* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 19 Outline The First Law of Thermodynamics - Help-A-Bull
Survey
Document related concepts
Transcript
Force and Potential Energy (3D) β’ Combing these in vector form, ππ ππ ππ π=β π+ π+ π ππ₯ ππ¦ ππ§ β’ We can write this more succinctly using the βdelβ operator. π π π π= π +π +π ππ₯ ππ¦ ππ§ β’ The force is the negative gradient of the potential. π = βππ Energy Diagram β’ We can glean a lot of information by looking at graph of the potential energy. Energy Diagram Example Chapter 7 Summary Potential Energy and Energy Conservation β’ Gravitational potential energy: πg = ππβ β’ Conservation of mechanical energy πΈtot = πΎ1 + πg1 = πΎ2 + πg2 = constant β’ Elastic potential energy: πel = 12ππ₯ 2 β’ Conservative forces β’ Potential energy, reversible, path-independent, zero closed loop β’ Conservation of energy: βπΎ + βπ + βπint = 0 β’ Force and potential energy: π = β β’ Energy diagrams β’ Stable minima and unstable maxima ππ π ππ₯ + ππ π ππ¦ + ππ π ππ§ Chapter 8 Outline Momentum, Impulse, and Collisions β’ Momentum β’ Impulse β’ Conservation of momentum β’ Vector components β’ Collisions β’ Elastic and inelastic β’ Center of mass β’ Rocket propulsion Momentum β’ Consider the case of a collision between two cars. β’ Using Newtonβs laws to find the resulting motion is difficult. β’ We do not fully know the exact forces involved. β’ We can deal with situations such as these by considering a new concept, momentum. β’ Newtonβs second law: ππ π π=π = ππ ππ‘ ππ‘ β’ We call the product of mass and velocity momentum, π. π = ππ Momentum β’ We therefore rewrite Newtonβs second law. The net force acting on a particle equals the time rate of change of momentum. ππ π= ππ‘ Impulse-Momentum Theorem β’ We have already considered a force applied over some distance (work). β’ What about a force applied for some time? This is called the impulse, π±. β’ First consider a constant force. π± = Ξ£π Ξπ‘ β’ But, Ξ£π = π«π π₯π‘ , so π± = Ξπ β’ This is called the impulse-momentum theorem. π± = π2 β π1 β’ The change in momentum over a time period is the impulse of the net force that acts on the particle during that interval. Impulse β’ In general, we express the impulse as the integral of the force over time. π‘2 π±= Ξ£π ππ‘ π‘1 β’ We can define an average force, πave , such that π± = πave Ξπ‘ Impulse Example Conservation of Momentum β’ Consider two bodies that interact with each other but nothing else. β’ In this system, there are no external forces, only internal forces. β’ This is an isolated system. β’ Each body exerts a force on the other with an equal magnitude but opposite direction. ππ΅ on π΄ + ππ΄ on π΅ = 0 πππ΄ πππ΅ π(ππ΄ + ππ΅ ) + = =0 ππ‘ ππ‘ ππ‘ β’ The total momentum of the system, π· = ππ΄ + ππ΅ is constant. Conservation of Momentum β’ If the net external force on a system is zero, the total momentum of the system is constant. β’ Conservation of momentum. β’ This is a fundamental principle. β’ Treat each vector component separately. π· = ππ΄ ππ΄ + ππ΅ ππ΅ + β― = constant Conservation of Momentum Example Types of Collisions β’ We define a collision to be any strong interaction between bodies that lasts for a relatively short time. β’ In an elastic collision, all of the forces between the colliding bodies are conservative, no mechanical energy is lost and the total kinetic energy is the same before and after. β’ In an inelastic collision, the internal forces are not all conservative, the total kinetic after the collision is less than before. β’ If the bodies stick together after the collision, it is a completely, or perfectly inelastic collision. β’ Regardless, momentum is conserved! Collision Example Elastic Collisions in One Dimension β’ For an elastic collision, both momentum and mechanical energy are conserved. β’ In one dimension, ππ΄ π£π΄1π₯ + ππ΅ π£π΅1π₯ = ππ΄ π£π΄2π₯ + ππ΅ π£π΅2π₯ 1 1 1 1 2 2 2 2 π π£ + π π£ = π π£ + π΄ π΅ π΄ π΄1π₯ π΅1π₯ π΄2π₯ 2 2 2 2ππ΅ π£π΅2π₯ β’ Given the masses and initial velocities, we can solve for the final velocities. β’ For the special case where one body is initially at rest, this reduces to: 2ππ΄ π£π΅2π₯ = π£π΄1π₯ ππ΄ + ππ΅ Collision Example #2