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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  o CxCx CyCy Components.

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Presentation on theme: "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  o CxCx CyCy Components."— Presentation transcript:

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2 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  o CxCx CyCy Components of total displacement R are thus: R x = A x + B x + C x = 0 + 4.0 km + 2.2 km = 6.2 km R y = A y + B y + C y = 2.6 km + 0 km + 2.2 km = 4.8 km Magnitude of R is: |R| = (R x 2 + R y 2 ) 1/2 = [(6.2 km) 2 + (4.8 km) 2 ] 1/2 = 7.8 km Direction of R found from trig. functions: sin  = 4.8 km/7.8 km   = sin –1 (4.8/7.8) = 38 o  RxRx RyRy R

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

4 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 – x 0 = ½(v 0 + v)t  x 2 – x 1 = ½(v 1 + v 2 )t  x 1 = 8100 km, x 2 = 384,000 km – 8100 km = 375,900 km, v 1 = 18.0 km/s, v 2 = 18.0 km/s  t = 2(x 2 – x 1 ) / (v 1 + v 2 ) = 20,433 s = t leg2  t total = t leg1 + t leg2 + t leg3 = 900 s + 20,433 s + 900 s = 22,233 s = 370.56 min = 6.18 hr

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

6 Example Problem #3.12 v 0x = v 0. v 0y = 0. x = x 0 + v 0x t. y = y 0 + v 0y t – 0.5gt 2. x y Let x 0 = 0 and y 0 = 0 What must v 0 be so that x = 1.75 m when y = –9.00 m?  Time it takes to travel 9.00 m vertically: y = – 0.5gt 2 = – 0.5(9.8 m/s 2 )t 2 = –9.00 m  t = 1.36 s  Speed to travel 1.75 m horizontally: x = v 0 t  v 0 = x / t = 1.75 m / 1.36 s = 1.29 m/s

7 Example Problem #3.34 a tan = 0.5 m/s 2 a rad (a) a rad = v 2 / R = (3 m/s) 2 / (14.0 m) = 0.643 m/s 2 a tan = 0.5 m/s 2  a = [(a rad ) 2 + (a tan ) 2 ] 1/2 = [(0.643 m/s 2 ) 2 + (0.5 m/s 2 ) 2 ] 1/2 = 0.814 m/s 2  Direction of a determined by: tan  = a rad / a tan   = tan –1 (0.643/0.5)   = 52.1 0 (up from horizontal) (b) v a 52.1 0

8 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/s 2. (b) m = 4.00 kg x y T mgNmNm N2:  F x = T = ma x = (4.00 kg)(2.50 m/s 2 ) = 10.0 N (c) M = 6.00 kg x y T MgNMNM F Net force points in the +x – direction (same direction as acceleration), making force F larger in magnitude

9 Example Problem #4.39 (continued) (d) M = 6.00 kg x y T MgNMNM F N2:  F x = F – T = Ma x F = Ma x + T = (6.00 kg)(2.50 m/s 2 ) + 10.0 N = 25.0 N

10 Example Problem #5.7 T By T Bx 40 0 Free–body diagram of wrecking ball: x y TATA TBTB mg mg = (4090 kg)(9.8 m/s 2 ) = 40000 N (a)  F y = 0  T By – mg = 0  T B cos40 0 – mg = 0  T B = mg / cos40 0 = 5.23  10 4 N (b)  F x = 0  T Bx – T A = 0  T B sin40 0 – T A = 0  T A = T B sin40 0 = 3.36  10 4 N

11 Example Problem #5.10 11 0 v = constant Piano moving at const. velocity  a = 0 (piano in equilibrium) Free–body diagram: 11 0 x y F N W N W WxWx WyWy In general:        F x = 0  F – W x = 0  F = W x  F = mg sin11 0 = (180 kg)(9.8 m/s 2 ) sin11 0 = 336.6 N x y

12 VI.B. Dynamics Problems 1.Dynamics problems involve bodies which have a nonzero acceleration 2.From Newton’s 2 nd Law: 3.In component form: 4.Summary of Problem–Solving Strategy: a)Similar to strategy given for statics problems (bodies in equilibrium) b)Exception: In Step #3, set as appropriate

13 Problem #5.16  Free–body diagram: W N WxWx WyWy  x y Apply Newton’s 2 nd Law in x – direction:   F x = ma x  W x = ma x  Wsin  = ma x  mg sin  = ma x  sin  = a x / g From 1 – D motion with constant acceleration:  v 2 = v 0 2 + 2a(x – x 0 )  a = (v 2 – v 0 2 ) / 2(x – x 0 )

14 Analysis of Swinging Pail of Water Top: vtvt +y vbvb Bottom: Free–body diagrams of water: mg FpFp (top) mg FpFp (bottom) Force exerted on water by pail at top:  F y = ma y = m(–v t 2 / r)  – F p – mg = m(–v t 2 / r)  F p = m (v t 2 / r) – mg

15 Analysis of Swinging Pail of Water Minimum value of v t for water to remain in pail:  Minimum force pail can exert is zero, so set F p = 0 and solve for minimum speed v t,min :  mg = m(–v t,min 2 / r)  v t,min 2 = rg  v t,min = (rg) 1/2 Force exerted on water by pail at bottom:  F y = ma y = m(v b 2 / r)  F p – mg = m(v b 2 / r)  F p = m(v b 2 / 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)

16 Example Problem #5.97 Free–body diagram of car: x y R W N f (a)  F x = ma x = m(v 2 /R)  f = m(v 2 /R)  R = mv 2 /f  f =  s N =  s W =  s mg  R = mv 2 /  s mg = v 2 /  s g = (35.8 m/s) 2 /(0.76)(9.8 m/s 2 ) = 171.7 m (about 563 ft.) (b) From above, v max 2 =  s gR  v max = (  s gR) 1/2 = [(0.20)(9.8 m/s 2 )(171.7 m)] 1/2 = 18.34 m/s = 41.0 mph (c) v max = (  s gR) 1/2 = [(0.37)(9.8 m/s 2 )(171.7 m)] 1/2 = 24.95 m/s = 55.8 mph  The posted speed limit is evidently designed for wet road conditions.

17 Example Problem #6.23 (a) Free–body diagram of 12–pack : x y Work–Energy Theorem: W tot = K 2 – K 1 = ½ m(v 2 2 – v 1 2 ) = ½ mv 2 2 (v 1 = 0  starts from rest)  W tot = W N + W mg + W F = W F  W F = Fs cos0 0 = Fs = (36.0 N)(1.20 m) = 43.2 J = ½ mv 2 2  v 2 = [2(43.2 J) / m] 1/2 = [2(43.2 J) / 4.30 kg] 1/2 = 4.48 m/s F = 36 N N mg fkfk (b) Free–body diagram of 12–pack : x y F = 36 N N mg

18 Example Problem #6.23 (continued)  W tot = W f + W F  W f = fs cos180 0 = –fs = – (  k N)s = –  k mgs = –(0.30)(4.30 kg)(9.8 m/s 2 )(1.20 m) = –15.17 J  W tot = –15.17 J + 43.2 J = 28.03 J  v 2 = [2(28.03 J) / 4.30 kg] 1/2 = 3.61 m/s

19 Example Problem f mg N (a) W tot = K 2 – K 1 = ½ m(v 2 2 – v 1 2 ) = –½ mv 0 2 (v 2 = 0)  W tot = W f + W N + W mg = W f  W f = fs cos180 0 = –fs = – (  k N)s = –  k mgs  –  k mgs = –½ mv 0 2  s = v 0 2 / 2  k g (b) For v 0 = 80.0 km/h = 22.2 m/s:  k = v 0 2 / 2gs = (22.2 m/s) 2 / 2(9.8 m/s 2 )(91.2 m) = 0.28  For v 0 = 60.0 km/h = 16.7 m/s: s = (16.7 m/s) 2 / 2(0.28)(9.8 m/s 2 ) = 50.8 m v0v0 v = 0

20 Example Problem #6.36 x = 0 x = – 0.025 m (Spring compressed) x = 0 (Spring relaxed) v2v2 (a) W = 1/2 kx 2 = 1/2 (200 N/m)(0.025 m) 2 = 0.062 J (b) W tot = K 2 – K 1 = 1/2 m(v 2 2 – v 1 2 ) = 1/2 mv 2 2 (block initially at rest when spring is compressed)  W tot = W N + W mg + W S = W S = 0.062 J  0.062 J = 1/2 mv 2 2  v 2 = [2(0.062 J) / m] 1/2 = [2(0.062 J) / (4.00 kg)] 1/2 = 0.177 m/s

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

22 1.Product of weight mg and vertical height y is defined as the gravitational potential energy U = mgy a)Has units of (kg)(m/s 2 )(m) = J 2.Work done by gravity can thus be interpreted as a change in the gravitational potential energy: Path independent – all that matters is change in vertical position IX.B. Gravitational Potential Energy (II.B.3)

23 Design of a Loop–the–Loop Roller Coaster Suppose we wish to design the following Loop–the–Loop roller coaster: 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.) 1 2 H – 2R y 2 = 2R y 1 =R Conservation of mechanical energy: 1/2mv 1 2 + mgy 1 = 1/2mv 2 2 + mgy 2  Assume that roller coaster starts from rest at top of hill. Then we have: mgH = 1/2mv 2 2 + mg(2R)  v 2 2 = 2mg(H – 2R) / m = 2g(H – 2R) H y 0

24 Design of a Loop–the–Loop Roller Coaster For car to make it safely over the loop: a car = a rad  g (remember water in bucket)  v 2 2 / R  g  2g(H – 2R) / R  g  H  R/2 + 2R, or H  5R/2.

25 Example Problem 45 0 20 m 30 0 20 m George 0 +y Conservation of mechanical energy: 1/2mv 1 2 + mgy 1 = 1/2mv 2 2 + mgy 2  y 1 = –(20 m)cos45 0 = –14.14 m  y 2 = –(20 m)cos30 0 = –17.32 m  v 1 = 0 (George starts from rest) y1y1 y2y2

26 Example Problem #7.12 (continued) So: 1/2 mv 2 2 = mgy 1 – mgy 2  v 2 2 = 2g(y 1 – y 2 )  v 2 = [2g(y 1 – y 2 )] 1/2  v 2 = [2(9.8 m/s 2 )(–14.14 m – (–17.32 m))] 1/2  v 2 = 7.89 m/s

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

28 Example Problem #7.43 x = 0 x 1 = –0.20 m Point 1 (Spring compressed) Point 2 (Block stopped) With friction doing work on the block, we have: ½ mv 1 2 + ½ kx 1 2 + W f = ½ mv 2 2 + ½ kx 2 2 000  W f = – ½ kx 1 2 = – ½ (100 N/m)(–0.20 m) 2 = –2 J Also, W f = – f s = – f (1.00 m) = –2 J  f = 2 N =  k N =  k mg   k = 2 N / mg = 2 N / (0.50 kg)(9.8 m/s 2 ) = 0.41 s = 1.0 m

29 Example Problem #7.73 Pt. A 30 0 From the Work – Energy Theorem: K A + U A,g + U A,el + W f = K B + U B,g + U B,el (Spring compressed) 000 Pt. B 30 0 v = 7 m/s L = 6 m  K B = ½ mv B 2 = ½ (1.50 kg)(7.00 m/s) 2 = 36.75 J  U B,g = mgH = mgLsin(30 0 ) = (1.50 kg)(9.8 m/s 2 )(6.00 m)sin(30 0 ) = 44.1 J H  W f = – f s = –  k NL = –  k mgcos(30 0 )L = – (0.50)(1.50 kg)(9.8 m/s 2 )cos(30 0 )(6.00 m) = – 38.19 J N mg f  U A,el = 36.75 J + 44.1 J – (–38.19 J) = 119.0 J  U A,el = K B + U B,g – W f

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

31 a)Limiting behavior: m 1 => 0: b)Limiting behavior: v 1 => 0: c)Limiting behavior: m 1 = m 2 : d)In inelastic collisions, the KE of system after collision < KE of system before collision (Why?) K.E. before collision = K 1 = 1/2 m 1 v 1i 2 K.E. after collision = K 2 = 1/2 (m 1 + m 2 )v f 2 = 1/2 (m 1 + m 2 )[m 1 / (m 1 + m 2 )] 2 v 1i 2 So K 2 / K 1 = m 1 / (m 1 + m 2 ) < 1 so K 2 < K 1 X. E. Collisions v f => 0 v f = 1/2 v 1i

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

33 1.Consider a head-on collision where one object is at rest at t=0: 2.Conservation of kinetic energy gives: 1/2 m A v A1 2 = 1/2m A v A2 2 + 1/2m B v B2 2 (X.F.1) 3.Conservation of momentum gives: m A v A1 = m A v A2 + m B v B2 (X.F.2) 4.Combining F.1 and F.2 yields: m B, v B1 = 0 m A, v A1 m A, v A2 m B, v B2 X. F. Elastic Collisions (X.F.3,4)

34 5.Limiting Case: m A = m B v A2 = 0 and v B2 = v A1 6.Limiting Case: m A << m B v A2  – v A1 v B2 << v A1 Similar to result of ping-pong ball (m A ) striking stationary bowling ball (m B ) 7.Limiting Case: m A >> m B v A2  v A1 v B2  2v A1 Similar to result of bowling ball (m A ) striking stationary ping-pong ball (m B ) X. F. Elastic Collisions

35 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: P x,initial = P x,final P y,initial = P y,final Note: True for elastic or inelastic collisions x m B, v B1 = 0 m A, v A1   m A, v A2 m B, v B2 X. F. Elastic Collisions (X.F.5,6)

36 Example Problem Conservation of Momentum: ice BEFORE ice AFTER “Odd – Job”x y 36.9 0 v hat v OJ x–direction: 0 = p hat,x + p OJ,x y–direction: 0 = p hat,y + p OJ,y We are interested in the horizontal recoil velocity of the bad guy, so from the x – component equation we have:  p OJ,x = – p hat,x  m OJ v OJ = – m hat v hat,x = – m hat v hat cos36.9 0  v OJ = – m hat v hat cos36.9 0 / m OJ = – (4.50 kg)(22.0 m/s)cos36.9 0 / 120 kg = – 0.66 m/s

37 Example Problem #8.20 BEFORE v A = 0 AB v B = 0 A B AFTER vB’vB’vA’vA’ (a) Conservation of momentum: P total,before = P total,after m A v A + m B v B = m A v A ’ + m B v B ’ 0 = m A v A ’ + m B v B ’  m A v A ’ = –m B v B ’ v A ’ = –m B v B ’ / m A = –[(3.00 kg)(1.20 m/s)] / 1.00 kg, = –3.6 m/s (b) Conservation of mechanical energy: K before + U el,before = K after + U el,after 1/2m A v A 2 + 1/2m B v B 2 + U el,before = 1/2m A v A ’ 2 + 1/2m B v B ’ 2 + U el,after U el,before = 1/2(1 kg)(–3.6 m/s) 2 + 1/2(3 kg)(1.2m/s) 2 = 8.6 J 00 0

38 Example Problem #8.50 (a) x cm = (m 1 x 1 + m 2 x 2 ) / (m 1 + m 2 ) = (1800 kg)(40.0 m) / 3000 kg = 24.0 m ahead of m 1 (or 16.0 m behind m 2 ) x = 0 (b) P = m 1 v 1 + m 2 v 2 = (1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s) = 5.04  10 4 kg  m/s (c) v cm = (m 1 v 1 + m 2 v 2 ) / (m 1 + m 2 ) = [(1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s)/3000 kg = 16.8 m/s

39 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). E i = mgh. E f = 1/2mv 2 + (1/2)I  2. But: I = 1/2mR 2, and  v/R So: mgh = 1/2mv 2 + (1/2)(1/2mR 2  v/R) 2 = (3/4)mv 2, or v = SQRT(4/3gh). Example Problem h

40 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 2 nd Law), find the final speed. v cm E 1 = mgh = mgsin  E 2 = 1/2mv 2 + 1/2I  2 = 1/2mv 2 + 1/2(2/5mR 2 )(v/R) 2 = 1/2mv 2 ( 1+ 2/5) = 7/10mv 2 = mgsin  ; v = 1.2(gsin  ) 1/2  h


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