These lecture notes cover 1.8, 1.9, 1.11, 1.12, and most of

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1 Part I: Introduction Reading assignent: Chapters 1 and 2 These lecture notes cover 1.8, 1.9, 1.11, 1.12, and ost of Change of Coordinates A k k j Let A be a vector with coponents (relative to Cartesian coordinate sstes O, O ): A = A î+a ĵ+a ˆk = A î +A ĵ +A ˆk = A. (1) i O O i j The sets of utuall orthogonal unit vectors (î, ĵ, ˆk) and (î, ĵ, ˆk ) are the bases for coordinate sstes O and O. Taking the dot product of A with each of the pried unit vectors, we get: A î = A = A (î î ) + A (ĵ î ) + A (ˆk î ) A ĵ = A = A (î ĵ ) + A (ĵ ĵ ) + A (ˆk ĵ ) A ˆk = A = A (î ˆk ) + A (ĵ ˆk ) + A (ˆk ˆk ) since î, ĵ, and ˆk are utuall orthogonal. In atri notation: A î î ĵ î ˆk î A A = î ĵ ĵ ĵ ˆk ĵ A A = T A, (2) A î ˆk ĵ ˆk ˆk ˆk A where A and A are the colun atri representations of the coponents of A ( A ) relative to the coordinate sstes O and O, and where T is the transforation atri. 1

2 Introduction 2 Note that while A = A (sae vector), A A (different coponents in the two coordinate sstes). The notation A is due to Dirac, and is referred to as the ket of A. Siilarl, a row atri representation of the coponents of A is the bra of A (where the a in bra is pronounced as in hat), written A = [A A A ]. In this notation, the bra-ket of A is: A A = [ ] A A A A A = A 2 + A 2 + A 2 = A A, A the inner product of A using the standard rules for atri ultiplication. The scalar products î î, î ĵ, etc., are the transforation coefficients, a.k.a. the directional cosines of the coordinate sstes O and O. Siilarl, we can take the dot products of A with the unpried unit vectors to get: A A = A î î ĵ î ˆk î A î ĵ ĵ ĵ ˆk ĵ A A = S A, (3) î ˆk ĵ ˆk ˆk ˆk A since î, ĵ, and ˆk are utuall orthogonal. Since the dot product coutes (e.g., î ĵ = ĵ î, etc.), S = T T, where T eans the transpose of T (coluns written as rows). Net, substituting equation (2) into equation (3) gives us: A = ST A ST = I, the identit atri. Siilarl, substituting equation (3) into equation (2) gives us TS = I, and the fact that ST = TS = I S = T 1.

3 Introduction 3 A atri whose inverse is also its transpose is called an orthogonal atri. Theore: The inner product of two vectors is independent of the coordinate sste chosen. Proof : Start b noting that A = ( A ) T = (T A ) T = ( A ) T T T = A T T using equation (2), the fact that the transpose of a ket is a bra, and that the transpose of a product is the product of the transposes in reversed order. Thus, A B = A T T T B = A T 1 T B = A B, proving the theore. In particular, A A = A A = A A = A 2, and the length of a vector does not depend upon the coordinate sste chosen to describe A; what should be an intuitivel obvious result. Eaple 1.1: Let O be O rotated about the ˆk ais b angle φ, and let O be O rotated about the ŷ ais b angle θ. Then, φ θ θ φ T 1 = î î = cosφ î ĵ = cos ( π 2 + φ) = sinφ î ˆk = î ˆk = 0 ĵ î = cos ( π 2 φ) = sinφ ĵ ĵ = cosφ ĵ ˆk = ĵ ˆk = 0 ˆk î = ˆk î = 0 ˆk ĵ = ˆk ĵ = 0 ˆk ˆk = ˆk ˆk = 1 cosφ sin φ 0 sinφ cosφ pivot

4 Introduction 4 Then, A = A A = T 1 A = A cosφ sinφ 0 A sin φ cosφ 0 A A For the second coordinate rotation, î î = cosθ ĵ î = 0 ˆk î = sinθ î ĵ = 0 ĵ ĵ = 1 ˆk ĵ = 0 î ˆk = sin θ ĵ ˆk = 0 ˆk ˆk = cosθ cosθ 0 sin θ T 2 = sin θ 0 cosθ Then, A = T 2 A = T 2 T 1 A, A cosθ cosφ cosθ sin φ sin θ A A = sinφ cosφ 0 A, A sin θ cosφ sin θ sinφ cosθ A after perforing the atri ultiplication. In general, after a succession of n rotations, A (n) = T n T n 1...T 2 T 1 A. Since atrices do not coute, T 2 T 1 T 1 T 2. Phsicall, this eans rotation about the - about about ais followed b rota- tion about the -ais is not the sae as rota- tion about the -ais fol- about about lowed b rotation about the -ais.

5 Introduction Vector derivatives Let A(t) = A (t)î + A (t)ĵ + A (t)ˆk be a vector epressed in ters of its Cartesian coordinates. Then, its derivative with respect to t is: d A dt 0 = da dt î + A dî dt + da dt ĵ + A dĵ dt + da dt ˆk dˆk + A dt = da dt î + da dt ĵ + da dt ˆk. 0 0 If t is tie, we often replace d dt with. Thus, d A dt = A = A î + A ĵ + A ˆk. If B is another vector and a a scalar, both t-dependent, one can show: d dt ( A + B) = d A dt + d B dt ; d dt (a A) = da dt A + a d A dt ; d dt ( A B) = d A dt B + A d B dt, d dt ( A B) = d A dt B + A d B dt, (order is uniportant); (order is iportant). Plane polar coordinates (2-D) j i θ r e θ e r The position vector, r, is given b: r = î + ĵ = rê r = ê r, where ê r is a standard notation for unit vectors. For Cartesian coordinates, î, ĵ, ˆk, is also standard,

6 Introduction 6 though we could have used ê, ê, ê just as well. Thus, v = r = ṙê r + r ê r. (1) Now, what is ê r? To aintain ê r a unit vector, dê r ê r and thus to ê θ. Therefore, dê r = (1) dθ ê θ ê r = θ ê θ. տ agnitude of êr de θ e θ dθ de r e r Substituting ê r into equation (1) gives: v = ṙê r }{{} radial + r θê θ. (2) }{{} transverse As for ê θ, note that dê θ and ê r are antiparallel, and we have, dê θ = dθ ê r ê θ = θ ê r. Thus, we can differentiate equation (2) to get: a = v = rê r + ṙ ê r + ṙ θê θ + r θê θ + r θ ê θ = rê r + ṙ θê θ + ṙ θê θ + r θê θ r θ 2 ê r = ( r r θ 2 ) ê }{{} r + (r θ + 2ṙ θ) ê }{{} θ. radial transverse The r θ 2 ter is the centripetal acceleration ( ve centre-seeking ), and the 2ṙ θ ter is a new ter, the so-called Coriolis acceleration. Clindrical coordinates (3-D) e eφ r = Rê R + ê ; (N.B.: ê = ˆk) v = Ṙê R + R φê φ + żê ; a = ( R R φ 2 )ê R + (R φ + 2Ṙ φ)ê φ + ê. k i r j φ Re R e e R

7 Introduction 7 These are essentiall polar coordinates (r R, θ φ) with a -ais added. Spherical coordinates (3-D) The triad ê θ, ê φ, ê r can be obtained b: 1. rotating î, ĵ, ˆk an angle φ about ˆk, î r cosφ sin φ 0 î θ ĵ = sinφ cosφ 0 ĵ ; k j ˆk ˆk i φ 2. then rotating î, ĵ, ˆk an angle θ about ĵ, î cosθ 0 sin θ î cosθ 0 sinθ cosφ sin φ 0 î ĵ = ĵ = sinφ cosφ 0 ĵ ˆk sinθ 0 cosθ ˆk sin θ 0 cosθ ˆk cosθ cosφ cosθ sin φ sin θ î ê θ = sinφ cosφ 0 ĵ = ê φ. sinθ cosφ sin θ sinφ cosθ ˆk ê r er e θ eφ ê r = sin θ cosφî + sin θ sin φĵ + cosθ ˆk; ê θ = cosθ cosφî + cosθ sin φĵ sin θ ˆk; ê φ = sin φî + cosφĵ. Fro these, we can find the tie-derivatives of the unit vectors: ê r = î( θ cosθ cosφ φ sin θ sin φ) + ĵ( θ cosθ sin φ + φ sinθ cosθ) ˆk θ sinθ = θ (î cosθ cosφ + ĵ cosθ sinφ ˆk sinθ) }{{} = θ ê θ + φ sin θ ê φ. Siilarl, ê θ + φ sinθ ( î sin φ + ĵ cosφ) }{{} ê φ ê θ = θ ê r + φ cosθ ê φ ; êφ = φ sin θ ê r φ cosθ ê θ.

8 Introduction 8 The last ter is proportional to ê θ and not ê φ, as given incorrectl in equation ( c) of the tet (a tpo new to edition 7). Thus, we can work out the kineatical quantities in spherical coordinates: r = rê r ; v = r = ṙê r + r θê θ + r φ sin θê φ ; a = ( r r φ 2 sin 2 θ r }{{ θ } 2 )ê r + (r θ r φ 2 sin θ cosθ + }{{}}{{} 2ṙ θ )ê θ centripetal centripetal Coriolis + (r φ sinθ + 2ṙ φ sin θ + 2r θ φ cosθ)ê }{{} φ. Coriolis Eercise: Derive v and a above. Eaple 1.2: Find the acceleration of an point P on the ri of the wheel, spinning about two aes as shown. With the origin O otionless at the centre, the coordinates of P are: (r, θ, φ) = (b, ω 1 t, ω 2 t). Thus, a P = ( bω 2 2 sin 2 (ω 1 t) bω 2 1)ê r bω 2 2 sin(ω 1 t) cos(ω 1 t)ê θ + 2bω 1 ω 2 cos(ω 1 t)ê φ, P ω 1 Q b O since ṙ = r = θ = φ = 0. At Q (top of the wheel) where θ = ω 1 t = n2π, a Q = bω 2 1êr + 2bω 1 ω 2 ê φ. ω2 1.3 Dnaics, and Newton s Three Laws of Motion In dnaics, we seek the kineatical quantities, r, v, and a, as a function of tie given the forces acting on an object. To this end, we have Newton s three laws of otion:

9 Introduction 9 1. With no eternal forces, an object oves with a constant velocit. 2. F = p, where F = Fet is the su of all eternal forces acting on a ass,, and where p = v is the linear oentu of. 3. For ever action, there is an equal and opposite reaction. All forces can be identified as a eber of an action-reaction pair. For constant ass, Newton s second law gives us a second order differential vector equation of the for: F( r, r, t) = d2 r dt 2, (1) which we solve to find r(t), and thus v = r and a = r. Conservation of linear oentu If asses, 1 and 2, with oenta p 1 and p 2, eperience no eternal forces, 0 = F = p = d dt ( p 1 + p 2 ) p 1 + p 2 = constant, where p is the total oentu of the sste. This is the law of conservation of oentu and is true regardless of an internal forces 1 and 2 a eert on each other. 1.4 Rectilinear otion Motion restricted to 1-D ( r ) is called rectilinear otion. The siplest case is F = ẍ = constant, where equation (1; 1.3) becoes: d 2 dt 2 = F = a = constant,

10 Introduction 10 which can be integrated easil: d 2 dt dt = adt d 2 dt = v(t) = v 0 + at, (2) where v 0 is the su of the constants of integration for both integrals. Integrating equation (2), we get: d dt dt = (v 0 + at)dt (t) = 0 + v 0 t at2, (3) where 0 is another constant of integration. Fro equation (2), t = (v v 0 )/a. Substitute this into equation (3) to get: 0 = v 0 v v 0 a + a 2 ( v v0 2a( 0 ) = 2v 0 v 2v v 2 2v 0 v + v0 2 = v 2 v0. 2 a ) 2 We thus obtain the failar equations of kineatics, valid for constant a: = 0 + v 0 t at2 ; v = v 0 + at; (4) v 2 = v a( 0 ). Free-bod Diagras A surprising aount of proble solving in phsics is algorithic. Students who struggle in undergraduate phsics are often those who fail to recognise, adopt, and properl appl these algoriths. Solving a dnaical proble is one such algorith. To appl Newton s second law properl, we ust identif all forces and their directions correctl. For this, we create a free-bod diagra: - siplif the sste as uch as possible (block dot; bea line);

11 Introduction 11 - draw a coordinate sste whose aes are aligned with a and/or the pluralit of forces; - draw all eternal forces as arrows pointing in their direction of action and anchored on the dot or where the act on the line; - include acceleration and/or velocit vectors off-dot. If the direction of F, a, or v is unknown, draw its coponents as arrows aligned with the positive coordinate aes. Label forces with their agnitudes, not as vectors. Thus, N for noral force, not N. Other forces: T for tension, f k for kinetic friction, f s for static friction, etc. If a force has a siple functional for (e.g., g, k, etc.), use this as a label. Wean ourself fro the high-school habit of labelling forces F N, F T, F g, etc. This convention will get ver awkward ver quickl. Eaple 1.3: What is the force on the hinge? - Represent bea with a line, since location of forces is iportant. - Anchor forces to the line where the act Mg acts at centre of ass, g and T at right end, H and H at left end. - Direction of hinge force unknown a priori draw two separate forces, H and H, pointing in the + direction of each ais. H H hinge Mg M α FBD α β T β g

12 Introduction 12 Eaple 1.4: A ass bounces up and down on a spring. - Spring force, F k = k, acts in - direction onl, but unknown if it acts F k k k g FBD a up or down align it with +-ais. - Label spring force with its functional for, k, including the. - include acceleration off-dot. Direction unknown align it with +-ais. Appling Newton s Second Law To write down Newton s Second Law fro the FBDs: - Break F = a into its coponents: F = a, etc. - Consider each coponent separatel: - forces on LHS, accelerations on RHS; - forces and accelerations + ais include as labelled; - forces and accelerations ais ultipl labels b 1. For eaple 1.3: / / F = H T cos(α + β) = 0 no acceleration F = H + T sin(α + β) g Mg = 0. H, H, and T all point in the + direction, and take the + sign. T, g, and Mg all point in the direction, and take the sign. Note we onl have two equations in three unknowns: H, H, and T. The third equation coes fro considering torques, which we ll see in Chapter 8.

13 Introduction 13 For eaple 1.4, there are no forces in the -direction. Thus, / F = k g = a Spring force points in + direction and is included as labelled, nael k. g points in the direction, and takes the sign. a points in the + direction and takes the + sign. Eaple 1.5: A block of ass slides down a plane inclined at an angle θ with a coefficient of kinetic friction µ k. a) Find a. Step 1: Free-bod Diagra. Identif relevant forces: g, N, and f k. - block reduced to a dot (C of M) - all forces anchored to the dot - forces point in direction of action, if known - forces labelled with agnitudes - a included off-dot - coordinate sste defined Step 2: Appl Newton s Second Law f k N θ θ g 0 v,a FBD θ a / / F = a g sin θ f k = a (5) F = a N g cosθ = 0 N = g cosθ (6) Note that N is not alwas g (as in this eaple)!! Step 3: Include phsical odels or constraints, if an: f k = µ k N (7)

14 Introduction 14 Step 4: Perfor algebra. Substitute equation (6) into equation (7): and then equation (8) into equation (5): f k = µ k g cosθ, (8) g(sin θ µ k cosθ) = a a = g(sin θ µ k cosθ). (9) µ k < tanθ, a > 0 ( speeds up as it goes downhill) Note that if µ k = tanθ, a = 0 ( oves with constant speed) µ k > tanθ, a < 0 ( slows down and eventuall stops) The angle θ k = tan 1 µ k is the angle of kinetic friction. It is the critical angle at which slides down the hill at constant speed. b) Find v after slides a distance l = 0, assuing θ > θ k and v 0 = 0. Since a is constant, we can use kineatics (equations 4): using equation (9). Thus, 0, starts fro rest v 2 = v a( 0) = 2g(sinθ µ k cosθ)l, v = 2gl(sinθ µ k cosθ), which akes sense onl if µ k < tanθ (and thus θ > θ k ). Aside: A reinder of the Coulob odel for static and kinetic friction Two solids in contact along a surface with a relative speed v rel eert a frictional force on each other. An epiricall deterined odel for this friction, valid for an surfaces (e.g., sooth table top) but not all (e.g., stick side of tape), is due to Coulob and given b:

15 Introduction 15 f = { fs µ s N, v rel = 0 (static friction); f k = µ k N, v rel > 0 (kinetic friction), (10) where N is the noral force between the objects, and where µ s and µ k are the coefficients of static and kinetic friction, respectivel. Direction of f s is opposite to the otion that surface would eperience in its absence. (Careful, this can be subtle! See car eaple below.) Direction of f k is opposite to the otion of the surface. In this odel, ver weak van der Waals forces between surface atos of two objects in contact cause surface drag when an eternal tangential force is applied. If v rel = 0, atos have tie to set up stronger bonds µ s > µ k. Note that f s µ s N, and not necessaril equal! Indeed, for a book sitting on a horiontal table top, f s = 0. (If it weren t, f s would be unbalanced and the book would accelerate!) Lift one end of the table slowl, and f s obligingl increases with the tilt to balance the coponent of gravit along the table top, keeping the book at rest. Once the tilt reaches a critical angle, θ s (angle of repose), f s is the aiu that surface can sustain, nael µ s N. It is onl when an object is on the verge of slipping that f s = µ s N. Beond θ s, f s no longer increases with tilt and gravit wins out, causing the book to slide. Mechanics θ

16 Introduction 16 Sliding eans v rel > 0 and the kinetic friction, f k = µ k N, takes over. Since µ k < µ s, the friction force drops suddenl fro f s, a, just before the book starts sliding, to f k, just after. As the table is tilted further, the noral force and thus f k decreases until, at θ = π/2, N = f k = 0. To quantif this eaple, consider the FBD first for v rel = 0: / g sin θ f s = 0; / N g cosθ = 0 f s = g sin θ and N = g cosθ. (11) f s,k N v a θ rel To find the angle of repose, set f s = f s, a = µ s N: g g sin θ s = µ s g cosθ s θ s = tan 1 µ s, analogous to the angle of kinetic friction, θ k, defined after equation (9). Net, consider the FBD for v rel > 0. Then, / g sinθ f k = a; / N g cosθ = 0 f k = µ k N = µ k g cosθ. (12) Cobining equations (11) and (12): f g f(θ) = { fs = g sin θ, θ θ s ; f k = µ k g cosθ, θ > θ s, µ sg µkg µkgcosθ gsinθ µsgcosθ plotted for µ s = (θ s = 30 ) and µ k = f (θ) θs = θ

17 Introduction 17 Eercise: Leave the table horiontal and, instead, appl a graduall increasing horiontal force, F, to the book. Plot (qualitativel) the frictional force, f, as a function of F applied. Eaple 1.6: A car accelerates fro rest without squealing the tires. What tpe of friction is between the tires and road, and what is its direction? a Solution: Tires that aren t squealed don t slip relative to the road, and the point of the tire in contact with the road reains at rest relative to the road even as the car oves forward. (Think about it: sliding tires leave skid arks which happens onl if the tires are squealed.) Thus, the friction is static friction. If the tires were to slip, the ecess spin would be clockwise and the botto of the tire would ove to the left (in diagra) relative to the road. This is the otion f s is resisting (not the otion of the car) f s points to the right. Indeed, the force eerted b the engine is not what accelerates the car, at least not directl. With no friction between the tires and the road, the wheels just spin and the car goes nowhere regardless of what the engine does. Since the engine applies its force to the ale (not the road!), it is an internal force to the car; onl eternal forces can cause acceleration. The force that accelerates the car is the friction force delivered b the road to the tires. Since the car accelerates to the right, f s ust point to the right, confiring our analsis above. f s

18 Introduction 18 Appling Newton s Third Law It is both siple and critical to identif action-reaction pairs correctl, et an people including professional phsicists will get it wrong. Action-reaction pairs satisf three siple rules: 1. alwas have equal agnitude, opposite direction 2. alwas act between two objects and are alwas of the sae tpe (e.g., gravit, noral, friction, etc.) 3. If A eerts action force on B, B eerts reaction force on A. Thus, both ebers of an action-reaction pair never appear on the sae FBD. All three rules can be applied siultaneousl b using the following neonic: If an action is described as: object A eerts a force of tpe X on object B, then the reaction is: object B eerts a force of (sae) tpe X on A. N g Eaple 1.7: An object is at rest on a table. If the action-force is the object s weight, what is the reaction force? Most students (and soe phsics professors!) will sa the noral force of the table acting upward on. Wrong!! - noral force is a residual of electroagnetic forces; weight is a gravitational force not the sae tpe! (violates rule 2) - both g and N act on the sae object! (violates rule 3)

19 Introduction 19 - of the three rules, the N g pair satisfies onl rule 1 not enough! To find the reaction force of g, use the neonic: Action force is the Earth eerting a gravitational force g downward on. Reaction force is eerting a gravitational force g upward on the Earth. This action reaction pair satisfies all three rules. In particular, onl g down acts on and appears on its FBD. The reaction force g up would appear on the FBD of the Earth (if one were needed). So what is the reaction of the table eerting a noral force N upward on? Using the neonic: eerts a noral force N downward on the table. - N upward acts on and appears on its FBD. - N downward acts on the table and does not appear on the FBD for (but would on the table s FBD if we were analsing the forces on it). Eaple 1.8: A horiontal force F is applied to sitting on top of M. The coefficient of static friction between and M is µ s, while M sits on a frictionless surface. For what value of F will start to slip? N F M f M on N M on a M F f on M g action reaction pairs N on M Mg Fro Newton s 3 rd Law, f M on = f onm = f s ; N M on = N onm = N M.

20 Introduction 20 Then, fro the FBDs, we can write down Newton s 2 nd Law: Then, for : / F = a F f s = a; (1) / F = a N M g = 0; (2) for M: / F = Ma f s = Ma; (3) / F = Ma N Mg N M = 0 (4) M(1) (3) M(F f s ) f s = 0 F = M + M f s. For aiu force, F a, set f s = µ s N M = µ s g, fro (2). Thus, F a = M + M µ sg. Eercise: Find the aiu acceleration. Teaser: What is wrong with the following arguent? Since for ever action there is an equal and opposite reaction, action-reaction forces ust alwas cancel, leaving no net force on an object. Thus, objects can never accelerate. 1.5 Position-dependent forces Suppose F(, ẋ, t) = F(). Then equation (1, 1.3) can be written: F() = ẍ = dẋ dt = d dẋ } dt{{ d} using chain rule F() = v dv d. Now, integrate between 0 and where v( 0 ) = v 0 and v() = v: 0 F( )d = 0 v dv d d = v v 0 v dv = 1 2 (v2 v 2 0 ) = K K 0,

21 Introduction 21 where K = 1 2 v2 is the kinetic energ of 1. Now, define work to be: W 0 F( )d W = K K 0. (1) This is the work-kinetic theore: the net work done on a sste b the su of all eternal forces is equal to the change in kinetic energ of that sste. Because F = F(), equation (1) is in principle integrable. In this case, there eists a function, U(), (potential energ) that satisfies: 0 F( )d F() = du() (2) d du ( ) U = d = du = U() + U 0 d 0, (3) U 0 where U 0 is the value of U when K = K 0. Coparing equations (1) and (3), we get: U + U 0 = K K 0 E K + U = K 0 + U 0, where E is the echanical energ of the sste. Conservation of energ in rectilinear otion: When F = F(), the echanical energ is conserved. Thus, K = 1 2 v2 = E U() and, v = d dt Solve for dt and integrate to get: t t 0 dt = t t 0 = = ± 2 ( ) E U(). (4) d 0 v( ) = ± d ( ). (5) 0 2 E U( ) 1 Fowles and Cassida use what ost outside the UK would consider non-standard noenclature for an of their phsical variables. Thus, the use T for kinetic energ, not K, τ for period, not T, N for torque, not τ, R for noral force, not (alwas) N, etc. I shall not follow their convention.

22 Introduction 22 U( ) classicall forbidden regions This gives t() which, in principle, can be inverted to give (t). E a b Note that equations (4) and (5) ake sense onl if U() < E. In the diagra, a particle with echanical energ E ust reain between points a and b (turning points) and in the classicall allowed region. In Quantu Mechanics, this requireent can be violated. Eaple 1.9: a) Find v() for a siple haronic oscillator, F() = k. Equation (2) U() = du() d = F() = k 0 k d = 1 2 k(2 2 0 ), k v where 0 is the position of unstretched spring and where U( 0 ) = 0. If we take 0 = 0, then U() = 1 2 k2 and equation (4) becoes: 2 v() = ± (E 1 2 k2 ). At aiu etension, set = where K = 0 E = 1 2 k2. Thus, 2 k k v() = ± 2 (2 2 ) = ± (2 2 ). The ± eans could be oving in either direction. b) Find (t). Let t 0 = 0 and ω = k/, the frequenc of the oscillator. Then, equation (5) becoes: t = ± 0 d ω ( 2 2 ) = 1 ω cos 1 (/ ). }{{} integral tables = cos(ωt) (t) = cos(ωt),

23 Introduction 23 where the + option is chosen b requiring = + at t = 0. c) Find v(t) and a(t). v(t) = d(t) = ω sin(ωt) dt a(t) = dv(t) = ω 2 cos(ωt) = ω 2 (t). dt Note that a = ω 2 = k/ surprisingl recovered the initial force function. a = F = k, where we have not 1.6 Velocit-dependent forces Suppose F(, v, t) = F(v) and let v(t 0 ) = v 0. Then equation (1, 1.3) F(v) = dv dt = d dv } dt{{ d} chain rule = v dv d v dv v 0 F(v ) = v v dv v 0 F(v ) = t t 0 dt = t t 0 (1) 0 d = 0. (2) where (t 0 ) = 0. Doing the integral in equation (1) gives us t(v) (and thus, v(t)), while equation (2) gives us (v) (and thus, v()). Eaple 1.10: A siple odel for viscosit is F(v) = bv with b constant. a) Find (t) if this is the onl force acting on an object of ass. Fro equation (1): t t 0 = v v 0 dv bv = ( ) v b ln v 0 v(t) = v 0 e b (t t 0). (3) Integrate equation (3) over t to get: t t 0 v(t )dt = t t 0 d dt dt = 0 d = 0

24 Introduction 24 b) Find v(). = t Fro equation (2), v 0 e b (t t 0 ) dt = v 0 t 0 b e b (t t 0 ) t = v 0( b e (t t0) 1 ) t 0 b (t) = 0 + v 0( b e (t t0) 1 ). (4) b 0 = v v 0 v dv b v = b v v 0 dv = b (v v 0). v() = v 0 + b ( 0). Alternatel, we can substitute equation (3) into equation (4) to get: = 0 + v b v 0 b v() = v 0 + b ( 0), as before.