Chapter 6 – Work, Power and Efficiency
... W = Eg2 – Eg1 W = ∆Eg From this we conclude that work done on an object results in a change in the potential energy of the object. Elastic Potential Energy Many objects can be stretched, compressed, bent or change shape when a force is applied and then return to their original shape after the force ...
... W = Eg2 – Eg1 W = ∆Eg From this we conclude that work done on an object results in a change in the potential energy of the object. Elastic Potential Energy Many objects can be stretched, compressed, bent or change shape when a force is applied and then return to their original shape after the force ...
Lecture Notes 2d order homogeneous DEs with constant coefficients
... with initial conditions for y(0),y ' (0)................yn-1(0) has a unique solution of the form : y( x) A 1 y 1( x) A 2 y 2( x) ......................................... A n y n( x) where the y i (x) am linearly independent solutions to the DE and the Ai are constants determined from the initi ...
... with initial conditions for y(0),y ' (0)................yn-1(0) has a unique solution of the form : y( x) A 1 y 1( x) A 2 y 2( x) ......................................... A n y n( x) where the y i (x) am linearly independent solutions to the DE and the Ai are constants determined from the initi ...
Torque
... Example 1: A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. ...
... Example 1: A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. ...
Document
... Our book states that in isolated systems momentum is conserved. Another way to state this is that we can account for all the parts involved in the problem. ΣF ...
... Our book states that in isolated systems momentum is conserved. Another way to state this is that we can account for all the parts involved in the problem. ΣF ...
Review sheet 4 Newton
... A ball is dropped from a person’s hand and falls to Earth. Identify an action-reaction pair, and compare the forces exerted by each object. a. The hand exerts a force on the ball; Earth exerts a force on the hand. b. Earth exerts a force on the ball; the hand exerts a force on Earth. c. Earth exerts ...
... A ball is dropped from a person’s hand and falls to Earth. Identify an action-reaction pair, and compare the forces exerted by each object. a. The hand exerts a force on the ball; Earth exerts a force on the hand. b. Earth exerts a force on the ball; the hand exerts a force on Earth. c. Earth exerts ...
What is angular momentum?
... is no net torque (the forces are internal)so L is a constant but the moment of inertia changes and since L = Iω as I decreases ω increases. So if I decreases by a factor of 2 then ω increases by a factor of 2. The kinetic energy is = Iω2 so if I decreases the kinetic energy increases by the amount o ...
... is no net torque (the forces are internal)so L is a constant but the moment of inertia changes and since L = Iω as I decreases ω increases. So if I decreases by a factor of 2 then ω increases by a factor of 2. The kinetic energy is = Iω2 so if I decreases the kinetic energy increases by the amount o ...
Rotational motion is all around us
... g. In Chapter 8, we saw that the net force acting on a system of particles is equal to the net external force acting on the system because the internal forces (those exerted by the particles within the system on one another) cancel in pairs. The treatment of internal torques exerted by the particles ...
... g. In Chapter 8, we saw that the net force acting on a system of particles is equal to the net external force acting on the system because the internal forces (those exerted by the particles within the system on one another) cancel in pairs. The treatment of internal torques exerted by the particles ...
L15 - unix.eng.ua.edu
... • full specification of microstate of the system is given by the values of all positions and all momenta of all atoms G = (pN,rN) ...
... • full specification of microstate of the system is given by the values of all positions and all momenta of all atoms G = (pN,rN) ...
Classical central-force problem
In classical mechanics, the central-force problem is to determine the motion of a particle under the influence of a single central force. A central force is a force that points from the particle directly towards (or directly away from) a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center. In many important cases, the problem can be solved analytically, i.e., in terms of well-studied functions such as trigonometric functions.The solution of this problem is important to classical physics, since many naturally occurring forces are central. Examples include gravity and electromagnetism as described by Newton's law of universal gravitation and Coulomb's law, respectively. The problem is also important because some more complicated problems in classical physics (such as the two-body problem with forces along the line connecting the two bodies) can be reduced to a central-force problem. Finally, the solution to the central-force problem often makes a good initial approximation of the true motion, as in calculating the motion of the planets in the Solar System.