from the practical and theoretical point of view: as solutions to theoretical and practical exercises are provided…
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Application of Rigid Bodies Dynamics to Snooker
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Rigid Bodies and Application to Snooker Introduction The study of rigid body dynamics allows us to understand the main principles and laws of mechanics on practice as it provides visual examples on the base of natural observations. Rigid body dynamics principles and laws allow us to simply a number of real life physical problems, which deal with motion of bodies and successfully solve them. None would argue that principles of rigid body dynamics are widely applied on practice in different spheres of engineering especially in mechanical engineering. Rigid body dynamics allows describing motion and interaction of different physical bodies. In this project a special case is devoted to practical problems of snooker. In this project I made a study of main laws and principles of rigid body dynamics from the practical and theoretical point of view: as solutions to theoretical and practical exercises are provided. It required review and deeper study of vector analysis and analytical geometry. Background Rigid body in mechanics is a system of material points, which doesn't change in time. So it's an idealized system for which the distance between its particles remains constant in time under any motion. Phenomenological mechanics considers rigid body to be a solid matter, in which particles are subjected to internal forces in the form of normal and tangent tensions. Such tensions are caused by external deformations. In case they're re no deformations, there are no tensions inside rigid body. In many cases deformations are so small that can be neglected. So such model is an idealized rigid body, which is not able to deform and even though internal tensions can take place because of external forces. Rigid body is a mechanical system with six degrees of freedom. In order to define the position of a rigid body it's enough to know the position of at least 3 points: A, B, C, which do not belong to one line. In order to prove that rigid body is described by six degrees of freedom we have to take point D. Distances AD, BD, CD are considered to be known as they remain unchangeable under any motion of a body. Position of these three points can be expressed through coordinates: XA, YA, ZA; XB, YB, ZB; XC, YC, ZC. These 9 coordinates form the following rations: (XA-XB)^2+(YA-YB)^2+(ZA-ZB)^2=AB^2=const (XC-XB)^2+(YC-YB)^2+(ZC-ZB)^2=CB^2=const (XA-XC)^2+(YA-YC)^2+(ZA-ZC)^2=AC^2=const Because the lengths of sides of triangle ABC remain the same. So only six coordinates are left independent - rigid body has 6 degrees of freedom. If the body has fixed points the number of freedoms degrees reduces. If rigid body is fixed in one point - it has 3 degrees of freedom, if rigid body can only rotate around one axis is has one degree of freedom, if a body can slide across axis it has two degrees of freedom. In order to understand how x(t) and R(t) change over time we should remind the following formulas: Resultant is v= V+ [w, R] (using vector properties). Kinetic energy of a rigid body is total kinetic energy of rotation plus total kinetic energy of motion: T=.5 mv^2 + 0.5Iw^2 Where I is moment of inertia of a rigid body (mass analogue for rotational motion) Moment of inertia is defined as I=miRi^2. Moment of inertia is additive so moment of inertia of a rigid body is formed from the sum of inertia moments of its parts. Any body, independently from rotational motion or rest has definite moment of inertia. Mass distribution in the limits of a body can be characterized by density: p=m/V So moment of inertia can be expressed as I=piRi^2Vi, if density is constant: I=pRi^2Vi In limit it can be expressed in the following integral: I=R^2dm=pR^2dV The inertia tensor is a set of nine values (which can be written in the form of 3X3 matrix), which shows the dependence of shape and distribution of mass in the rigid body caused by its rotational motion. It's often explained as a scaling factor between angular momentum and angular velocity.1 Inertia tensor matrix has the following structure and its components are calculated according to the rules provided below: where: Ixx = Iyy = Izz = Ixy = Iyx = Iyz = Izy = Ixz = Izx = are integrals over the volume of the body. The solutions for simple form bodies such as cube, cylinder, disk, sphere, etc, are trivial and these integrals are solved easily. In case of a complex body special numerical methods should be applied. Physics of snooker The use of rigid body dynamics can be presented on application of principles of mechanics to snooker. Snooker is an excellent example for demonstration of collisions, conservation of momentum, impulse and energy. The principle of impulse conservation is the main principle, which is used in the game when a player has to make estimation what force to apply in order to hit the ball most effectively. Besides, physics of billiards is very interesting in terms of modeling more complicated chaotic processes of different nature based on collisions and disorder. Physics studies the field of billiards, which studies different chaotic processes and models them on the base of the simplified pool models. These methods are widely used in statistical physics and analytical mechanics. In terms of understanding of rigid body dynamics pool is one of the most visual and simple examples as nearly everybody is well acquainted with this game and has a clear understanding about the way balls behave on the table. Understandably in order to describe snooker game and motion of balls we need to study and describe their trajectories. Because we can idealize motion of balls we can neglect a number of factors which are always taken into consideration in other life-time dynamic systems: air resistance, transformation of kinetic energy to heat (for the most types of collisions), scratches and other microscopic details which may effect the shape of balls and surface. For a direct question problem such conditions are accepted by in many cases for the problems of real time modeling they may result in errors in the output. For example if one makes a review of the most of pool computer simulations it will become obvious that in many cases motion of balls and the results of their collisions contradict laws of physics. In order to describe pool game we should introduce vectors of their motion into 2-dimension space (XY) and describe their motion through these vectors. Because the motion of ball also includes rotational motion we should introduce the vector of angular velocity, which can be found from vector of velocity and radius vector of the ball. =[r,v] the direction of this vector will be perpendicular to both vector r and vector v. The methods to calculate vector product and its qualities are provided and proved in (Exercise 2.2.1). Resultant velocity of the ball is given by the formula: v= V + [, R]. In order to include rotation velocity to the total kinetic energy a mass analogue for rotational motion is used which is called moment of inertia. Moment of inertia can be calculated as a sum of elementary masses of a moving system multiplied on the distance to the center of mass of the system. For the general case it's often easier to express moments in terms of tensors than in terms of inertia moments. Tensor properties and methods for tensor evaluations are shown in exercises 2.1.1 -2.1.4. It's also shown how to evaluate tensor of a ball as moment of inertia of a ball is used in evaluations of the ball's motion properties. In a number of pool problems the motion of a ball is considered to be linear and is slows only under the influence of a frication force. Friction force is the only force, which is applied to moving ball in simplified pool problems. It's proportional to the force of gravity: f=mg and is explained by the interaction of the molecules of the moving ball with molecules of the table surface. Coefficient of the friction force depends only upon the nature of the material and is depend from the moving object. Friction force can be expressed in terms of velocity through the following formula: f=-v/|v|, where minus shows that it's directed against the direction of the moving object. In exercise 2.3.1 it was proved that friction force is a constant. Besides, using conservation laws in exercise 2.3.2 it was shown that path of the bal can be expressed in terms of it's initial speed and friction coefficient only. In exercise 2.3.3 it was shown how conservation of momentum could be applied in order to choose the most optimal strategy in snooker when hitting balls. Besides it was shown that hitting different points on the surface of the ball in relation it's center of mass can result different types of its motion; uniform and motion with acceleration which is also important. Rigid bodies Exercises Conclusion This project makes a review of the most important rigid body dynamics problems and covers some aspects of rigid body dynamics applications. In this particular case application to snooker. Using fundamental laws of dynamics it becomes possible to model snooker game and develop the most optimal tactics based on laws of physics. Conservation laws allow to calculate needed hitting force, and allow to model different variants of snooker ball motion (uniform, accelerating and slowing) using only few dynamics rules (conservation laws and Steiner's theorem). Bibliography: Goldstein, Classical Mechanics Landau, L. Mechanics Read More
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