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Classical Mechanics Lecture 8

Midterm November 8. Classical Mechanics Lecture 8. Today's Examples: a) Potential Energy b) Mechanical Energy. Summary Unit 8: Potential & Mechanical Energy. Energy Conservation Problems in general. For systems with only conservative forces acting. E mechanical is a constant.

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Classical Mechanics Lecture 8

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  1. Midterm November 8 Classical Mechanics Lecture 8 Today's Examples: a) Potential Energy b) Mechanical Energy

  2. Summary Unit 8: Potential & Mechanical Energy

  3. Energy Conservation Problems in general For systems with only conservative forces acting Emechanical is a constant

  4. Energy Conservation Problems in general • conservation of mechanical energy can be used to “easily” solve problems. • Define coordinates: where is U=0? • Identify important configurations i.e where potential is minimized U=0. • Identify important configurations, i.e starting point where mass is motionless K=0 (for conservative forces) ALWAYS! • Problem usually states the configurations of interest!

  5. Pendulum Problem Using Conservation of Mechanical energy Using Work Formalism Conserve Energy from initial to final position

  6. Pendulum Problem Don’t forget centripetal acceleration and force…required to maintain circular path. Tension is …”what it has to be!” At bottom of path:

  7. Pendulum Problem

  8. Pendulum Problem Kinetic energy of mass prior to string hitting peg is conserved. Set h=0 to be at bottom equilibrium position

  9. Pendulum Problem Radius for centripetal acceleration has been shortened to L/5 !

  10. Loop the Loop To stay on loop, the normal force, N, must be greater than zero.

  11.  Mass must start higher than top of loop

  12. Gravity and Springs…Oh my! Conservation of Energy Coordinate System Initial configuration Final configuration Solve Quadratic Equation for x!

  13. Pendulum 2 Given speed and tension at certain point in pendulum trajectory can use conservation of mechanical energy to solve for L and m….

  14. Pendulum 2

  15. Pendulum 2

  16. Pendulum 2

  17. Classical Mechanics Lecture 9 Today's Concepts: a) Energy and Friction b) Potential energy & force

  18. Main Points

  19. Main Points

  20. Macroscopic Work done by Friction

  21. Macroscopic Work done by Friction

  22. Macroscopic Work: Applied to big (i.e. macroscopic) objectsrather than point particles (picky detail) We call it “macroscopic” to distinguish it from “microscopic”. This is not a new idea – it’s the same “work” you are used to.

  23. f f

  24. “Heat” is just the kinetic energy of the atoms!

  25. Incline with Friction:Newton’s law and kinematics

  26. Incline with Friction: Work-Kinetic Energy

  27. Checkpoint A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and onto a floor where friction causes it to stop a distance r from the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is mk. What is the macroscopic work done on the block by friction during this process? A) mgHB) –mgHC) mkmgDD) 0 m H D Mechanics Lecture 9, Slide 34

  28. Work by Friction :Wfriction<0 must be negative N1 N2 mg mmg mg Motionless box released on frictionless incline which then slides on horizontal surface with friction to a stop. m H

  29. CheckPoint What is the macroscopic work done on the block by friction during this process? A) mgHB) –mgHC) mkmgDD) 0 B) All of the potential energy goes to kinetic as it slides down the ramp, then the friction does negative work to slow the box to stop C) Since the floor has friction, the work done by the block by friction is the normal force times the coefficient of kinetic friction times the distance. m H D

  30. Checkpoint A block of mass m, initially held at rest on a frictionless ramp a vertical distance Habove the floor, slides down the ramp and onto a floor where friction causes it to stop a distance Dfrom the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is mk. What is the total macroscopic work done on the block by all forces during this process? A) mgHB) –mgHC) mkmgDD) 0 m H D Mechanics Lecture 9, Slide 37

  31. CheckPoint What is the total macroscopic work done on the block by all forces during this process? A) mgHB) –mgHC) mkmgDD) 0 B) work= change in potential energy C) The only work being done on the object is by the friction force. D) total work is equal to the change in kinetic energy. since the box starts and ends at rest, the change in kinetic energy is zero. m H D

  32. Force from Potential Energy:1Dstart here

  33. Force from Potential Energy

  34. Potential Energy vs. Force

  35. Potential Energy vs. Force

  36. Potential Energy vs. Force

  37. Potential Energy vs. Force

  38. Clicker Question U(x) x (c) (d) (a) (b) Suppose the potential energy of some object U as a function of x looks like the plot shown below. Where is the force on the object zero? A) (a) B) (b) C) (c) D) (d)

  39. Clicker Question U(x) x (c) (d) (a) (b) Suppose the potential energy of some object U as a function ofx looks like the plot shown below. Where is the force on the object in the +x direction? A) To the left of (b) B) To the right of (b) C) Nowhere

  40. Clicker Question U(x) x (c) (d) (a) (b) Suppose the potential energy of some object U as a function of x looks like the plot shown below. Where is the force on the object biggest in the –xdirection? A) (a) B) (b) C) (c) D) (d)

  41. Equilibrium

  42. Equilibrium points

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