Chap. 4: Work and Energy. R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265

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1 Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics Group, B I T S Pilani September 7, 2012

2 Outline 1 Work Energy Theorem 2 Potential Energy 3 Non Conservative Force 4 Conservation Laws and Particle Collisions

3 The Fundamental Problem of Mechanics Given all forces, solve to find r(t). F = m a Problem! What we know is F(x) and not F(t). Solve half the problem! m d v dt = F( r)

4 Equation of Motion in One Dimension m d2 x = F(x) or mdv dt2 dt = F(x) Integrate with respect to x xb dv xb m x a dt dx = F(x)dx x a Change variable from x to t using differentials: ( ) dx dx = dt = vdt dt

5 Equation of Motion in One Dimension m d2 x = F(x) or mdv dt2 dt = F(x) Integrate with respect to x xb dv tb m x a dt dx=m dv t a dt vdt tb ( d 1 =m )dt dt 2 v2 = 12 t a mv2 t a = 1 2 mv2 b 1 2 mv2 a t b

6 Equation of Motion in One Dimension m d2 x = F(x) or mdv dt2 dt = F(x) Work-Energy Theorem: 1 2 mv2 b 1 2 mv2 a = xb x a F(x)dx

7 Equation of Motion in One Dimension m d2 x = F(x) or mdv dt2 dt = F(x) Work-Energy Theorem: Motion in 3-D 1 2 mv2 b 1 2 mv2 a = b a F d r

8 How Useful is Work-Energy Theorem Evaluation of b a F d r is tentamount to knowing the trajectory beforehand! Two cases where Work-Energy Theorem is useful. Conservative Forces: For many interesting forces b a F d r depends only on end-points! Constrained Motion: External Constraints pre-determine the trajectory.

9 How Useful is Work-Energy Theorem Regardless of whether forces are conservative or not the Work- Energy theorem is always true

10 Prob.4.5 Mass m whirls on a frictionless table, held to circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from l 1 to l 2. Show that the work done in pulling the string equals the increase in kinetic energy of the mass.

11 Prob.4.5 Mass m whirls on a frictionless table, held to circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from l 1 to l 2. Show that the work done in pulling the string equals the increase in kinetic energy of the mass. The fundamental mystery here is... how can you pull radially and still end up changing the angular velocity of the mass.

12 Solution 4.5 For cicular motion: F r = m v2 1 l 1 = m l 1 ω 2 1 If F r is increased, m will move to smaller r: Evaluate: W = F r (r) = m v2 (r) r l2 l 1 F r (r)dr = = mrω 2 (r) l2 l 1 mrω 2 (r)dr

13 a = ( r r θ 2 )ˆr + (r θ + 2ṙ θ) }{{} =0 r ω + 2ṙω=0 1dω ω ω(r) ω 1 dr dt = 21 r dt dω r ω = 2 dr l 1 r [ θ = ω] ˆθ ω(r) = l 2 ω 1 1 r 2 W = l2 l2 F r (r)dr = mrω 2 (r)dr l 1 l 1

14 W= ml 4 1 ω2 1 l2 dr l 1 r = ml4 1 ω2 1 = 1 ( ) l 4 2 m 1 ω2 1 l 2 l 2 1 ω2 1 2 ( 1 l l 2 1 ) W= 1 2 m[ (l 2 ω 2 ) 2 (l 1 ω 1 ) 2] [ ω 2 l 2 2 = ω 1l 2 1 =K 2 K 1 ]

15 Prob. 4.7 A ring of mass M, hangs from a thread, and two beads of mass m slide on it without friction. The beads are released simultaneously from the top of the ring and slide down opposite side. Show that the ring will start to rise if m > 3M/2, and find the angle at which this occurs.

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17 Forces are different for cosθ > 2/3 and cosθ < 2/3

18 Forces are different for cosθ > 2/3 and cosθ < 2/3

19 cosθ, N and T vs. θ (m = 2M)

20 N and T vs. θ: Magnified view

21 2Ncosθ vs. θ

22 Accelaration of the ring vs. θ

23 Prob A block of mass M on a horizontal frictionless table is connected to a spring (spring constant k). The block is set in motion so that it oscillates about its equilibrium position with a certain amplitude A 0. The period of motion is T 0 = 2π M/k.

24 Prob (a) A lump of sticky putty of mass m is dropped onto the block. the putty sticks without bouncing. The putty hits M at the instant when the velocity of M is zero. Find 1 The new period 2 The new amplitude 3 The change in mechanical energy of the system (b) Repeat part a, assuming that the sticky putty hits M at the instant when M has its maximum velocity.

25 Potential Energy For a conservative Force Field: rb r a F d r=func. of ( r b ) func. of ( r a ) W ba = U( r b ) + U( r a ) W ba =K b K a W ba = U b + U a = K b K a [Work energyth.] U a + K a = U b + K b = E

26 What Does Potential Energy Tell us About Force? xb U b U a = F(x)dx x a U(x + x) U(x) = U U= x+ x x F(x)dx U F(x)(x + x x) = F(x) x F(x) U or F(x) lim x 0 = du dx x

27 Potential Energy Determins Stability of a System Harmonic Oscillator: U = kx 2 /2

28 Potential Energy Determins Stability of a System Simple Pendulum: U(θ) = mgl(1 cosθ) and du dθ = mglsinθ

29 Potential Energy Determins Stability of a System Simple Pendulum: U(θ) = mgl(1 cosθ) and du dθ = mglsinθ

30 Potential Energy Determins Stability of a System General Stability Criteria:

31 Potential Energy Determins Stability of a System General Stability Criteria:

32 Potential Energy Determins Stability of a System General Stability Criteria:

33 Rock me Spin me yet I Stand Tall: The Amazingly Stable Teeter-Toy

34 Rock me Spin me yet I Stand Tall: The Amazingly Stable Teeter-Toy

35 Rock me Spin me yet I Stand Tall: The Amazingly Stable Teeter-Toy

36 Rock me Spin me yet I Stand Tall: The Amazingly Stable Teeter-Toy

37 A Sports Car Stability requires low center of mass and hence the peculiar design of a sports car.

38 Non-Conservative Forces b W total ba = a b W total ba = F = F c + F nc a F d r F c d r + b a F nc d r K b K a = U b + U a + W nc ba K b + U b (K a + U a ) = W nc ba E b E a = W nc ba

39 Power: Time rate of doing Work Units: P = dw dt = F d r dt = F v [S.I.] 1 W=1J/s [CGS]1 erg/s = 10 7 W [English] 1 hp=550 lb ft/s 746W

40 Power: Time rate of doing Work Typical Power Consumption Man running upstairs: 1/2 to 1 hp for 30 s A husky man can do work over a period of 8 hours only at a rate of 0.2 hp 1000 Kcal Per person energy use: India (0.7 kw), Germany (6 kw), USA (11.4 kw)

41 Prob A 160 lb man leaps into the air from a crouching position. His center of gravity rises 1.5 ft before he leaves the ground, and it then rises 3 ft to the top of his leap. What power does he develop assuming that he pushes the ground with constant force?

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43 Sol.4.18 P=W/T (W = work done by N) W=N 1.5 (c.g. rises by 1.5ft) N=mg + ma or N = a a= v2 [ 2s = 64 v = 2gs = = 8 ] 3 N=480 lb W = 720 lb.ft T = v/a = 3/8 P = W/T = 3325lb.ft/s 6hp

44 Prob In the preceeding problem take F = F 0 cosωt where F 0 is the peak force, and the contact with ground ends at ωt = π/2. Find the peak power that the man develops during the jump.

45 Sol.4.19 P(t)=N(t)v(t) [N(t) = F(t)] N(t) mg=ma(t) m v(t) 0 dv= t 0 (F 0 cosωt mg)dt v(t)= F 0 mω sinωt gt [F 0,ω?] x(t)= F 0 mω 2 (1 cosωt) 1 2 gt2

46 Sol.4.19 Boundary conditions: v(t = π/2ω)=8 3 = F 0 mω gπ 2ω x(t = π/2ω)=1.5 ft = F 0 mω gπ2 2 8ω 2

47 Sol.4.19 ω = 9.96s 1 F 0 = 832 lb t = π 2ω = 0.16s P(t) = F(t)v(t) F 0 F 0 sin2ωt 2mω }{{} 2mgωtcosωt }{{} 1 2 A reasonable approximation: F 0 >> mg then 1 st >> 2 nd

48 Sol.4.19 P(t) F2 0 2mω sin2ωt For P max. : dp dt =0 dp dt =F2 0 m cos2ωt = 0 ( P max. = P t= π = F2 0 4ω 2mω sin 2ω π ) 4ω = F2 0 2mω

49 Sol.4.19 Check: d 2 P dt 2 = t= π 4ω m 2ωsin2ωt F 2 0 t= π 4ω < 0

50 Sol.4.19

51 Sol.4.19

52 What do collisions teach us? F = m a is a double-edge sword Collisions studies gave most profound knowledge about fundamental Physics Constraints from energy and momentum conservation severe enough to extract vital information about scattering

53 Geneva

54 USA

55 Classical Collisions A + B C + D Mass is conserved: m A + m B m C + m D Momentum is conserved: p A + p B = p C + p D K.E. may or may not be conserved

56 Classical Collisions A + B C + D Types of Collisions sticky: K.E. decreases K A + K B > K C + K D Explosive: K.E. increases K A + K B < K C + K D Elastic: K.E. is conserved K A + K B = K C + K D

57 Prob A small ball of mass m is placed on top of the a superball of mass M, and the two balls are dropped to the floor from height h.how high does the small ball rise after the collision? Assume that the collision is perfecly elastic, and that m << M.

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59 Sol Let v 1 and v 2 be the initial velocities of m and M before collision and v 1 and v 2, after collision. In a two body one dimensional collision: v 1 =(m M)v 1 + 2Mv 2 m + M v 2 =(M m)v 2 + 2mv 1 m + M Here: v 1 = 2gh, v 2 = 2gh, and M >> m

60 Sol v 1 = 3M 2gh M = 3 2gh If m rises to height h after collision: 1 2 m 1v 2 1 = m 1gh h = v 2 1 2g = 9h

61 Problem 4.9 A superball of mass m bounces back and forth between two surfaces with speed v 0. Gravity is neglected and collisions are perfectly elastic. (a) Find the average force F on each wall. (b) If one surface is moving uniformly toward the other with speed V << v 0, the bounce rate will increase due to shorter distance between collisions, and because the ball s speed increases when it bounces from the moving surface. Find the F in terms of separation of surfaces, x. (c) Show that work needed to push the surface from l to x equals gain in kinetic energy of the ball.

62 Problem 4.29

63 Sol (a) Momentum transfer to wall in 1 bounce: m(2v 0 ) No. of bounce per unit time: v 0 /2l v 0 F = 2mv 0 2l = mv2 0 l (b) F(x) = mv(x)2 x Integrate a(x) = dv(x) dt to find v(x).

64 Sol (b) v after each bounce v = 2V (Sling-shot effect) This change happens in time t = 2x/v dv dt =Vv x dv t = v 0 v v 0 Vdt x x = l v(x) = v 0l x F = mv(x)2 x dx x = ml2 v 2 0 x 3

65 Sol (c) Work done in moving the surface: W= x l = ml2 v x F(x ) dx = ml 2 v 2 dx 0 l x [ 3 1 x 1 ] 2 l 2 = ml2 v x 2 2 mv2 0 = 1 2 mv(x)2 1 2 mv2 0 =Change in K.E. of ball

66 Sling-Shot (Gravity-Assist) and NASA s Cassini

67 Collisions and Center of Mass Coordinates How does it help? Total momentum in C-system is zero Initial and final velocities lie in the same plane Each particle is scattered through the same angle θ in the plane of scattering. For elastic collisions, the speed of each particle is same before & after the collision

68 Collisions and Center of Mass Coordinates m 1 & m 2 having velocity v 1 & v 2. V = m 1v 1 + m 2 v 2 m 1 + m 2

69 Collisions and Center of Mass Coordinates v 1c =v 1 V = m 2 m 1 + m 2 (v 1 v 2 ) v 2c =v 2 V = m 1 m 1 + m 2 (v 1 v 2 )

70 Collisions and Center of Mass Coordinates The momenta in C-system: p 1c =m 1 v 1c = m 1m 2 m 1 + m 2 (v 1 v 2 ) =µv p 2c = µv 0=p 1c + p 2c

71 C-System Plane of Scattering m 1 and m 2 striking with v 1 & v 2. Initial velocities in L & C-system.

72 C-System Plane of Scattering m 1 and m 2 striking with v 1 & v 2. Final velocities in L & C-system (The plane in general is different)

73 C-System Plane of Scattering C-system plane of scattering. For elastic scattering, v 1c = v 1c and v 2c = v 2c. Velocity vectors simple rotate.

74 If m 2 is at rest V= m 1 m 1 + m 2 v 1 v 1c =v 1 V = m 2 m 1 + m 2 v 1 v 2c = V = m 1 m 1 + m 2 v 1

75 If m 2 is at rest V= m 1 m 1 + m 2 v 1 v 1c =v 1 V = m 2 m 1 + m 2 v 1 v 2c = V = m 1 m 1 + m 2 v 1