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Example Problem #1.38

Example Problem #1.38. For components of C , we have: C x = (3.1 km)(cos45 o ) = 2.2 km C y = (3.1 km)(sin45 o ) = 2.2 km. 3.1 km. C y. R. R y. 45 o. q. C x. R x. Components of total displacement R are thus:

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Example Problem #1.38

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  1. Example Problem #1.38 • For components of C, we have: • Cx = (3.1 km)(cos45o) = 2.2 km • Cy = (3.1 km)(sin45o) = 2.2 km 3.1 km Cy R Ry 45o q Cx Rx • Components of total displacement R are thus: • Rx = Ax + Bx + Cx = 0 + 4.0 km + 2.2 km = 6.2 km • Ry = Ay + By + Cy = 2.6 km + 0 km + 2.2 km = 4.8 km • Magnitude of R is: • |R| = (Rx2 + Ry2)1/2 = [(6.2 km)2 + (4.8 km)2]1/2 = 7.8 km • Direction of R found from trig. functions: • sinq = 4.8 km/7.8 km  q = sin–1(4.8/7.8) = 38o

  2. Example Problem #2.33 • (a) max. speed attained when acceleration = 0  constant speed achieved (aqua region) • v = v0 + at  look at times when t = 0 (v0 = 0) and t = t1 (v = max. and const.) • v = at1 a = 20.0 m/s2  t1 = 15 min = 900 s • v = (20.0 m/s2)(900 s) = 18,000 m/s = 18.0 km/s • (b) 1st leg of journey (represented by red line in graphs): x – x0 = ½(v0 + v)t  x0 = 0, v0 = 0 • x = x1 = vt/2  v = 18.0 km/s  t = 900 s  x1 = 8100 km 3rd leg of journey (yellow line): x – x0 = ½(v0 + v)t  x3 – x2 = ½(v2 + v3)t  v2 = 18.0 km/s, v3 = 0, t = 900 s  x3 – x2 = 8100 km  leg 1 + leg 3 = 16,200 km

  3. Example Problem #2.33 • (b continued) leg 2 distance = 384,000 km – 16,200 km = 367,800 km • fraction of total distance = leg 2 dist. / total dist. = 367,800 km / 384,000 km = 0.958 • (c) Find time ship was traveling at constant speed (leg 2, aqua region): • x – x0 = ½(v0 + v)t  x2 – x1 = ½(v1 + v2)t • x1 = 8100 km, x2 = 384,000 km – 8100 km = 375,900 km, v1 = 18.0 km/s, v2 = 18.0 km/s • t = 2(x2 – x1) / (v1 + v2) = 20,433 s = tleg2 • ttotal = tleg1 + tleg2 + tleg3 = 900 s + 20,433 s + 900 s = 22,233 s = 370.56 min = 6.18 hr

  4. +y 443 m • y = y0 + v0t – (1/2)gt2 • – 443 m = 0 + 0 – (1/2) gt2 • 886 m / g = t2 t2 = 90.4 s2 • t =  9.5 s  t = 9.5 s (neg. value has no physical meaning) Free Fallin’ How long would it take for Tom Petty to go “Free-Fallin’” from the top of the Sears Tower? How fast will he be moving just before he hits the ground? • v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s (negative sign means downward direction)

  5. y x Let x0 = 0 and y0 = 0 Example Problem #3.12 v0x = v0. v0y = 0. x = x0 + v0xt. y = y0 + v0yt – 0.5gt2. What must v0 be so that x = 1.75 m when y = –9.00 m?  Time it takes to travel 9.00 m vertically: y = – 0.5gt2 = – 0.5(9.8 m/s2)t2 = –9.00 m  t = 1.36 s  Speed to travel 1.75 m horizontally: x = v0t  v0 = x / t = 1.75 m / 1.36 s = 1.29 m/s

  6. (a) arad = v2 / R = (3 m/s)2 / (14.0 m) = 0.643 m/s2 atan = 0.5 m/s2  a = [(arad)2 + (atan)2]1/2 = [(0.643 m/s2)2 + (0.5 m/s2)2]1/2 = 0.814 m/s2 arad atan = 0.5 m/s2  Direction of a determined by: tan q = arad / atan  q = tan–1 (0.643/0.5)  q = 52.10 (up from horizontal) (b) a 52.10 v Example Problem #3.34

  7. y (b) m = 4.00 kg N2: S Fx = T = max = (4.00 kg)(2.50 m/s2) = 10.0 N T x mg Nm (c) M = 6.00 kg Net force points in the +x – direction (same direction as acceleration), making force F larger in magnitude T F y Mg NM x Example Problem #4.39 (a) Since the two crates are connected by the light rope, they move together with the same acceleration of 2.50 m/s2.

  8. Example Problem #4.39 (continued) (d) M = 6.00 kg T F y Mg NM x N2: S Fx = F – T = Max F = Max + T = (6.00 kg)(2.50 m/s2) + 10.0 N = 25.0 N

  9. Free–body diagram of wrecking ball: TB y TBy TA 400 TBx x mg Example Problem #5.7 mg = (4090 kg)(9.8 m/s2) = 40000 N (a) S Fy= 0  TBy – mg = 0  TBcos400 – mg = 0  TB = mg / cos400 = 5.23  104 N (b) S Fx= 0  TBx – TA = 0  TBsin400 – TA = 0  TA = TBsin400 = 3.36  104 N

  10. Free–body diagram: In general: N N y x q F Wy y 110 q x W q 900 - q 110 W Wx Example Problem #5.10 v = constant Piano moving at const. velocity  a = 0 (piano in equilibrium) 110 S Fx = 0  F – Wx = 0  F = Wx  F = mg sin110 = (180 kg)(9.8 m/s2) sin110 = 336.6 N

  11. VI.B. Dynamics Problems • Dynamics problems involve bodies which have a nonzero acceleration • From Newton’s 2nd Law: • In component form: • Summary of Problem–Solving Strategy: • Similar to strategy given for statics problems (bodies in equilibrium) • Exception: In Step #3, set as appropriate

  12. y Free–body diagram: N x Wy q W Wx Problem #5.16 q • Apply Newton’s 2nd Law in x – direction: • S Fx = max  Wx = max  Wsinq = max • mg sinq = max  sinq = ax / g • From 1 – D motion with constant acceleration: • v2 = v02 + 2a(x – x0)  a = (v2 – v02) / 2(x – x0)

  13. Fp Fp (top) (bottom) mg mg Analysis of Swinging Pail of Water Top: +y Bottom: vt vb Free–body diagrams of water: Force exerted on water by pail at top: S Fy = may = m(–vt2 / r)  – Fp – mg = m(–vt2 / r)  Fp = m (vt2 / r) – mg

  14. Analysis of Swinging Pail of Water Minimum value of vt for water to remain in pail:  Minimum force pail can exert is zero, so set Fp = 0 and solve for minimum speed vt,min: 0 - mg = m(–vt,min2 / r)  vt,min2 = rg  vt,min = (rg)1/2 Force exerted on water by pail at bottom: S Fy = may = m(vb2 / r)  Fp – mg = m(vb2 / r)  Fp = m(vb2 / r) + mg Remember that centripetal force is not an external force acting on a body – it is just the name of the net force acting on a body undergoing circular motion (so there is no arrow for centripetal force on a free–body diagram)

  15. N f W y Example Problem #5.97 Free–body diagram of car: R x (a) S Fx = max = m(v2/R)  f = m(v2/R)  R = mv2/f  f = msN = msW =msmg  R = mv2/msmg = v2/msg = (35.8 m/s)2/(0.76)(9.8 m/s2) = 171.7 m (about 563 ft.) (b) From above, vmax2 = msgR  vmax = (msgR)1/2 = [(0.20)(9.8 m/s2)(171.7 m)]1/2 = 18.34 m/s = 41.0 mph • (c) vmax = (msgR)1/2 = [(0.37)(9.8 m/s2)(171.7 m)]1/2 = 24.95 m/s = 55.8 mph • The posted speed limit is evidently designed for wet road conditions.

  16. N F = 36 N mg y (b) Free–body diagram of 12–pack : N F = 36 N fk mg x y Example Problem #6.23 (a) Free–body diagram of 12–pack : x Work–Energy Theorem: Wtot = K2 – K1 = ½ m(v22 – v12) = ½ mv22 (v1 = 0  starts from rest)  Wtot = WN + Wmg + WF = WF  WF = Fs cos00 = Fs = (36.0 N)(1.20 m) = 43.2 J = ½ mv22  v2 = [2(43.2 J) / m]1/2 = [2(43.2 J) / 4.30 kg]1/2 = 4.48 m/s

  17. Example Problem #6.23 (continued)  Wtot = Wf + WF  Wf = fs cos1800 = –fs = – (mkN)s = –mkmgs = –(0.30)(4.30 kg)(9.8 m/s2)(1.20 m) = –15.17 J  Wtot = –15.17 J + 43.2 J = 28.03 J  v2 = [2(28.03 J) / 4.30 kg]1/2 = 3.61 m/s

  18. N v0 v = 0 f mg Example Problem (a) Wtot = K2 – K1 = ½ m(v22 – v12) = –½ mv02 (v2 = 0)  Wtot = Wf + WN + Wmg = Wf  Wf = fs cos1800 = –fs = – (mkN)s = –mkmgs  –mkmgs = –½ mv02 s = v02 / 2mkg (b) For v0 = 80.0 km/h = 22.2 m/s: mk = v02 / 2gs = (22.2 m/s)2 / 2(9.8 m/s2)(91.2 m) = 0.28  For v0 = 60.0 km/h = 16.7 m/s: s = (16.7 m/s)2 / 2(0.28)(9.8 m/s2) = 50.8 m

  19. x = – 0.025 m Example Problem #6.36 v2 x = 0 x = 0 (Spring compressed) (Spring relaxed) (a) W = 1/2 kx2 = 1/2 (200 N/m)(0.025 m)2 = 0.062 J (b) Wtot = K2 – K1 = 1/2 m(v22 – v12) = 1/2 mv22 (block initially at rest when spring is compressed)  Wtot = WN + Wmg + WS = WS = 0.062 J  0.062 J = 1/2 mv22  v2 = [2(0.062 J) / m]1/2 = [2(0.062 J) / (4.00 kg)]1/2 = 0.177 m/s

  20. (IX.B.1) (IX.B.2) IX.B. Gravitational Potential Energy 1. Consider the fall of Tom Petty from the Sears Tower again • Once Tom Petty begins to fall, gravity does work on him, accelerating him toward the ground • Work done by gravity on Tom Petty: • More generally: y Free – body diagram (neglect air resistance): h (y1) W = Mg 0 (y2)

  21. Path independent – all that matters is change in vertical position IX.B. Gravitational Potential Energy • Product of weight mg and vertical height y is defined as the gravitational potential energy U = mgy • Has units of (kg)(m/s2)(m) = J • Work done by gravity can thus be interpreted as a change in the gravitational potential energy: (II.B.3)

  22. 1 2 H – 2R y y2 = 2R y1 = H 0 Design of a Loop–the–Loop Roller Coaster Suppose we wish to design the following Loop–the–Loop roller coaster: R What is the minimum value of H such that the roller coaster cars make it safely around the loop? (Assuming cars fall under influence of gravity only.) • Conservation of mechanical energy: 1/2mv12 + mgy1 = 1/2mv22 + mgy2 • Assume that roller coaster starts from rest at top of hill. Then we have: mgH = 1/2mv22 + mg(2R) • v22 = 2mg(H – 2R) / m = 2g(H – 2R)

  23. Design of a Loop–the–Loop Roller Coaster • For car to make it safely over the loop: acar = arad g (remember water in bucket) • v22 / R  g • 2g(H – 2R) / R  g • H  R/2 + 2R, or H  5R/2.

  24. 0 +y y1 y2 Example Problem George 20 m 20 m 450 300 • Conservation of mechanical energy: 1/2mv12 + mgy1 = 1/2mv22 + mgy2 • y1 = –(20 m)cos450 = –14.14 m y2 = –(20 m)cos300 = –17.32 m v1 = 0 (George starts from rest)

  25. Example Problem #7.12 (continued) • So: • 1/2 mv22 = mgy1 – mgy2 • v22 = 2g(y1 – y2) v2 = [2g(y1 – y2)]1/2 v2 = [2(9.8 m/s2)(–14.14 m – (–17.32 m))]1/2 v2 = 7.89 m/s

  26. pt. 2 y2 = 22.0 m y1 = 0 Ug = 0 pt. 1 Example Problem #7.18 (a) Conservation of mechanical energy: K1 + U1,g + U1,el = K2 +U2,g + U2,el  K1 = 0 (pebble initially at rest)  U1,g = 0 (by choice)  K2 = 0 (pebble at rest at max. height)  U2,el = 0 (slingshot in relaxed position)  U1,el = U2,g = mgy2 = (0.01 kg)(9.8 m/s2)(22.0 m) = 2.16 J (b) U1,el = 2.16 J = U2,g = mgy  y = 2.16 J / mg = 2.16 J / (0.025 kg)(9.8 m/s2) = 8.82 m (c) No air resistance, no deformation of the rubber band

  27. s = 1.0 m Point 2 (Block stopped) 0 0 0 x1 = –0.20 m Example Problem #7.43 x = 0 Point 1 (Spring compressed) With friction doing work on the block, we have: ½ mv12 + ½ kx12 + Wf = ½ mv22 + ½ kx22  Wf = – ½ kx12 = – ½ (100 N/m)(–0.20 m)2 = –2 J Also, Wf = – f s = – f (1.00 m) = –2 J  f = 2 N = mkN = mkmg  mk = 2 N / mg = 2 N / (0.50 kg)(9.8 m/s2) = 0.41

  28. v = 7 m/s Pt. B L = 6 m H 300 300 0 0 0 • UB,g = mgH = mgLsin(300) = (1.50 kg)(9.8 m/s2)(6.00 m)sin(300) = 44.1 J N f • Wf = – f s = – mkNL = – mkmgcos(300)L = – (0.50)(1.50 kg)(9.8 m/s2)cos(300)(6.00 m) = – 38.19 J mg Example Problem #7.73 (Spring compressed) Pt. A From the Work – Energy Theorem: KA + UA,g + UA,el + Wf = KB + UB,g + UB,el  UA,el = KB + UB,g – Wf • KB = ½ mvB2 = ½ (1.50 kg)(7.00 m/s)2 = 36.75 J • UA,el = 36.75 J + 44.1 J – (–38.19 J) = 119.0 J

  29. b) vf? Conservation of momentum: X. E. Collisions • Inelastic collisions a) Classic example: car crash where cars stick together after collision (X.E.1,2) (X.E.3) (X.E.4)

  30. X. E. Collisions vf => 0 vf => 0 • Limiting behavior: m1 => 0: • Limiting behavior: v1 => 0: • Limiting behavior: m1 = m2: • In inelastic collisions, the KE of system after collision < KE of system before collision (Why?) • K.E. before collision = K1 = 1/2 m1v1i2 • K.E. after collision = K2 = 1/2 (m1 + m2)vf2 = 1/2 (m1 + m2)[m1 / (m1 + m2)]2v1i2 • So K2 / K1 = m1 / (m1 + m2) < 1 so K2 < K1 vf = 1/2 v1i

  31. X. E. Collisions 2. Elastic collisions • Forces between colliding bodies are conservative • Kinetic energy is conserved (may be temporarily converted to elastic potential energy) • Momentum is conserved

  32. mB, vB1 = 0 mA, vA2 mB, vB2 mA, vA1 X. F. Elastic Collisions • Consider a head-on collision where one object is at rest at t=0: • Conservation of kinetic energy gives: 1/2 mAvA12 = 1/2mAvA22 + 1/2mBvB22 (X.F.1) 3. Conservation of momentum gives: mAvA1 = mAvA2 + mBvB2 (X.F.2) 4. Combining F.1 and F.2 yields: (X.F.3,4)

  33. X. F. Elastic Collisions 5. Limiting Case: mA = mB vA2 = 0 and vB2 = vA1 6. Limiting Case: mA << mB vA2 – vA1 vB2 << vA1 Similar to result of ping-pong ball (mA) striking stationary bowling ball (mB) 7. Limiting Case: mA >> mB vA2  vA1 vB2  2vA1 Similar to result of bowling ball (mA) striking stationary ping-pong ball (mB)

  34. mB, vB1 = 0 mA, vA2 q j mA, vA1 mB, vB2 X. F. Elastic Collisions 8. “Grazing” collision: In this case, we need to remember to take both magnitude and direction into account a) Use the component form of conservation of momentum: Px,initial = Px,final Py,initial = Py,final Note: True for elastic or inelastic collisions x (X.F.5,6)

  35. y 36.90 x vhat “Odd – Job” Example Problem vOJ Conservation of Momentum: ice ice BEFORE AFTER x–direction: 0 = phat,x + pOJ,x y–direction: 0 = phat,y + pOJ,y We are interested in the horizontal recoil velocity of the bad guy, so from the x – component equation we have:  pOJ,x = – phat,x mOJvOJ = – mhatvhat,x = – mhatvhat cos36.90  vOJ = – mhatvhat cos36.90 / mOJ = – (4.50 kg)(22.0 m/s)cos36.90 / 120 kg = – 0.66 m/s

  36. A A B B 0 0 0 Example Problem #8.20 vA = 0 vB = 0 vA’ vB’ BEFORE AFTER (a) Conservation of momentum: Ptotal,before = Ptotal,aftermAvA + mBvB = mAvA’ + mBvB’ 0 = mAvA’ + mBvB’  mAvA’ = –mBvB’ vA’ = –mBvB’ / mA = –[(3.00 kg)(1.20 m/s)] / 1.00 kg, = –3.6 m/s (b) Conservation of mechanical energy: Kbefore + Uel,before = Kafter + Uel,after 1/2mAvA2 + 1/2mBvB2 + Uel,before = 1/2mAvA’2 + 1/2mBvB’2 + Uel,after Uel,before = 1/2(1 kg)(–3.6 m/s)2 + 1/2(3 kg)(1.2m/s)2 = 8.6 J

  37. x = 0 Example Problem #8.50 (a) xcm = (m1x1 + m2x2) / (m1 + m2) = (1800 kg)(40.0 m) / 3000 kg = 24.0 m ahead of m1 (or 16.0 m behind m2) (b) P = m1v1 + m2v2 = (1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 5.04  104 kgm/s (c) vcm = (m1v1 + m2v2) / (m1 + m2) = [(1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s)/3000 kg = 16.8 m/s

  38. Example Problem Consider a solid disk rolling without slipping down an inclined plane. What is its velocity at the bottom of the plane if it starts from rest (~no friction). Ei = mgh. Ef = 1/2mv2 + (1/2)Iw2. But: I = 1/2mR2, and w = v/R So: mgh = 1/2mv2 + (1/2)(1/2mR2)(v/R)2 = (3/4)mv2, or v = SQRT(4/3gh). h

  39. Example Problem A uniform solid ball of mass m and radius R rolls without slipping down a plane inclined at an angle q. Using dynamics (Newton’s 2nd Law), find the final speed. vcm h q E1 = mgh = mgsinq E2 = 1/2mv2 + 1/2Iw2= 1/2mv2 + 1/2(2/5mR2)(v/R)2 = 1/2mv2( 1+ 2/5) = 7/10mv2 = mgsinq; v = 1.2(gsinq)1/2

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