Transcript Document
Physics 103 – Introduction to Physics I Motion Forces Energy 1 First Dimensions Units Precision Coordinate Systems Vectors Kinematics Motion Variables Constant Acceleration 2 Dimensions The dimension of a physical quantity specifies what sort of quantity it is— space, time, energy, etc. We find that the dimensions of all physical quantities can be expressed as combinations of a few fundamental dimensions: length [L], mass [M], time [T], and either electric charge [Q] or electrical current [A]. E M L T 2 For example, energy - 2 . The physical quantity speed has dimensions of L T . 3 Units International System (SI) The units of the fundamental dimensions in the SI are dimension SI cgs Customary [L] meter (m) second (s or sec) kilogram (kg) centimeter (cm) second foot (ft) second gram (g) slug or pound-mass [T] [M] The SI units will be introduced as we go along. 4 Unit Conversions We might measure the length of an (American) football field with a meter stick and a yard stick. We’d get two different numerical values, but obviously there is one field with one length. We’d say that 100 yards 91.44 meters. In other words, 100 yards 1.0 91 .44 meters Suppose we wish to convert 2 miles into meters. [2 miles = 3520 yards.] The units cancel or multiply just like common numerical factors. Since we want to cancel the yards in the numerator, the conversion factor is written with the yards in its denominator. 5280 feet 1 yard 91.44 m eters 2 m iles 1 m ile 3 feet 100 yards 3218.688 m eters Since each conversion factor equals 1, the physical measurement is unchanged, though the numerical value is changed. Note: the units are a part of the measurement as important as the number. They must always be kept together. 5 Precision & Significant Digits Instruments cannot perform measurements to arbitrary precision. A meter stick commonly has markings 1 millimeter (mm) apart, so distances shorter than that cannot be measured accurately with a meter stick. We report only significant digits—those whose values we feel sure are accurately measured. There are two basic rules: (i) the last significant digit is the first uncertain digit and (ii) when combining numbers, the result has no more significant digits than the least precise of the original numbers. A third rule is, the exercises and problems in the textbook assume there are 3 significant digits. Therefore, we never include more than 3 significant digits in our numerical results, no matter that the calculator displays 8 or 10 or more. 6 The uncertainty in a numerical value may be expressed in terms of a tolerance, as 23.273 0.005 Alternatively, the uncertainty can be shown in scientific notation simply by the number of digits displayed in the mantissa. 1.5 103 2 digits, the 5 is uncertain. 1.50 103 3 digits, the 0 is uncertain. 756. 37.2 0.83 2.5 796.53 800 8.0 102 3.563 3.20 11.4016 11.4 1.14101 5.6 17.59291886 18 1.8 101 or is it 1.76101 [Notice the ambiguity. Do we speak of the number of significant digits, or of the relative “place” of the uncertain digit? That is, should it be 18 or 17.6?] 7 Coordinate Systems We measure locations in space relative to a coordinate system. Firstly we select the origin of coordinates, and then the directions of orthogonal axes. Since the directions shown by orthogonal axes are mutually perpendicular, components along different axes are independent of each other. The commonly used two-dimensional coordinate systems are the Cartesian and the plane polar systems. 8 The three dimensional Cartesian coordinate system is comprised of three mutually perpendicular, straight axes, commonly denoted x, y, & z or ˆi , ˆj , & kˆ . [We’ll talk about those hat-things later.] The spherical polar coordinate system is comprised of a radius and two angles, as shown in the figure. Notice how the polar coordinates are defined in terms of the Cartesian system. Any point in space can be uniquely specified by listing three numerical coordinates. 9 Vectors As used in Physics, a scalar is a quantity that has only one property—a magnitude. Energy, speed, temperature, and mass are scalar quantities. A vector is a quantity that has two properties—a magnitude and a direction. Displacement, velocity, acceleration, and force are vector quantities. In text or equations, vectors are denoted with either a line or an arrow on top, thusly: A or A In diagrams, a vector is represented by an arrow. In text books, vectors are often denoted by bold-faced letters: A . Weirdly, University Physics uses both bold-face & an arrow! A is not the same as A! . . . . 10 The directions defined by the Cartesian coordinate axes are symbolized by unit vectors, ˆi , ˆj , kˆ . A set of unit vectors that define a coordinate system are called a basis set. Components Two dimensional: Vx V cos V y V sin V Vx2 V y2 tan 1 Vy Vx a aˆ a A unit vector is a vector of magnitude 1. E.g., , where is the a magnitude of the vector a . Often, the magnitude of a vector is indicated by the letter without the arrow on top: a a . A A 11 An arbitrary vector can be written as a sum of the basis set unit vectors. A Axˆi Ay ˆj Az kˆ Direction cosines Ax cos A Ay cos A Az cos A emweb.unl.edu 12 Adding vectors The sum of two vectors is also a vector. A B C Drawn to scale. Ax Bx C x Ay By C y Az Bz C z C C xˆi C y ˆj C z kˆ A vector may be multiplied by a scalar. This affects the magnitude of the vector, but does not affect its direction. The exception to this rule is multiplication by –1. That leaves the magnitude unchanged, but reverses the direction. 13 Vector Products scalar (or dot) product—result is a scalar; the operation is symbolized by a dot. A B AB cos Ax Bx Ay By Az Bz The angle is the angle from A to B . Note: A B B A and A B C A B A C . Vector (or cross) product—result is a another vector; the operation is symbolized by a cross, . A B C C C AB sin , direction perpendicular to both A and B according to the right-hand-rule. [Use the three-finger version.] iˆ C A B Ax Bx ˆj Ay By kˆ Az iˆAy Bz By Az ˆj Ax Bz Bx Az kˆAx By Bx Ay Bz 14 Kinematics Simply describe the motion of an object. 15 Motion Variables r The displacement vector, , The average velocity during the time r interval t is defined to be v . t It’s the time-rate-of-change in the displacement. In terms of vector components, we’d write x y z . v x , v y , v z points from the origin to the present location of the particle. r If a particle is at 1 at time t t1 r and at 2 at some later time t t ,2 then we say the change in displacement is r r2 r1 . Likewise, the elapsed time or time interval is t t 2 t1 . Similarly, the average acceleration is a v v2 v1 t t2 t1 t . t The instantaneous velocity is defined to be r dr v lim t dt . t 0 . The instantaneous acceleration is v dv a lim t dt t 0 t t, r , v , a 16 Constant Acceleration vx vx 2 vx1 a x t t 2 t1 vx 2 vx1 a x t 2 t1 t2 t1 x2 x1 2 vx x2 x1 v x 2 v x1 v x 2 v x1 v x22 v x21 ax x2 x1 2 x2 x1 2 v x 2 v x1 v x22 v x21 2a x x2 x1 vx x x2 x1 t t2 t1 x2 x1 vx t2 t1 v x vx 2 vx1 x2 x1 2 and vx 2 vx1 t 2 x2 x1 t 2 t1 2 t1 x1 If a x is constant, then also vx vx 2 vx1 . 2 v x1 a x t 2 t1 v x1 t2 t1 2 1 2 x2 x1 v x1 t 2 t1 a x t 2 t1 2 Commonly, t1 0, t2 t, x1 xo , x2 x, vx1 vxo , vx 2 vx We have four equations that each relate three of the motion variables. 17 Space-time Mathematically we can treat time and space on the same footing. The displacement vector in space-time has 4 components. r ct , x , y , z The scaling factor c is needed to make the units of all 4 components the same, e.g., meters. The geometry of space-time is not Euclidian, but is non-Euclidian. Therefore, ˆi ˆi ˆj ˆj kˆ kˆ 1; ˆt ˆt 1 18 Example: a train traveling on a straight and level track starting from rest; ends at rest. time, t (seconds) acceleration, ax (m/s2) 0 - 10 2 10 – 40 0 40 - ? -4 What is the total displacement? Segment 1: We are given the acceleration, elapsed time and initial velocity — vxo = 0 m/s. x x o v xo t 1 a x t 2 2 1 2 x1 0 m 0 m/s 10 s 2 m/s 2 10 s 100 m 2 Segment 2: To find the total displacement at the end of the second segment, we need the velocity component at the end of the first segment. vx1 vxo ax t 0 m/s 2 m/s2 10 s 20 m/s vx 2 vx1 20 m/s since ax 0 1 1 2 x2 x1 v x1t a x t 2 100 m 20 m/s 30 s 0 m/s 2 30 s 700 m 2 2 Segment 3: For this segment, we know x2, vx2, vx3, and ax, but not t . v x23 v x22 2 a x x3 x 2 2 v x23 v x22 0 20 m /s x3 x 2 700 m 750 m 2 ax 2 4 m /s2 19 Example: A hot air balloon is rising at a constant speed of 5 m/s. At time zero, the balloon is at a height of 20 m above the ground and the passenger in the balloon drops a sandbag, which falls freely straight downward. We observe that ay 9.8 m/s2 . What are the height of the sandbag and its velocity as functions of time? 1 m 1 m 2 2 y y o v yo t a y t 20m 5 t 9.8 2 t 2 s 2 s m m v y v yo a y t 5 9.8 2 t s s Free Fall! What is the y-component of the sandbag’s velocity when it hits the ground? 2 v y2 v yo 2 a y y yo 2 v y2 v yo 2a y y y0 5 m /s 2 9.8 m /s2 0 m 20 m 417 m 2 / s 2 2 v y 417 m 2 / s 2 20.4 m /s. (We know it' s going down.) 20 How long does that take? v y v yo a y t t v y v yo ay 20.4 m/s 5 m/s 2.59 s 2 9.8 m/s Alternative solution for the elapsed time: 1 y yo v yo t a y t 2 2 0 m 20 m 5 m/s t t 1 9.8 m/s2 t 2 2 5 m/s 25 m 2 / s 2 4 4.9 m/s2 20 m 9.8 m/s2 5 20.42 s 2.59 s or - 1.57 s 9.8 21 Projectile Motion Constant acceleration, in two dimensions. m ˆ ˆ a 0i 9.8 2 j s 22 Vector equations: 1 2 r ro vo t a t 2 Component equations: v vo a t v xo vo coso v yo vo sin o x xo vxot 1 2 1 y yo v yot a y t yo v yot ( g ) t 2 2 2 v x v xo a x t v xo Notice: The ycomponent of a is ay = -g. v y v yo a y t v yo ( g ) t 23 Example: How long does it take to reach maximum height, ymax? At maximum height, vy = 0 m/s v y v yo a y t t v y v yo ay 0 m /s 40 m /s sin53o 3.26 s 9.8 m /s What is the maximum height? v y2 v yo2 2a y y yo y yo v y2 v yo2 2a y 0 m /s 40 m/s sin53 0m 2 9.8 m/s 2 2 2 o 2 52 .1 m 24 When is the projectile at y = 25m? 1 2 y yo v yot a y t 2 1 2 a y t v yot yo y 0 2 2v yo 2 yo y 2 t t 0 ay ay t 2 6.52 s t 5.10 s 2 0 s 2 t (6.52 s) (6.52 s) 2 4 5.1 s 2 2 0.910s and 5.61s 25 What are the velocity components then, at t = 0.910 s and t = 5.61 s? v x v xo a x t vo cos o 40 m /s cos53o 24.1 m /s v y v yo a y t vo sin o ( g ) t 40 m /s sin 53o (9.8 m /s2 ) t 31.9 m /s 9.8 m /s2 t time (s) 0.910 5.61 velocity (m/s) v x 24.1 v y 23.0 vx 24.1 v y 23.0 26 Example: How far does the object travel in the x-direction? 1 2 x xo vox t a x t 2 We need to know the elapsed time, t. The total elapsed time is the time it takes to go up plus the time it takes to come down. Previously, we found that the time to reach maximum height was t = 3.26 s. 1 The total time, then, is 2x3.26s = 6.52 s. [Verify with 0 ymax 0 t 9.8 t 2 .] 2 1 2 x xo vox t a x t 0 vox t 0 2 m x vox t 24.1 6.52s 157m s 27 Example: What are the velocity & position components at t = 3 seconds? v x v xo a x t 20 m/s 0 m/s2 3s 20 m/s v y v yo a y t 0 m/s (9.8 m/s2 ) 3s 29.4 m/s x xo v xot 0 m 20 m/s 3s 60m 1 2 1 2 y yo v yot a y t 0 m 0 m/s 3s 9.8 m/s2 3s 44.1 m 2 2 28 29 Uniform Circular Motion Curvilinear motion – not in a straight line. Envision an object having, at the moment, a velocity subject to an acceleration a . . v , and We might decompose the acceleration into components parallel to and perpendicular to the velocity vector. The parallel acceleration component affects the speed of the object, while the perpendicular component affects the direction of the velocity vector, but does not change its magnitude. At any instant, the velocity to tangent to the curve. 30 Circular motion Uniform circular motion refers to motion on a circular path at constant speed. While the magnitude of the velocity is constant, the velocity vector is not constant. The same is true of the acceleration vector—its magnitude is constant but its direction is not. However, the acceleration is always directed toward the center of the circular path. The component of acceleration parallel to the velocity vector is zero. The acceleration component directed toward the center of the circle is called the centripetal acceleration. 31 Let the origin be at the center of the circle, as shown. 32 Consider two successive displacement and velocity vectors. r r2 r1 v v2 v1 v v1 v2 By the definition of uniform circular motion, r r1 r2 arad v t In the limit as t 0 , v v & r r v r . Both are isosceles triangles, with the same angle. v r v 1 vr vr v 2 arad t t r rt r . The centripetal or radial acceleration is always v 2 on a circular arc of radius r. r 33 Second Dynamics Newton’s “Laws” Energy Momentum Conservation 34 Dynamics Relationships among Motion and Force and Energy. 35 Newton’s “Laws” of Motion “An object in uniform motion remains in uniform motion unless it is acted upon by an external force.” [In this context, uniform motion means moving with constant velocity.] “The change in motion of an object is directly proportional to the net external force.” . “For every action, there is an equal and opposite reaction.” 36 37 A force is an external influence that changes the motion of an object, or of a system of objects. We find that there are four fundamental forces in nature, gravity, electromagnetic force, and the strong and weak nuclear forces. All particles of matter interact through one or more of these four fundamental forces. All other types of forces that we might give a name to are some manifestation of one of the fundamental forces. Dimensions of force are F M 2L T The SI unit of force is the Newton (N). 1 N = kg m/s2 38 Fundamental concepts: i) Space and time ii) Matter and energy Macroscopic objects—collections of many atoms & molecules. Molecules—combinations of several atoms; chemical substance. Atoms—combinations of protons, neutrons & electrons; chemical element. Subatomic particles—protons, neutrons, electrons, et al. A particle is an idealized object that has no shape or internal structure. Any object may be treated as if it were a particle depending on the context. 39 Two of the attributes of matter are i) resists changes in its motion—matter has inertia, and ii) a force acts between any two pieces of matter—material objects or particles exert forces on each other. The quantitative measure of inertia is called the inertial mass of a particle. Imagine two particles exerting equal and opposite forces on each other. We observe their accelerations. a2 m1 a1 m2 dp F dt We write Newton’s 2nd “Law” in mathematical form: The quantity p mv 2 1 v c2 called the m om entum . 40 Classical assumptions: i) time is independent of space and is absolute. ii) 3-d space is Euclidian—”flat.” Unless a particle is moving at a very great speed, its momentum is approximately p mv Further, if the particle’s mass is unchanging, then dp dv F m dt dt 41 42 Equilibrium Should the vector sum of all forces acting on an object be equal to zero, then dp 0 dt and the object is said to be in Static equilibrium Dynamic equilibrium p0 p0 43 Isolated body diagram(s) An isolated body diagram is a sketch of the object only, with arrows indicating each force acting only on that object. 44 Action & Reaction Force is an interaction between two material objects. E.g., there is a gravitational interaction between the Earth and the Moon. They exert forces on each other of equal magnitudes but opposite directions. N and N form an action- reaction pair. W and N are not an action- reaction pair. N N N W 0 45 Newton’s Universal “Law” of Gravitation Any two objects exert gravitational forces on each other, equal in magnitude and opposite in direction. M 1M 2 Fg 12 G rˆ 2 r12 Take care with the directions. The unit vector rˆ points from M1 to M2. The gravitational force on M2 is in the rˆ direction, toward M1. 46 Gravitational Field Let’s say M1 is at the origin of coordinates. The presence of M1 gives rise to a gravitational field that extends outward into space. M1 g G 2 rˆ r An object of mass M2 located at r experiences a gravitational force. M 1M 2 Fg M 2 g G rˆ 2 r In the context of the 2nd “Law” The acceleration due to gravity is M 1M 2 ˆ Fg G r M2g 2 r M1 ar a g G 2 r 47 Near the Earth’s surface, a g G M1 ME m G 9 . 8 r122 RE2 s2 Near the surface of another body, such as the Moon or Mars, the acceleration due to gravity is different, not 9.8 m/s2. 48 Weight Weight is the term we use to refer to the force of gravity near the Earth’s surface, or near a planetary body’s surface, or near a moon’s surface, etc. Fg mg We do not measure weight of an object directly. Instead, we place the object on a scale. The number we read off of the scale is actually the contact force exerted upward by the scale on the object. If the object is in equilibrium, then we infer that the weight has the same magnitude. F y N Fg 0 N Fg 49 Suppose the object is not in equilibrium. F y N Fg m A N m A Fg N m A m g Suppose A = -g. Then N = 0. The object is in free fall, but not weightless. The term weightless is a misnomer. 50 Friction Friction always opposes the motion of an object, or what’s called the object’s impending motion. decompose Kinetic or Dynam icFriction Ff K N Static Friction Ff S N Fx Wx N x Ffx max F cos30o 0 0 N max Fy Wy N y Ffy may F sin 30o mg N 0 0 N mg F sin 30 o a x 1 F cos 30 o mg F sin 30 o m 51 Example: Two objects N1x W1x Ff 1x Ma1x 0 0 Ff 1 Ma1x N1y W1y Ff 1y Ma1y N1 Mg 0 0 Ff 1 1 N1 Ff 2 2 N 2 Fx N2 x W2 x Ff 2 x R f 1x R1x ma2 x Fx 0 0 Ff 2 Ff 1 0 ma2 x Fy N 2 y W2 y Ff 2 y R f 1y R1y ma2 y Fy N 2 mg 0 0 N1 0 52 Cords & Tension T1 mg T W ma Ty Wy may T mg may 0 T mg 2T2 T1 W2 Ma2 2T2 T1 Mg 0 T2 1 T1 Mg 1 m M g 2 2 The ideal cord is massless, non-stretchable and perfectly flexible. This means that it can sustain tension, but cannot resist compression along its length. It means also that the tension in the cord is the same throughout its entire length (as long as we ignore friction). 53 Example: F W T T left right ma Wx Tleftx Trightx 0 0 T cos T cos 0 Wy Tlefty Trighty No help. mg 0 mg T sin T sin 0 T 2 sin 54 Pulleys Apply Newton’s 2nd “Law” to the pulley and to the hanging mass. W p T2 T2 T1 0 T2 T1 W p 2 T1 Mg 0 T1 Mg 55 Case Studies in Applying Newton’s 2nd “Law.” Circular motion Inclined plane Restoring forces—spring & pendulum Systems of objects 56 57 Circular Motion ferris wheel Tr Wr m ar Tt Wt m at v2 Tr m g cos m r Tt m gsin m at 180o 0o v2 Tr m g( 1 ) m r 58 driving ‘round a curve v2 N x Wx F fx m ax N sin 0 F f cos m ar N sin s N cos m R example mg s 0, R 50m, v 13.4 m/s cos s sin mg v2 sin 0 m cos 0 R 2 2 2 v sin v 13 . 4 m /s m cos gR 9.8 m /s2 50 m R t an1 0.366 N y Wy Ffy may N cos mg Ff sin 0 N mg sin s cos cos s sin 20o 59 circular orbit Universal “Law” of Gravitation Fg G Mm r2 F m a Newton’s 2nd “law” Fg m ar Mm v2 G 2 m r r Orbital speed v GM r 60 Inclined Plane N W ma N x Wx max 0 Wx max N y Wy may N Wy may 0 Wx W sin m gsin Wy W cos m g cos N W F ma N x Wx Fx max 0 Wx F max mgsin F max N y Wy Fy may N Wy 0 may N mgcos 0 61 Restoring Force Hooke’s “Law” linear restoring force Fs k o Fs k x xo F ma Fs W ma Fs m g 0 k o m g 0 mg o k Typically, we place the origin at the resting length of the spring. 62 Pendulum F m a T W ma Radial and tangential components. vt2 vt2 Tr Wr m ar T W cos m T m g cos m r Tt Wt mat 0 mgsin mat at g sin 63 System of Objects T W1 m1a1 T m1g m1a1 T N W2 m2a2 Tx N x Wx m2a2 x T 0 W2 sin m2a2 x Ty N y Wy m2a2 y 0 N W2 cos 0 a1 a2 x T m1 g a1 a1 a2 x a1 1 W2 sin T m2 Newton’s “Laws” apply to each object as well as to the system as a whole. 1 W2 sin m1 g a1 a1 1 m2 g sin m1 g m2 m1 m2 64 “Atwood’s Machine” T2 W2 m2a2 T2 m2 g m2a2 T1 W1 m1a1 T1 m1g m1a1 a1 a2 a T1 T2 T T m1 g m1a T m 2 g m2 a a T m1a m1 g m1 m2 m1 g m2 m1 m2 m1 2m1m2 g m1 g g m2 m1 m2 m1 65 Reference Frames The position vector of the point P is written down with respect to two different reference frames. r R r r ut r r r ut The motion variables, as measured by observers in different frames. dr dr dR v v u dt dt dt du 0 dt a a 66 r r ut v v u du a a dt Apply Newton’s 2nd “Law” in the two reference frames. F m a T W ma ma F T W ma 0 x’: y’: T sin W sin 0 x: T sin 0 max 0 T cos W cos 0 y: T cos W 0 du dt du W W m dt ax Now, bring them together: The perceived weight is different. 67 An Accelerated Reference Frame 2 du d R 2 A dt dt In the elevator Outside the elevator Fc mg 0 Fc mg mA mg m g m A The reading on the scale is F’c. The observer in the elevator interprets that as his/her weight. 68 Energy Work Kinetic Energy Potential Energy 69 Physical Work W F ds 2 1 x2 y2 z2 x1 y1 z1 W Fx dx Fy dy Fz dz F L M L T 2 dimensions 2 The SI unit of work is the Joule (J). m2 1 J 1 kg 2 s 70 constant force W F s Fs cos W Ff s Ff s cos180o Ff s 71 dW F ds Fx dx force varies with position Wnet Fx x x2 W F x dx x1 x2 k 2 k 2 k 2 W Fdx kxdx x x1 x 2 2 2 x1 2 x1 x1 x2 x2 72 Power Power is the rate at which work is done. If W is the work done during an elapsed time, t , then the average power during that interval is The SI unit for power is the Watt: 1W 1 J s W P F v t . A kilowatt is 1000 Watts. The electric bills often mention kilowatt-hours. That’s one kilowatt times one hour = 3.6x106 Joules. Imagine a locomotive engine dragging a train along a straight track at a constant speed of 20 m/sec. Let’s say the engine exerts a force of 105 N and pulls the train 100 m. The locomotive engine expends P F v Fv 105 N 20m / sec 2 106W or P F d / t 105 N 100m / 5 sec 2 106 W 73 Kinetic Energy An increment of work is done during an incremental displacement. We assume that the applied force is constant during the incremental displacement. vx vx v2 x v1x 1 1 1 W Fx x m x m t m v22x v12x mv 22x mv12x t t 2 2 2 2 Work-Energy Theorem Kinetic energy W 1 2 1 mv 2 x mv12x K 2 2 1 2 K mv 2 74 Examples: Wg Fg r Fgx x Fgy y 0 mgy Wg mg y K v 1 2 1 2 mv mv o 2 2 2 1 2 m g y m vo m 2 Keep in mind: y y yo Energy is a scalar; it has no directional information. 75 Every force acting does some work. The block slides up the Incline from xo to x. Wg Mg x xo sin W f F f x xo K N x xo WN N x xo cos90o 0 WT T x xo cos0 o T x xo W Wg W f WN WT W K 1 2 1 2 mv mv o 2 2 76 Potential Energy Conservative forces are those for which the work done during a displacement is independent of the path followed. Call the work done by a conservative force, WC. W F ds 2 1 The potential energy function is defined thusly: WC U x2 y2 z2 x1 y1 z1 WC Fx dx Fy dy Fz dz For x-components To derive the potential energy function for a specified force, we evaluate the work. x2 x2 x1 x1 U 2 Fx dx U 1 Fx dx x2 WC Fx dx U U 2 U 1 x1 Because we are interested in potential energy changes, we can set the zero of potential energy for convenience. 77 Conversely, we can derive the force components from the potential energy function: U x , y , z x Fy U x , y , z y Fz U x , y , z z F Fxˆi Fy ˆj Fz kˆ Fx For a spring: For uniform gravity: Gradient operator F U x , y , z ˆi U ˆj U kˆ U x y z Fx d 1 2 kx kx dx 2 Fy d m gy m g dy 78 WC U Spring: 1 2 1 2 1 2 WC Fx dx F cos180 dx kxdx kx k 0 kx 2 2 2 0 0 0 x x x 0 1 2 U s kx 2 79 Gravity: y2 y2 y1 y1 WC Fy dy mgdy mgy 2 mgy 1 U 2 U1 U g mgy 80 Mechanical Energy E K U Wtotal Wother Wc K Wtotal Wother U K Wother K U E If only conservative external forces are acting, then the total mechanical energy of a system is conserved. E Wother 0 81 Gravity and spring restoring forces are conservative. Friction is non-conservative. Let us say that a number of forces do work on an object. Wtotal Wother WC K U I K U I 0 E I 0 The I is the change in internal energy of the object. Typically it is manifested as an increase in temperature. E.g., friction causes a decrease in mechanical energy & an increase in internal energy. 82 Near the Earth’s surface, K 2 U 2 K1 U 1 1 2 1 2 m v2 m gy2 m v1 m gy1 2 2 Spring K 2 U 2 K1 U 1 1 2 1 2 1 2 1 2 m v2 kx2 m v1 kx1 2 2 2 2 83 Recall the “Law” of Gravitation Gm1 m2 Fg r2 11 G 6.67 10 r 12 m1 m2 F12 Fg N m2 kg 2 Gm1m2 r122 GM earth m GM earth m g g 2 2 Rearth Rearth 84 Gravitational potential energy r2 Gm1m2 Gm1m2 Gm1m2 U (r ) Wg Fg dr dr 2 r r1 r2 r1 r1 Gm1m2 U (r ) r Notice that we are setting U = 0 at r = ∞. r2 Let’s take a closer look at our (+/-) signs. The force is in the –r direction. The displacement is in the +r direction, and r is always (+). The anti-derivative gives a (-) sign. The W is negative change in U. Hence, U is (-). 85 Escape from Earth Define “escape” 1 Gm1m2 1 Gm1m2 m2 v22 m2 v12 2 r2 2 r1 00 vo2 r1 Rearth , v1 vo , v2 0, & r2 1 2 GM earth m m vo 2 Rearth 2GM earth Rearth Nm 2 24 2 6.67 10 5 . 97 10 kg 2 kg 4 m 4 m iles 1 . 12 10 2 . 49 10 6.38106 m s hour 11 vo 2GM earth Rearth Notice the direction does not matter! (Assuming the direction is not straight down.) Nor does the mass m. Compare the escape velocity with the orbital speed of the space shuttle in a circular orbit of altitude 325 km. It’s about 7.6x103 m/s (17000 mph). 86 Impulse & Momentum Go back to the definition of acceleration. Impulse v x a x t vx mt a x mt t Fx t m vx 2 m vx1 J x Fx t p x In general J p Fdt Conservation of momentum,if F 0, thenp 0. 87 A ball bounces straight off a wall. p x mvx 2 mvx1 Let’s say that m = 0.5 kg, vx1 = 40 m/s, and vx2 = -20 m/s. Then the change in momentum is m m kg m p x mvx 2 mvx1 0.5 kg - 20 40 30 s s s This is the impulse on the ball! The ball exerts an equal and opposite impulse on the wall. If the impact lasts t 0.01 s , then the average force on the wall is Fx 1 1 kg m p x 30 3000 N t 0.01 s s 88 One dimensional elastic collision p1 p2 m Av A1 mB vB1 m Av A2 mB vB 2 K1 K 2 1 1 1 1 m Av A21 mB vB21 m Av A2 2 mB vB2 2 2 2 2 2 Inelastic collision—kinetic energy is not conserved. K2 K1 I 89 Two dimensional elastic collision p1 p2 m Av Ax1 mB vBx1 m Av Ax 2 mB vBx2 m Av Ay1 mB vBy1 m Av Ay 2 mB vBy 2 K1 K 2 1 1 1 1 m Av A21 mB vB21 mAv A2 2 mB vB2 2 2 2 2 2 90 “Perfectly” inelastic collision p1 p2 m Av Ax1 mB vBx1 m A mB vx 2 m Av Ay1 mB vBy1 m A mB v y 2 p1 p2 mA mB v1 mAvA2 mB vB 2 91 92 Center of Mass rcm m r ii m i m v d ii vcm rcm dt mi vcm mi mi vi P The total momentum of a system of particles is equal to the total mass of the system times the velocity of the center of mass. P Mvcm If no net external force acts on any part of the system, then P is constant, and so is v . cm The individual parts of the system may exert forces on each other, but those do not affect the motion of the center of mass. 93 On the other hand, if one or more external forces acts on the system, then vcm is not constant. The sum of all forces acting on all parts of the system is F F F m a M a ext int i i cm Because of Newton’s 3rd “Law”, Fint . 0 Consequently, the center of mass of a system of particles moves like a particle of mass M. . Fext Macm 94 Rockets & Rain Drops Suppose the total mass of a moving object is not constant. Say the net external force acting on an object (such as a rocket or a rain drop) is Fext . Assume that during a short time interval, t , the Fext is approximately constant. Then the impulse delivered to the mass, m, is Fextt p Further suppose that during that interval t p mv2 mv2 m mv1 p mv2 v1 mv2 v1 v v2 v1 V v2 v2 , the . mass changes by an amount m . We may as well just let m + Δm be m at this point. Fextt m mv mV p mv1 v v1 m V v p mv mv mV p mv mv mV m mv mV Fextt p mv mV v m dv dm Fext m V m V t t dt dt 95 Recap: v V is the velocity of the object (rocket or rain drop), is the velocity of the m relative to the object, and dm is the absolute value of the time rate of change in the mass of the object. dt Actually, we have to be careful of the directions of things. As derived here, if m is leaving the object, then the object is losing mass and v is in the opposite direction as V . . Consider a rocket in the absence of gravity or any other external force. dv dm 0m V dt dt dm dv m V dt dt dv V dm dt m dt In real life, there is always gravity, friction, air resistance, etc. 96 Third Rotation Vibration Wave Propagation 97 Rotation Rigid body Arc length, radians There are 2 radians in 360o. s s r or in radians r name definition angular displacement, angular velocity component, A rigid body is one in which all the rij are constant. angular acceleration, d dt d d 2 2 dt dt 98 Equations of rotational motion 1 2 o o t t 2 o t 2 o2 2 o o s r 2 v t r a t r vt2 ar r 2 r Example: t 3 s o 0 234 (radians) radians 108 s constant o unknown o 2 60 radians radians 20 t t t 3 s2 s2 234 0 radians radians 78 t 3 s s 99 Moment of Inertia I mi ri2 I r 2 Volume dm r dxdydz 2 Volume The dimensions of moment of inertia are [M][L]2. 100 101 Rotational Dynamics Angular kinetic energy & angular momentum Kr 1 1 1 mi vi2 mi ri 2 2 I 2 2 2 2 ˆi Lrp x m vx ˆj kˆ y m vy z m vz torque dL d r p dp dr dp r p r 0 r F dt dt dt dt dt dL r F dt L Li ri pi i mii mii2 I Rigid body: dL d I I dt dt 102 103 Unwinding Consider an ideal cord wound around a solid cylinder of radius R = 0.5 m and mass M = 10 kg. The cylinder is set on a horizontal axis and mass of m = 4 kg is hung on the free end of the cord. What’s the torque experienced by the cylinder and what’s the acceleration downward of the mass, m? F y ma y T mg ma Fy 0 a N Mg T 0 I T R I a R a R TR 2 TR 2 2T I TR 0 a TR R I I M 0.5MR 2 T m( a g ) 0 T m ( g a ) a R 2 m g a 1 m a g 4.36 2 M M s 1 2m 1 T m g g 21.8 N M 1 2m a 1 a 1 m I I MR 2 10kg 0.5m(4.36 2 ) 10.9 Nm R 2 R 2 s 104 Roll down an incline F ma Mg sin f Mg sin N Max N Mg cos 0 N Mg cos a x g sin cos I (We'll computetorquesabout thecenterof mass.) N g f I 0 0 f R MgR cos 5 2 g cos R 2 MR 2 5 Now, if the ball is to roll without slipping, what must be true? The friction must be just right such that a x R 1 1 2 2 Kinetic energy: K 2 Mv 2 I 2 No slipping. . . . K v 1 1 1 12 Mv 2 MR 2 Mv 2 2 25 R 2 5 105 Gyroscope Pulley Lo Loiˆ L riˆ Mgkˆ 0iˆ Nkˆ t L rMgˆj t 106 107 108 Static Equilibrium A uniform beam of length r = 4 m and mass 10 kg supports a 20 kg mass as shown. The beam in turn is supported by a taut wire. What’s the tension in the wire? The beam is in static equilibrium. Therefore, the sum of forces on the beam is zero, and the sum of torques exerted by those forces is zero. Fb is the weight of the beam itself. Fx T sin 60o 0 T 4m sin 67o 20kg g 4m sin 53o 10kg g Fy T y Fgy Fby 0 Fy T cos60 20kg g 10kg g 0 o T g b 0 Fx Tx Fgx Fbx 0 4m sin 53o 0 2 3.68m T 626.13Nm 156.53Nm 0 T 213N Fx T sin 60o 213N 0.866 184N Fy 30kg g T cos60o 294N 213N 0.5 188N 109 Oscillation yt A sin2ft A is the amplitude, the maximum displacement either side of equilibrium. f is the frequency of oscillation, in cycles/second (Hz). is a phase factor, which depends on the initial y at t = 0. T 1 f 2 f radians/s y A sin t dy v A cos t dt dv a 2 A sin t dt For an object bouncing on a spring: F ma ky m 2 y k m 110 Mechanical energy of an oscillator Again, think of an object oscillating horizontally on a spring. The mechanical energy is 1 2 1 2 m v ky 2 2 1 1 1 1 2 E m 0 kA2 mA k 0 2 2 2 2 1 1 1 2 E m v2 ky 2 mA 2 2 2 E Notice E is proportional to the square of the amplitude. 111 pendulum m g sin I m 2 m g sin m 2 g sin Assume the oscillations are small sin Physical Pendulum I m g sin I d 2 m g 2 I dt m g I g d 2 g 2 dt g 112 113 Wave Propagation A wave is a disturbance in an elastic medium which travels, or propagates through the medium. The wave is intangible. The medium itself does not travel, but only oscillates back and forth. So there is not a net transport of matter from place to place. However, a wave transports energy from place to place, through the medium. Waves come in many forms, all with certain common properties. There are waves in a plucked string, seismic waves, sound waves, electromagnetic waves. These are different sorts of disturbances propagating in different sorts of media. In this course, we will consider the common properties. 114 115 Wave Motion x t yx, t Af T y is the displacement from equilibrium at position x and at time t. f is an unspecified function. i) wave speed is a property of the medium. ii) shape of the wave pulse is unchanged as it travels iii) two or more wave pulses that exist at the same place & time in a medium add—superimpose. 116 Harmonic wave – a wave of a particular shape that repeats itself. It’s periodic. x t x y x, t A cos 2 A cos 2f t T c k 2 2f yx, t A coskx t Each point in the medium (x) is displaced from equilibrium (y). As time passes, the pattern is shifted by a phase factor ct ; the wave pattern moves through the medium. 117 superposition reflection “Standing waves” yx, t y1 x, t y 2 x, t yx, t A coskx t A coskx t yx, t 2 A sin kxsin t cosa b cosa cosb sin a sin b 118 Stretched string c F m L Only vibrations that “fit” in the length of the string will persist. This is an example of resonance. Every physical system has “natural” modes at which it will vibrate. The natural modes depend on the physical properties of the system: mass, elasticity, size. We saw this same phenomenon with the spring and the pendulum. 119 “Beats” In this case, two waves are traveling in the same direction, but with slightly different frequencies. y y1 y 2 y Acos2 f1t cos2 f 2 t f f2 f f2 y 2 A cos 2 1 t cos 2 1 t 2 2 120 Spectrum 2 y Ai cos x 2f i t i i 121 . Energy While the medium in which the wave propagates does not flow from one place to another, the wave disturbance nonetheless carries energy from one place to another. Each mass element, dm, of the medium executes simple harmonic motion. K is the restoring force constant. It’s related to the frequency by 1 1 K E Ky 2 dm v2 2 2 dm 1 1 E dm 2 A 2 cos2 kx t dm A2 2 sin 2 kx t 2 2 1 E dm 2 A 2 cos2 kx t sin 2 kx t 2 Over one cycle, the cosine-squared and sine-squared average to 1 2 . The total mass of the medium spanning one cycle (or one wavelength) is , where is the mass per unit length of the medium. E 2 f 2 2 f 2 2 2 f 2 A2 f 2 2 2 In terms of the wave speed, c, E 2 c f A .The energy flux is the power transported through the medium by the wave: P fE 2 2 c f 2 A2 The intensity is the power pr unit area through which the power is transported: I P 2 2 c f 2 A 2 a , were is the mass per unit volume. 122 Pressure waves--Sound Compression, or longitudinal waves. Medium oscillates parallel to direction of propagation. Pressure amplitude, yp. Speed of sound waves depends on density, pressure, temperature & elasticity of the medium. Doppler effect. . . . deciBels. . . 123 124 The “Laws” of Thermodynamics 125