Comprehensive Review. Mahapatra

Transcription

1 Comprehensive Review Mahapatra

2 Ch2 : Equations of motion for constant acceleration x(t) x 0 v 0 t 1 2 at 2 v(t) v 0 at a(t) = a It follows from the one above if you take the derivative wrt time Just says that acceleration is constant in time; it is not useful in terms of algebra. From now on writing x or v implies x(t) or v(t) so as not to write (t) all the time. Algebra reminder: if you have a set of equations and a set of unknown variables their number have to be the same to be able to solve it. We have TWO independent equations and 6 variables 2a( x) V 2 f x, x 0,v,v 0,t,a - V 2 0

3 Ch 3: Ball Dropping a y = g (downwards) a x = 0 Constant for both cases!!! v x = 0 v x >0

4 Ch 3: Firing up in the air at an angle A ball is fired up with velocity V o and angle θ o. What is the range? y-component y y 0 0 v 0 y v 0 sin v y a y g t x-component x x 0 R? v x v 0 cos t Time of flight can be obtained by the x-component x x 0 R v x v 0 cos t R v 0 cos y y 0 v 0y t 1 2 a yt 2 R 0 v 0 sin 1 2 v 0 cos 2 g R v 0 cos tan 1 2 g R v 2 0 cos 2 R 2sin cos v 2 0 g sin(2) v 2 0 g

5 As a ship is approaching the dock at 45.0 cm/s, an important piece of landing equipment needs to be thrown to it before it can dock. This equipment is thrown at 15.0 m/s at 60.0 o above the horizontal from the top of a tower at the edge of the water, 8.75 m above the ship's deck. For this equipment to land at the front of the ship, at what distance from the dock should the ship be when the equipment is thrown? Air resistance can be neglected. y-component y y 0 h v 0 y v 0 sin v y x-component x x 0 R v x v 0 cos t T a y g t h y y 0 v 0y t 1 2 a y t 2 h v 0 sint 1 2 gt 2 t 2 2v 0 sin t 2h g g 0 R v 0 cost D R v B T R t 2v 0 sin 4v 2 0 sin 2 g g 2 2 8h g T

6 Ch3: Relative velocity in one dimension If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as v P/A. If P is moving relative to frame B and frame B is moving relative to frame A, then the x-velocity of P relative to frame A is v P/A-x = v P/B-x + v B/A-x.

7 Ch 3: Relative Velocity: Boat on the River You want to cross the river so that the boat gets exactly from A to B. The river has a current v C =4 km/h. Your boat s speed in still water is v B =20km/h? What is the angle you should aim at to do that? How long would it take? vb

8 Ch 4: Newton s Second Law The acceleration is in the SAME direction as the NET FORCE This is a VECTOR equation It applies for EACH direction If I have a NET force, what is my acceleration? More force more acceleration More mass less acceleration Vector Equation F ma F x Weight ma x :, Fy ma W mg y Notice that the 2 nd law also implies the 1 st law, if a=0 then the sum of all forces on an object is zero

9 Newton s Third Law Whenever one object exerts a force on a second object, the second exerts an equal and opposite force on the first object OR To every action there is an equal and opposite reaction

10 2016 Pearson Education Inc. Newton s third law

11 Skater Skater pushes on a wall The wall pushes back Equal and opposite force The push from the wall is a force Force provides an acceleration She flies off with some non-zero speed

12 F Groundon theperson Walking - F Person on theground She pushes on the ground and the ground PUSHES her forward Equal and opposite force

13 Newton s Third Law Objects at rest An apple on a table or a person in a chair there will be the weight (mass pulled downward by gravity) and the normal force (the table or chair s response).

14 Goals for Chapter 5 To use and apply Newton s Laws RM1 Connect these laws to Kinematics To study friction and fluid resistance To consider forces in circular motion First in Equilibrium (Newton s First Law) Then, in non-equilibrium (Second Law)

15 Slide 14 RM1 Rupak Mahapatra, 9/26/2016

16 1-D equilibrium Consider an athlete hanging on a massless rope.

17 Figure 5.3

18 A crate on an inclined plane An object on an inclined plane will have components of force in x and y space.

19 Box on an inclined plane A box with mass m is placed on a frictionless incline with angle and is allowed to slide down. a) What is the normal force? b) What is the acceleration of the box? c) How long does it take to fall if the ramp has a length L? F N mg cos mg sin w B =mg x : mgsin ma x y : F N mgcos 0 a x gsin

20 Person in an elevator Moving in an arbitrary direction

21 Friction Two types of friction: 1. Kinetic: The friction force that slows things down 2. Static: The force that makes it hard to even get things moving

22 Static Friction This is more complicated For static friction, the friction force can vary F Friction m Static F Normal Example of the refrigerator: If I don t push, what is the static friction force? What if I push a little?

23 Kinetic Friction For kinetic friction, it turns out that the larger the Normal Force the larger the friction. We can write F Friction = m Kinetic F Normal Here m is a constant Warning: THIS IS NOT A VECTOR EQUATION!

24 Box on an inclined plane 2 Same problem as before, but now friction is not negligible and the coefficient of friction is m. The inclined plane is adjustable and we change from 0 to 90 degrees. Mass is known and is equal to m. Calculate and draw a graph of: Friction force as a F N function of Acceleration of the box as a function of F f <μf N mg cos w B =mg If > max x : mgsin m K F N ma x If static x : mgsin F f 0 y : F N mgcos 0 mg sin F f mgsin m s F N mgsin max m s mgcos max mgsin max tan max m s y : F N mgcos 0 mgsin m k mgcos ma x

25 Ch5: The dynamics of uniform circular motion In uniform circular motion, both the acceleration and force are centripetal. When going in a circle the choice of coordinates is NOT optional. One of the axis has to point towards the center and in this case F r ma c m v2 R

26 Hanging ball going on circles For the situation below, if only L and the angle are known, (a) What is the tension in the cord? (b) What is the time it takes the ball to go around once? F F r y F F T T 2 v sin m R cos mg 0 2 v m Lsin F T mg cos mg cos sin m 4 2 Lsin T 2 T 2 L cos g

27 Rounding unbanked curve: maximum speed Given a coefficient of static friction, μ s, what is the maximum speed the car can go around the curve without skidding? F r F f m v2 R F y F N mg 0 F f max m s F N m s F N m v 2 max R v max m s gr

28 Rounding banked curve: maximum speed What is the angle you need to go around the curve at speed v without friction? F r F N sin m v2 R F r F N cos mg 0 F N mg cos mg v2 sin mgtan m cos R v grtan

29 Ch 6: Work done by a constant force Work done by a constant force on an object is W by F r F r s Fscos Fs Force on object Displacement of the object The maximum work (efficiency) done by a force is achieved when the force is in the direction of the displacement (it can be positive or negative) If the force is perpendicular to the displacement then it does not work (e.g. Normal forces do no work)

30 Sign of work: positive, negative or zero

31 Stepwise solution of work done by several forces Example 6.2 W T F T scos (5000N)(20m)cos(36.9 o ) 80kJ W f F f scos (3500N)(20m)cos(180 o ) 70kJ W mg F mg scos mgscos(90 o ) 0 W N F N scos F N scos(90 o ) 0 W tot W FT W f W mg W N 10kJ

32 Ch 6: Work-Energy Theorem The next idea couples kinematics (changes in velocity of an object) and Netwon s second law of motion (total force on an object leading to an acceleration) to the total work done on an object. The work done by the net (total) force on an object is equal to the change in the objects kinetic energy W total KE final KE initial where KE 1 2 m v2 We first show this for constant forces but we will see later that this is true for any type of forces (that is why we call it a theorem) F net ma According to Newton s 2 nd law (here the forces are constant and we take the x-direction to point along the direction of the net force (sum of all forces) on the object. Then we can use the 1-D kinematic equation v f 2 v i 2 2a(x f x i ) v f 2 v i 2 2as v f 2 v i 2 2 F net m s 1 2 m v 2 f 1 2 m v 2 i F net s F net s 1 2 m v 2 f 1 2 m v 2 i W net KE f KE i

33 The stretch of a spring and the force that caused it The force applied to an ideal spring will be proportional to its stretch. The graph of force on the y axis versus stretch on the x axis will yield a slope of k, the spring constant. Force due to a spring is: F spring kx x is positive for stress and negative for compression W spring b a (kx )dx 1 2 kb2 1 2 ka2

34 Power Power is the rate at which work is done. Average power is: Instantaneous power is: The SI unit of power is the watt (1 W = 1 J/s), but another familiar unit is the horsepower (1 hp = 746 W) Pearson Education Inc.

35 Ch 7: Conservation of Mechanical Energy For some types of problems, Mechanical Energy is conserved (more on this next week) E.g. Mechanical energy before you drop a brick is equal to the mechanical energy after you drop the brick K 2 +U 2 = K 1 +U 1 Conservation of Mechanical Energy E 2 =E 1

36 Problem Solving What are the types of examples we ll encounter? Gravity Things falling Springs Converting their potential energy into kinetic energy and back again Gravity: E = K + U = ½mv 2 + mgy Spring: E = K + U = ½mv 2 + ½kx 2

37 What s the speed in a vertical circle? Refer to Example 7.4 and Figure 7.9. KE top UE top KE bott UE bott 0 mgh 1 2 m v2 0 mgr 1 2 m v2 v 2gR

38 Work done by a spring Figure 7.13 below shows how a spring does work on a block as it is stretched and compressed Pearson Education Inc.

39 Problem 7.46: Energy + circular motion A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle. F N - + mg r : mg F N m v2 R g v 2 min R v 2 min gr (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)? (b) If the car starts at height h= 4.00 R and the radius is R = 20.0 m, compute the radial acceleration of the passengers when the car is at point C, which is at the end of a horizontal diameter. What is the minimum velocity so at B we are going around a CIRCLE? You will feel like you are flying and not touching the track? Now we know the velocity (or KE) we need at B so we can use conservation of energy (remember F N does no work so W other =0) to get it E A E B 1 2 m v 2 A mgy A 1 2 m v 2 B mgy B 0 mgh 1 mgr mg2r 2 h 5 2 R

40 Non/Conservative Forces W total KE final KE initial ( UE final UEinitial ) K 2 +U 2 = K 1 +U 1 For some types of problems, Mechanical Energy is conserved. Example: gravity. Work done depends NOT on the path, but only on the final and initial points. For other types of problems, some part of the energy is lost heat, sound and such. Example is friction. The work done due to friction takes away mechanical energy and hence doesn t conserve mechanical energy

41 Revisiting the work energy theorem The work energy theorem says the total work is equal to the change in KE W net KE 2 KE 1 On the other hand, we have seen that the work due to gravity ONLY DEPENDS ON THE INITIAL AND FINAL POINT OF THEIR PATH, NOT ON THE ACTUAL PATH. These type of forces (of which gravity is one) are called conservative forces. Let s break the total work done into two parts, the one done by the conservative forces and the ones done by non-conservative forces (e.g. friction) W conserv W non conserv KE 2 KE 1 UE 2 UE 1 W non conserv KE 2 KE 1 W non conserv E 2 E 1 where E KE UE If the work done by the non-conservative forces is zero then the total energy is conserved. This is a very powerful tool!!

42 Bring together two potential energies and friction Example 7.9 What is the spring constant needed? W non conserv E 2 E 1 W non conserv F f s (17,000N)(2m) E m v 2 1 mgy kx (2000kg)(4 m/s) 2 (2000kg)g(2m) 0 E m v 2 2 mgy kx k(2m) k k N/m

43 Ch8: Momentum and Newton s second law The momentum of a particle is the product of its mass and its velocity: Newton s second law can be written in terms of momentum: 2016 Pearson Education Inc.

44 Ch 8: Conservation of Momentum For a closed system (no external forces), by Newton s 3 rd law, F=0 Conservation of Momentum r m v r m ' i i i i i v i ' Sum of all Sum of all momentum before = momentum after True in X and Y directions separately!

45 Impulse: change in momentum r F d P r dt r p impulse r F net dt r F net ave t This is considered typically during a collision (short time) or explosion. p p 1x 2x (0.1Kg)(35m/s) 3.5Kg m/s (0.1Kg)(30m/s) 3.0Kg m/s p x p 2x p 1x 6.5Kg m/s 75mi/hr=35 m/s m=0.1 Kg 60 mi/hr=30 m/s Typical time spent with raquet: 0.01 s Force =6.5/0.01 = 650 N Feels like lifting 150 lb weight in one hand

46 Types of collisions according to energy before and after the collision Definitions: Elastic collision = TOTAL kinetic energy is conserved Inelastic collision = TOTAL kinetic energy is not conserved. Keep in mind Momentum is ALWAYS conserved in a collision Total Energy may or may not. Q: where does the energy go?

47 Elastic compared to inelastic

48 Objects colliding along a straight line (use component notation) x : m A v A1x m B v B1x m A v A 2x m B v B 2x x : (0.5)(2.0) (0.3)(2) (0.5) v A 2x (0.3)(2) v A 2x 0.40 m/s

49 A two-dimensional collision Robot A has a mass of m A, initially moves at v A1 parallel to the x-axis. After the collision with B, which has a mass of m B, robot A is moving at v A2 in a direction that makes and angle of degrees. What is the final velocity of B? KE Before KE After 1/ 2mA v A1 21/ 2mA v A2 2 1/ 2mB vb22 Before After x : m A v A1 0 m A v A 2 cos m B v B 2x y : 0 0 m A v A 2 sin m B v B 2y x : v B 2x m A v A1 m A v A 2 cos m B y : v B 2y m A v A 2 sin m B tan v B 2y v B 2x m A v A 2 sin m A v A1 m A v A 2 cos

50 Ch 9: Rotational Kinematics Take our results from linear physics and do the same for angular physics Analogue of Position Velocity Acceleration Force Mass Momentum Energy Chapters 1-3 Chapters 4-7

51 Rotational Motion Will do Chapters 9 and 10 in four combined lectures Start with Fixed Axis motion The relationship between linear and angular variables Rotating and translating at the same time Eventually move from kinematics to dynamics just like earlier this semester

52 Axis of Rotation: Definitions Pick a simple place to rotate around Call point O the Axis of Rotation Same as picking an origin Pick a fixed direction x Measure position as the angle w.r.t. direction of x

53 Velocity and Acceleration Define the angular velocity t or d dt : radians/ sec Define d dt as or the angular acceleration d 2 dt 2 radians/ sec 2

54 Uniform Angular Acceleration Derive the angular equations of motion for constant angular acceleration 0 0 t 1 t t

55 Pick reference point C Back to the Wheel Wheel is rotating but not moving, ground moves with speed V Use angular velocity : Velocity of P w.r.t. C = -R Same as velocity of ground w.r.t. bike Then velocity of C (and bike) w.r.t. the ground = +R=V Also, velocity of P w.r.t. ground is zero: = -R +R = 0 Sum of rotation and translation, just like boat and river problem before

56 Bicycle comes to Rest A bicycle with initial linear velocity V 0 decelerates uniformly (without slipping) to rest over a distance d. For a wheel of radius R: a) What is the angular velocity at t 0 =0? b) Total revolutions before it stops? c) Total angular distance traversed by wheel? d) The angular acceleration? e) The total time until it stops?

57 Chapter 10 Dynamics of Rotational Motion PowerPoint Lectures for University Physics, 14th Edition Hugh D. Young and Roger A. Freedman Lectures by Jason Harlow 2016 Pearson Education Inc.

58 Learning Goals for Chapter 10 Looking forward at what is meant by the torque produced by a force. how the net torque on a body affects the body s rotational motion. how to analyze the motion of a body that both rotates and moves as a whole through space. how to solve problems that involve work and power for rotating bodies. how the angular momentum of a body can remain constant even if the body changes shape Pearson Education Inc.

59 Three ways to calculate torque 2016 Pearson Education Inc.

60 Torque as a vector Torque can be expressed as a vector using the vector product. If you curl the fingers of your right hand in the direction of the force around the rotation axis, your outstretched thumb points in the direction of the torque vector Pearson Education Inc.

61 Torque and angular acceleration for a rigid body The rotational analog of Newton s second law for a rigid body is: Loosening or tightening a screw requires giving it an angular acceleration and hence applying a torque Pearson Education Inc.

62 2016 Pearson Education Inc.

63 2016 Pearson Education Inc.

64 2016 Pearson Education Inc.

65 2016 Pearson Education Inc.

66 Rigid body rotation about a moving axis The kinetic energy of a rotating and translating rigid body is K = 1/2 Mv cm2 + 1/2 I cm 2. There are two parts: 1. Motion of the CM (Center of Mass) 2. Rotation about the CM 2016 Pearson Education Inc.

67 Rolling without slipping The motion of a rolling wheel is the sum of the translational motion of the center of mass plus the rotational motion of the wheel around the center of mass. The condition for rolling without slipping is 2016 Pearson Education Inc.

68 2016 Pearson Education Inc.

69 2016 Pearson Education Inc.

70 Combined translation and rotation: dynamics The acceleration of the center of mass of a rigid body is: The rotational motion about the center of mass is described by the rotational analog of Newton s second law: This is true as long as the axis through the center of mass is an axis of symmetry, and the axis does not change direction Pearson Education Inc.

71 2016 Pearson Education Inc.

72 Work in rotational motion A tangential force applied to a rotating body does work on it Pearson Education Inc.

73 Work and power in rotational motion The total work done on a body by the torque is equal to the change in rotational kinetic energy of the body, and the power due to a torque is: When a helicopter s main rotor is spinning at a constant rate, positive work is done on the rotor by the engine and negative work is done on it by air resistance. Hence the net work being done is zero and the kinetic energy remains constant Pearson Education Inc.

74 Dynamics Recap in One Slide Force applied on a lever arm produces a torque: Like 2 nd law shows F=ma, this produces an angular acceleration We used this to solve many rotation problems, including translation + rotation

75 Angular momentum For a rigid body rotating around an axis of symmetry, the angular momentum is: For any system of particles, the rate of change of the total angular momentum equals the sum of the torques of all forces acting on all the particles: 2016 Pearson Education Inc.

76 Angular momentum The angular momentum of a rigid body rotating about a symmetry axis is parallel to the angular velocity and is given by 2016 Pearson Education Inc.

77 2016 Pearson Education Inc.

78 Conservation of angular momentum When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved) Pearson Education Inc.

79 2016 Pearson Education Inc.

80 Supernova and Neutron stars Twice as big as Sun, but as small as campus perimeter 2016 Pearson Education Inc.

81 Conditions for equilibrium For an extended body to be in static equilibrium, two conditions must be satisfied. The first condition is that the vector sum of all external forces acting on the body must be zero: The second condition is that the sum of external torques must be zero about any point: 2016 Pearson Education Inc.

82 2016 Pearson Education Inc.

83 2016 Pearson Education Inc.

84 2016 Pearson Education Inc.

85 2016 Pearson Education Inc.