Beginning a business - Gyro Grill Essay Example | Topics and Well Written Essays - 500 words

Beginning a business - Gyro Grill - Essay Example Even if an individual has unique culinary skills, he or she needs correct guidance to start a business, which has the probability to evolve in the long-term. This section of the paper attempts to examine various factors that need consideration, when starting a new business. Formulation of a business plan is imperative. For the Gyro Grill to start, an individual will need a rental space, some start up financial help in the form of loan from the bank and employees to work in the restaurant. Unless a proper business plan is not formulated, the bank will not consider giving a particular business a loan. A written business plan is essential, which lists down the aims and the objectives of the business, the amount of growth the business may experience in the coming years and the talents of the people, belonging to the family (Miller, 1978). Since Gyro Grill is a family business, an individual can state the number of people in the family who have gotten a culinary diplomas, the number of youngsters who have done proper business planning while studying and old people with an experience and knowledge of the business settings. It is also important to state which type of food Gyro Grill will serve, especially the specialty foods, which are necessary (Miller, 1978 ). If the business plan lives up to the bank’s requirements, then the family business will easily get a loan from the bank. However, Gyro Grill should not completely depend upon the bank for a loan. It must have its own financial resources, in the form of accumulated savings of the family. Initially, the business can employ people from the family, since it will have a smaller group of customers. The business needs registration with the law, as well. The Gyro Grill may act as an independent business, instead of a franchise from another restaurant. Therefore, the business will need effective marketing tools to spread the word of a new

Rigid Bodies and Application to Snooker Assignment

Rigid Bodies and Application to Snooker - Assignment Example 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. 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. ... (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