1. Physics 121 Mechanics Lecture notes are posted on www.physics.byu.edu/faculty/chesnel/physics121.aspx Instructor Karine Chesnel April 14, 2009 Final Review

2. Have you evaluated the class on Route Y? YES! Click! Do not forget to evaluate this class online until April 15 th Thank you! Quiz # 40 Final review04/14/09

3. Physics 121 Winter 2009 * Today * last lecture Last assignment: Online homework 24: (70pts) until midnight today! http://gardner.byu.edu/121w2/homework.html For any question after this class: Karine Chesnel N319 ESC 801-422-5687 kchesnel@byu.edu I will be available until Friday April 24 th

4. Physics 121 Winter 2009 - section 2 Class Average First test 83 /100 Second test 68/100 Third test 74/100 Test average 75/100 (30%) Final (20%) Homework 85/100(25%) Labs 96/100(15%) quizzes 113/100(10%) Prepare well for the FINAL test! Try to increase your average test score Class statistics

5. Final exam Friday April 17 through Wednesday April 22 At the testing center : 8 am – 9 pm (Mo- Fri) 10 am – 4 pm (Sat) Closed Book Only bring: - Phys 121 (Chesnel) Memorization sheets (5pages) - Math reference sheet - Pen / pencil - Calculator - your CID No time limit Multiple Choice Questions: 30 questions

6. Final Review ch 1 – ch 15 Part I: KinematicsCh. 1- 4 Part II: Laws of Motion (material points)Ch. 5- 8 Part III: Laws of Motion (Solids rotation)Ch. 9 - 13 Part IV: Oscillatory motionCh. 15

7. Part I Kinematics Ch. 2 Motion in one dimension Position, velocity, acceleration Case of constant velocity Case of constant acceleration Free falling motion Ch. 4 Motion in two dimensions Position, velocity, acceleration Case of constant acceleration Projectile motion Circular motion (uniform & non-uniform) Tangential and radial acceleration Ch. 3 Vectors Coordinate systems Algebra Ch. 1 Physics & measurments Standard units Dimensional analysis

8. Kinematics The displacement is the difference between two positions x1x1 x2x2  x= x 2 – x 1 Average acceleration (Instantaneous) acceleration Average velocity (Instantaneous) velocity The speed is the amplitude of the velocity Final review04/14/09

9. Motion under constant velocity V = constant x(t) t tt xx 0 x0x0 = Slope  x/  t Position at t=0 x(t) = x 0 + v.t Final review04/14/09

10. Motion under constant acceleration a = constant a(t) t 0 v(t) = v 0 + a.t v(t) t tt vv 0 v0v0 V (t) is linear x(t) t 0 x0x0 x(t) = x 0 +v 0.t+ ½ a.t 2 x (t) is parabolic Application: Free Fall a = g Final review04/14/09

11. Vectors components Cartesian to polar conversion in 2D x y (0,0) A=(x,y)   x y Final review04/14/09

12. Projectile Motion r(t) = r 0 + V 0 t + ½ g t 2 r0r0 V0V0 y x (0,0) a = g = (0, -g) V2V2 V1V1 V3V3 x(t) = x 0 + V 0,x t y(t) = y 0 + V 0,y t - ½ gt 2 v(t) = v 0 + gt Uniform velocity along x Free falling along y vertical horizontal Final review04/14/09

13. Projectile V0V0 y x (0,0) a = g V max Performances H Hits the ground R R = V 0 2 sin (2  /g The particle is projected with a speed V 0 at angle  The maximal height is The horizontal range is H = (V 0 sin  ) 2 / 2g  Review Ch.4 Motion in two dimensions1/27/09 Final review04/14/09

14. Uniform Circular Motion Angular speed is constant:  = .t Velocity: V x = - R 0  sin (  t  V y = R 0  cos (  t  Position x = R 0 cos (  t  y = R 0 sin (  t  Acceleration a x = - R 0   cos (  t  a y = - R 0   sin (  t  a = -   r |V| = R 0  The acceleration is centripetal. Its magnitude is |a| = R 0   (0,0) x y  R0R0 V a Final review04/14/09

15. Tangential and radial acceleration General case V1V1 V2V2 V3V3 a a a V is tangential to the trajectory The components of the acceleration in the Frenet frame are: Tangential The sign tells if the particle speeds up or slows down a t = dV/dt centripetal Always directed toward the center of curvature R radius of curvature a c = V 2 /R Final review04/14/09

16. Generalization position r (t) velocity First derivative acceleration Second derivative First integration Second integration a(t) Final review04/14/09

17. Quiz # 41 Two racquet balls are thrown in the air at the same time from the same height H. One ball (yellow) is thrown at some angle , with a vertical velocity V 0,y, and horizontal velocity V 0,x. The other ball (blue) is thrown vertically with the same vertical velocity V 0,y Which ball will hit the floor first? AThe blue ball BThe yellow ball CBoth of them V 0,y V0V0 x(t) = x 0 + V 0,x t y(t) = y 0 + V 0,y t - ½ gt 2 For the yellow ball the motion will be given by For the blue ball the motion will be given by y(t) = y 0 + V 0,y t - ½ gt 2 Both balls touch the ground at the same time! y H Final review04/14/09

18. Part II Laws of motion Ch. 6 Newton’s laws applications Circular Motion Drag forces and viscosity Friction Fictitious forces Ch. 8 Conservation of Energy Mechanical energy Conservation of energy Ch. 7 Work and energy Work Kinetic energy Potential energy Work- kinetic energy theorem Ch. 5 The Laws of Motion Newton’s first law Newton’s second law Newton’s third law

19. Summary of the Laws of Motions First Law: Principle of Inertia In a inertial frame, an isolated system remains at constant velocity or at rest Second Law: Forces and motion In an inertial frame the acceleration of a system is equal to the sum of all external forces divided by the system mass Third Law: Action and reaction If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1. F1F1 F2F2 Final review04/14/09

20. Forces of Friction F < f s,max = m s N m s is called the coefficient of static friction Static regime f k = m k N m k is called the coefficient of kinetic friction Kinetic regime F f Static regimeKinetic regime Two regimes Final review04/14/09

21. Work and kinetic energy Defining the kinetic energy Using Newton’s second law Work- Kinetic energy theorem Final review04/14/09

22. Mechanical energy We define the mechanical energy E mech as the sum of kinetic and potential energies Sometimes, the work of non conservatives forces (friction, collision) is transformed into internal energy thus Final review04/14/09

23. Closed System with conservative forces only F cons There are no non-conservative forces working The mechanical energy is constant The mechanical energy is conserved between initial and final points Final review04/14/09

24. Examples of Potential energy x L0L0 Spring Gravity g h Gravitational field m r M Final review04/14/09

25. Quiz # 42 ` F mgmg N F This man is pushing this box of 85kg on the carpet at a constant speed. How is the magnitude of the force he needs to apply? ALarger than the weight of the box BSame than the weight of the box CSame than the friction force DLarger than the friction force ENone of the answers According to Newton’s law Here the velocity is constant, so Thus and on the other hand This is true both vertically and horizontally f Final review04/14/09

26. Part III Laws of motion for Solids Ch. 10 Rotation of solid Rotational kinematics Rotational and translational quantities Rolling motion Torque Ch. 12 Static equilibrium and elasticity Rigid object in equilibrium Elastic properties of solid Ch. 11 Angular momentum Angular momentum Newton’s law for rotation Conservation of angular momentum Precession motion Ch. 9 Linear Momentum & collision Linear momentum Impulse Collisions 1D and 2D Ch. 13 Universal gravitation Newton’s law of Universal gravitation Gravitational Field & potential energy Kepler’s laws and motion of Planets

27. Linear Momentum & Impulse The Newton’s second law can now be written as The linear momentum of a particle is the product of its mass by its velocity Units: kg.m/s or N.s For an isolated system The impulse is the integral of the net force, during an abrupt interaction in a short time According to Newton’s 2nd law: Final review04/14/09

28. Collisions 1. Conservation of linear momentum 2. Conservation of kinetic energy (2) Elastic collision (1) V 1,i V 2,i V 1,f V 2,f Inelastic collision: change in kinetic energy Perfectly inelastic collision: the particles stick together Final review04/14/09

29. Collisions 1D If one of the objects is initially at rest: Combining (1) and (2), we get expression for final speeds: V 1,i V 1,f V 2,f V 2,i Collisions 2D 3 equations 4 unknow parameters V 1,i x y V 1,f V 2,f   Final review04/14/09

30. Solid characteristics M OC The center of mass is defined as: C O dm r   FaM C   The moment of inertia of the solid about one axis: I I’ D Final review04/14/09

31. Rotational kinematics Solid’s rotation Angular position  Angular speed  Angular acceleration  Linear/angular relationship Velocity Acceleration Tangential Centripetal For any point in the solid Rotational kinetic energy Final review04/14/09

32. Motion of rolling solid P  C R Non- sliding situation The kinetic energy of the solid is given by the sum of the translational and rotational components: K solid = K c + K rot If all the forces are conservative: Final review04/14/09

33. Torque & angular momentum We define the torque F   The angular momentum is defined as Deriving Newton’s second law in rotation angular momentum Linear momentum prL    When a force is inducing the rotation of a solid about a specific axis: For an object in pure rotation Final review04/14/09

34. Solving a problem Static equilibrium Define the system Locate the center of mass (where gravity is applied) Identify and list all the forces Apply the equality Choose a convenient point to calculate the torque (you may choose the point at which most of the forces are applied, so their torque is zero) List all the torques applied on the same point. Apply the equality Final review04/14/09

35. Gravitational laws Any object placed in that field experiences a gravitational force Any material object is producing a gravitational field M r urur m FgFg The gravitational field created by a spherical object is centripetal (field line is directed toward the center) The gravitational potential energy is UgUg 0 r Final review04/14/09

36. Kepler’s Laws “The orbit of each planet in the solar system is an ellipse with the Sun as one focus ” First Law “The line joining a planet to the sun sweeps out equal areas during equal time intervals as the planet travels along its orbit.” Second Law cst m L dt dA  2 0 “The square or the orbital period of any planet is proportional to the cube of the semimajor axis of the orbit” Third Law Final review04/14/09

37. Satellite Motion T sat =T Earth = 1 day Geostationary orbit Satellite speed (1) From Newton’s law (2) Escape speed The mechanical energy of The satellite on orbit is Final review04/14/09

38. Quiz # 43 A planet has a mass twice the mass of the Earth and a diameter 0.7 times the Earth diameter. What would be the weight, in Newtons, of a 82kg person standing at the surface of this planet? A 334 N B 803 N C 3280N D 483 N E 4830 N The weight of this person at the surface of this planet is Compared to the weight on earth 9.8 m/s 2 Gravitational field on Earth (Equivalent “mass’ on earth: 334kg!) Final review04/14/09

39. Part IV Oscillatory motion (ch15) Harmonic equation and solutions Energy of harmonic oscillator Spring motion Pendulum motions Damped oscillation Forced oscillation

40. x Harmonic motion Spring Equation of the motion is Harmonic equation A general solution to this harmonic equation is Amplitude angular frequency Phase constant (phase at t=0) with Unit = rad/s frequency Unit = Hz period Unit = s F Final review04/14/09

41. Position, velocity and acceleration Position, velocity and acceleration are all sinusoidal functions T x(t) t Position Velocity Acceleration t V(t) Phase quadrature t a(t) Opposite phase Final review04/14/09

42. Energy 0t U K 0x U K Final review04/14/09

43. The pendulum (punctual)  L m 0 The equation for motion is The solution for the angle position is with The oscillation frequency does not depend on the mass m The period of the oscillation is Final review04/14/09

44. Torsional pendulum A torsional pendulum uses the torque induced by torsion to oscillate 0 (rest)   For an angular displacement , the torque is The equation for the motion is then A general solution is: with L  According to Newton’s second law Final review04/14/09

45. Damped oscillations If the oscillator is moving in a resistive medium: friction, viscosity…. The oscillation will be damped. L0L0 x 0 F A Compressed The expression of the spring force is Applying the Newton’s second law Equation of the motion The expression of resistive force is Damping coefficient Final review04/14/09

46. Damped oscillations with The general solution for the motion is T x(t) t Underdamped x(t) t overdamped Critical x(t) t Final review04/14/09

47. Forced oscillations A general solution to this equation is AmplitudeFrequency forced Phase constant (phase at t=0) An external force is applied to the system, forcing the oscillation to a frequency  If any, resistive force isThe spring force is The external force is Final review04/14/09

48. Forced oscillations A general solution to a forced oscillation motion with and Resonance In absence of resistive forces, the amplitude of the oscillation is amplified to infinity when the force frequency   approaches the proper frequency  0 Low damping b No damping b=0    Large damping b Final review04/14/09

49. GOOD LUCK With the FINALS ! To contact me: Karine CHESNEL kchesnel@byu.edu Office: N319 ESC 801 – 422 – 5687 Final review04/14/09