# Dynamics_Part10.pdf

## Civil Engineering 2120 with Bowman at Marquette University *

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Engineering Mechanics - Dynamics Chapter 15 Given energy W A h 1 2 m A v A1 2 = momentum m A − v A1 m A v A2 m P v P2 += restitution ev A1 v A2 v P2 −= energy 1 2 m P v P2 2 1 2 kδ st 2 + 1 2 k δδ st +() 2 W P δ−= v A1 v A2 v P2 δ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A1 v A2 , v P2 , δ,()= v A2 v P2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 13.43− 27.04− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = δ 2.61 ft= Problem 15-62 The man A has weight W A and jumps from rest onto a platform P that has weight W P . The platform is mounted on a spring, which has stiffness k. If the coefficient of restitution between the man and the platform is e, and the man holds himself rigid during the motion, determine the required height h of the jump if the maximum compression of the spring becomes δ. Given: W A 100 lb= W P 60 lb= δ 2ft= k 200 lb ft = g 32.2 ft s 2 = e 0.6= Solution: m A W A g = m P W P g = δ st W P k = Guesses v A1 1 ft s = v A2 1 ft s = v P2 1− ft s = h 21 ft= Given energy W A h 1 2 m A v A1 2 = momentum m A − v A1 m A v A2 m P v P2 += restitution ev A1 v A2 v P2 −= 341 Engineering Mechanics - Dynamics Chapter 15 energy 1 2 m P v P2 2 1 2 kδ st 2 + 1 2 kδ 2 W P δδ st −()−= v A1 v A2 v P2 h ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A1 v A2 , v P2 , h,()= v A2 v P2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 7.04− 17.61− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = h 4.82 ft= Problem 15-63 The collar B of weight W B is at rest, and when it is in the position shown the spring is unstretched. If another collar A of weight W A strikes it so that B slides a distance b on the smooth rod before momentarily stopping, determine the velocity of A just after impact, and the average force exerted between A and B during the impact if the impact occurs in time Δt. The coefficient of restitution between A and B is e. Units Used: kip 10 3 lb= Given: W B 10 lb= W A 1lb= k 20 lb ft = g 32.2 ft s 2 = a 3ft= b 4ft= Δt 0.002 s= e 0.5= Solution: Guesses v A1 1 ft s = v A2 1 ft s = v B2 1 ft s = F 1lb= Given W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A1 W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 += ev A1 v B2 v A2 −= 342 Engineering Mechanics - Dynamics Chapter 15 W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A1 FΔt− W A g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A2 = 1 2 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 2 1 2 ka 2 b 2 + a− () 2 = v A1 v A2 v B2 F ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A1 v A2 , v B2 , F,()= v A2 42.80− ft s = F 2.49 kip= *Problem 15-64 If the girl throws the ball with horizontal velocity v A , determine the distance d so that the ball bounces once on the smooth surface and then lands in the cup at C. Given: v A 8 ft s = g 32.2 ft s 2 = e 0.8= h 3ft= Solution: t B 2 h g = t B 0.43 s= v By1 gt B = v By1 13.90 ft s = v By2 ev By1 = v By2 11.12 ft s = t C 2v By2 g = t C 0.69 s= dv A t B t C +()= d 8.98 ft= Problem 15-65 The ball is dropped from rest and falls a distance h before striking the smooth plane at A. If the coefficient of restitution is e, determine the distance R to where it again strikes the plane at B. Given: h 4ft= c 3= 343 Engineering Mechanics - Dynamics Chapter 15 e 0.8= d 4= g 32.2 ft s 2 = Solution: θ atan c d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 36.87 deg= v A1 2gh= v A1 16.05 ft s = v A1n v A1 cos θ()= v A1t v A1 sin θ()= v A2n ev A1n = v A2t v A1t = v A2x v A2n sin θ() v A2t cos θ()+= v A2x 13.87 ft s = v A2y v A2n cos θ() v A2t sin θ()−= v A2y 2.44 ft s = Guesses t 1s= R 10 ft= Given R cos θ() v A2x t= R− sin θ() g− 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 2 v A2y t+= R t ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find Rt,()= t 0.80 s= R 13.82 ft= Problem 15-66 The ball is dropped from rest and falls a distance h before striking the smooth plane at A. If it rebounds and in time t again strikes the plane at B, determine the coefficient of restitution e between the ball and the plane. Also, what is the distance R? Given: h 4ft= c 3= g 32.2 ft s 2 = t 0.5 s= d 4= Solution: θ atan c d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 36.87 deg= v A1 2gh= v A1 16.05 ft s = v A1n v A1 cos θ()= v A1t v A1 sin θ()= v A2t v A1t = 344 Engineering Mechanics - Dynamics Chapter 15 Guesses e 0.8= R 10 ft= v A2n 1 ft s = v A2x 1 ft s = v A2y 1 ft s = Given v A2n ev A1n = v A2x v A2n sin θ() v A2t cos θ()+= v A2y v A2n cos θ() v A2t sin θ()−= R cos θ() v A2x t= R− sin θ() g− 2 t 2 v A2y t+= e R v A2n v A2x v A2y ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find eR, v A2n , v A2x , v A2y ,()= R 7.23 ft= e 0.502= Problem 15-67 The ball of mass m b is thrown at the suspended block of mass m B with velocity v b . If the coefficient of restitution between the ball and the block is e, determine the maximum height h to which the block will swing before it momentarily stops. Given: m b 2kg= m B 20 kg= e 0.8= v b 4 m s = g 9.81 m s 2 = Solution: Guesses v A 1 m s = v B 1 m s = h 1m= Given momentum m b v b m b v A m B v B += restitution ev b v B v A −= energy 1 2 m B v B 2 m B gh= v A v B h ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A v B , h,()= v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.55− 0.65 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = h 21.84 mm= *Problem 15-68 The ball of mass m b is thrown at the suspended block of mass m B with a velocity of v b . If the time of impact between the ball and the block is Δt, determine the average normal force exerted on the block 345 Engineering Mechanics - Dynamics Chapter 15 during this time. Given: kN 10 3 N= m b 2kg= v b 4 m s = g 9.81 m s 2 = m B 20 kg= e 0.8= Δt 0.005 s= Solution: Guesses v A 1 m s = v B 1 m s = F 1N= Given momentum m b v b m b v A m B v B += restitution ev b v B v A −= momentum B 0 FΔt+ m B v B = v A v B F ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A v B , F,()= v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.55− 0.65 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = F 2.62 kN= Problem 15-69 A ball is thrown onto a rough floor at an angle θ. If it rebounds at an angle φ and the coefficient of kinetic friction is μ, determine the coefficient of restitution e. Neglect the size of the ball. Hint: Show that during impact, the average impulses in the x and y directions are related by I x = μΙ y . Since the time of impact is the same, F x Δt = μF y Δt or F x = μF y . Solution: ev 1 sin θ() v 2 sin φ()= v 2 v 1 e sin θ() sin φ() ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = [ 1 ] + → ()mv 1 cos θ() F x Δt− mv 2 cos φ()= [ 2 ] F x mv 1 cos θ() mv 2 cos φ()− Δt = (+↓) mv 1 sin θ() F y Δt− m− v 2 sin φ()= 346 Engineering Mechanics - Dynamics Chapter 15 F y mv 1 sin θ() mv 2 sin φ()+ Δt = [ 3 ] Since F x = μF y , from Eqs [2] and [3] mv 1 cos θ() mv 2 cos φ()− Δt μ mv 1 sin θ() mv 2 sin φ()+() Δt = v 2 v 1 cos θ() μ sin θ()− μ sin φ() cos φ()+ = [ 4 ] Substituting Eq. [4] into [1] yields: e sin φ() sin θ() cos θ() μ sin θ()− μ sin φ() cos φ()+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Problem 15-70 A ball is thrown onto a rough floor at an angle of θ. If it rebounds at the same angle φ , determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is e. Hint: Show that during impact, the average impulses in the x and y directions are related by I x = μI y . Since the time of impact is the same, F x Δt = μF y Δt or F x = μF y. Solution: ev 1 sin θ() v 2 sin φ()= v 2 v 1 e sin θ() sin φ() ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = [ 1 ] + → ()mv 1 cos θ() F x Δt− mv 2 cos φ()= [ 2 ] F x mv 1 cos θ() mv 2 cos φ()− Δt = (+↓) mv 1 sin θ() F y Δt− m− v 2 sin φ()= F y mv 1 sin θ() mv 2 sin φ()+ Δt = [ 3 ] 347 Engineering Mechanics - Dynamics Chapter 15 Since F x = μF y , from Eqs [2] and [3] mv 1 cos θ() mv 2 cos φ()− Δt μ mv 1 sin θ() mv 2 sin φ()+() Δt = v 2 v 1 cos θ() μ sin θ()− μ sin φ() cos φ()+ = [ 4 ] Substituting Eq. [4] into [1] yields: e sin φ() sin θ() cos θ() μ sin θ()− μ sin φ() cos φ()+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Given θ 45 deg= φ 45 deg= e 0.6= Guess μ 0.2= Given e sin φ() sin θ() cos θ() μ sin θ()− μ sin φ() cos φ()+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = μ Find μ()= μ 0.25= Problem 15-71 The ball bearing of weight W travels over the edge A with velocity v A . Determine the speed at which it rebounds from the smooth inclined plane at B. Take e = 0.8. Given: W 0.2 lb= θ 45 deg= e 0.8= v A 3 ft s = g 32.2 ft s 2 = Solution: Guesses v B1x 1 ft s = v B1y 1 ft s = v B2n 1 ft s = v B2t 1 ft s = t 1s= R 1ft= Given v B1x v A = v A tRcos θ()= 1− 2 gt 2 R− sin θ()= v B1y g− t= v B1x cos θ() v B1y sin θ()− v B2t = 348 Engineering Mechanics - Dynamics Chapter 15 ev B1y − cos θ() v B1x sin θ( )−()v B2n = v B1x v B1y v B2n v B2t t R ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v B1x v B1y , v B2n , v B2t , t, R,()= v B1x v B1y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3.00 6.00− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = t 0.19 s= R 0.79 ft= v B2n v B2t ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.70 6.36 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = v B2n v B2t ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 6.59 ft s = *Problem 15-72 The drop hammer H has a weight W H and falls from rest h onto a forged anvil plate P that has a weight W P . The plate is mounted on a set of springs that have a combined stiffness k T . Determine (a) the velocity of P and H just after collision and (b) the maximum compression in the springs caused by the impact. The coefficient of restitution between the hammer and the plate is e. Neglect friction along the vertical guideposts A and B. Given: W H 900 lb= k T 500 lb ft = W P 500 lb= g 32.2 ft s 2 = h 3ft= e 0.6= Solution: δ st W P k T = v H1 2gh= Guesses v H2 1 ft s = v P2 1 ft s = δ 2ft= Given W H g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v H1 W H g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v H2 W P g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v P2 += ev H1 v P2 v H2 −= 349 Engineering Mechanics - Dynamics Chapter 15 1 2 k T δ st 2 1 2 W P g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v P2 2 + 1 2 k T δ 2 W P δδ st −()−= v H2 v P2 δ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v H2 v P2 , δ,()= v H2 v P2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 5.96 14.30 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = δ 3.52 ft= Problem 15-73 It was observed that a tennis ball when served horizontally a distance h above the ground strikes the smooth ground at B a distance d away. Determine the initial velocity v A of the ball and the velocity v B (and θ) of the ball just after it strikes the court at B. The coefficient of restitution is e. Given: h 7.5 ft= d 20 ft= e 0.7= g 32.2 ft s 2 = Solution: Guesses v A 1 ft s = v B2 1 ft s = v By1 1 ft s = θ 10 deg= t 1s= Given h 1 2 gt 2 = dv A t= ev By1 v B2 sin θ()= v By1 gt= v A v B2 cos θ()= 350 Engineering Mechanics - Dynamics Chapter 15 v A t v By1 v B2 θ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A t, v By1 , v B2 , θ,()= v A 29.30 ft s = v B2 33.10 ft s = θ 27.70 deg= Problem 15-74 The tennis ball is struck with a horizontal velocity v A , strikes the smooth ground at B, and bounces upward at θ = θ 1 . Determine the initial velocity v A , the final velocity v B , and the coefficient of restitution between the ball and the ground. Given: h 7.5 ft= d 20 ft= θ 1 30 deg= g 32.2 ft s 2 = Solution: θθ 1 = Guesses v A 1 ft s = t 1s= v By1 1 ft s = v B2 1 ft s = e 0.5= Given h 1 2 gt 2 = dv A t= v By1 gt= ev By1 v B2 sin θ()= v A v B2 cos θ()= v A t v By1 v B2 e ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A t, v By1 , v B2 , e,()= v A 29.30 ft s = v B2 33.84 ft s = e 0.77= Problem 15-75 The ping-pong ball has mass M. If it is struck with the velocity shown, determine how high h it rises above the end of the smooth table after the rebound. The coefficient of restitution is e. 351 Engineering Mechanics - Dynamics Chapter 15 Given: M 2gm= a 2.25 m= e 0.8= b 0.75 m= θ 30 deg= g 9.81 m s 2 = v 18 m s = Solution: Guesses v 1x 1 m s = v 1y 1 m s = v 2x 1 m s = v 2y 1 m s = t 1 1s= t 2 2s= h 1m= Given v 1x v cos θ()= avcos θ()t 1 = v 1y gt 1 v sin θ()+= v 2x v 1x = ev 1y v 2y = bv 2x t 2 = hv 2y t 2 g 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 2 2 −= v 1x v 1y v 2x v 2y t 1 t 2 h ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v 1x v 1y , v 2x , v 2y , t 1 , t 2 , h,()= v 1x v 1y v 2x v 2y ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 15.59 10.42 15.59 8.33 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ m s = t 1 t 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.14 0.05 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ s= h 390 mm= *Problem 15-76 The box B of weight W B is dropped from rest a distance d from the top of the plate P of weight W P , which is supported by the spring having a stiffness k. Determine the maximum compression imparted to the spring. Neglect the mass of the spring. 352 Engineering Mechanics - Dynamics Chapter 15 Given: W B 5lb= W P 10 lb= g 32.2 ft s 2 = k 30 lb ft = d 5ft= e 0.6= Solution: δ st W P k = v B1 2gd= Guesses v B2 1 ft s = v P2 1 ft s = δ 2ft= Given W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B1 W B g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 W P g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v P2 += ev B1 v P2 v B2 −= 1 2 kδ st 2 1 2 W P g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v P2 2 + 1 2 kδ 2 W P δδ st −()−= v B2 v P2 δ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v B2 v P2 , δ,()= v B2 v P2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.20− 9.57 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = δ 1.31 ft= Problem 15-77 A pitching machine throws the ball of weight M towards the wall with an initial velocity v A as shown. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall and (c) the distance d from the wall to where it strikes the ground at C. Given: M 0.5 kg= a 3m= v A 10 m s = b 1.5 m= e 0.5= θ 30 deg= g 9.81 m s 2 = 353 Engineering Mechanics - Dynamics Chapter 15 v Bx1 1 m s = v Bx2 1 m s = v By1 1 m s = v By2 1 m s = h 1m= d 1m= t 1 1s= t 2 1s= Given v A cos θ()t 1 a= bv A sin θ()t 1 + 1 2 gt 1 2 − h= v By2 v By1 = v A sin θ() gt 1 − v By1 = dv Bx2 t 2 = hv By2 t 2 + 1 2 gt 2 2 − 0= v A cos θ() v Bx1 = ev Bx1 v Bx2 = v Bx1 v By1 v Bx2 v By2 h t 1 t 2 d ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v Bx1 v By1 , v Bx2 , v By2 , h, t 1 , t 2 , d,()= v Bx1 v By1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 8.81 m s = v Bx2 v By2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.62 m s = d 3.96 m= Problem 15-78 The box of weight W b slides on the surface for which the coefficient of friction is μ k . The box has velocity v when it is a distance d from the plate. If it strikes the plate, which has weight W p and is held in position by an unstretched spring of stiffness k, determine the maximum compression imparted to the spring. The coefficient of restitution between the box and the plate is e. Assume that the plate slides smoothly. 354 GuessesSolution: Engineering Mechanics - Dynamics Chapter 15 Given: W b 20 lb= W p 10 lb= μ k 0.3= k 400 lb ft = v 15 ft s = e 0.8= d 2ft= g 32.2 ft s 2 = Solution: Guesses v b1 1 ft s = v b2 1 ft s = v p2 1 ft s = δ 1ft= Given 1 2 W b g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2 μ k W b d− 1 2 W b g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v b1 2 = W b g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v b1 W b g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v b2 W p g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v p2 += ev b1 v p2 v b2 −= 1 2 W p g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v p2 2 1 2 kδ 2 = v b1 v b2 v p2 δ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v b1 v b2 , v p2 , δ,()= v b1 v b2 v p2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 13.65 5.46 16.38 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ft s = δ 0.456 ft= Problem 15-79 The billiard ball of mass M is moving with a speed v when it strikes the side of the pool table at A. If the coefficient of restitution between the ball and the side of the table is e, determine the speed of the ball just after striking the table twice, i.e., at A, then at B. Neglect the size of the ball. 355 Engineering Mechanics - Dynamics Chapter 15 Given: M 200 gm= v 2.5 m s = θ 45 deg= e 0.6= Solution: Guesses v 2 1 m s = θ 2 1 deg= v 3 1 m s = θ 3 1 deg= Given evsin θ() v 2 sin θ 2 ()= v cos θ () v 2 cos θ 2 ()= ev 2 cos θ 2 () v 3 sin θ 3 ()= v 2 sin θ 2 () v 3 cos θ 3 ()= v 2 v 3 θ 2 θ 3 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v 2 v 3 , θ 2 , θ 3 ,()= v 2 v 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.06 1.50 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = θ 2 θ 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 31.0 45.0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ deg= v 3 1.500 m s = *Problem 15-80 The three balls each have the same mass m. If A is released from rest at θ, determine the angle φ to which C rises after collision. The coefficient of restitution between each ball is e. Solution: Energy 0 l 1 cos θ()−()mg+ 1 2 mv A 2 = v A 2 1 cos θ()−()gl= Collision of ball A with B: mv A 0+ mv' A mv' B += ev A v' B v' A −= v' B 1 2 1 e+()v' B = Collision of ball B with C: 356 Engineering Mechanics - Dynamics Chapter 15 mv' B 0+ m v'' B m v'' C += ev' B v'' C v'' B −= v'' C 1 4 1 e+() 2 v A = Energy 1 2 m v'' c 2 0+ 0 l 1 cos φ()−()mg+= 1 2 1 16 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 e+() 4 2( ) 1 cos θ()−()1 cos φ()−()= 1 e+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 1 cos θ()−()1 cos φ()−= φ acos 1 1 e+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 1 cos θ()−()− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Problem 15-81 Two smooth billiard balls A and B each have mass M. If A strikes B with a velocity v A as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e. Neglect the size of each ball. Given: M 0.2 kg= θ 40 deg= v A 1.5 m s = e 0.85= Solution: Guesses v A2 1 m s = v B2 1 m s = θ 2 20 deg= Given M− v A cos θ() Mv B2 Mv A2 cos θ 2 ()+= ev A cos θ() v A2 cos θ 2 ()v B2 −= v A sin θ() v A2 sin θ 2 ()= v A2 v B2 θ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , θ 2 ,()= θ 2 95.1 deg= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.968 1.063− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = 357 Engineering Mechanics - Dynamics Chapter 15 The two hockey pucks A and B each have a mass M. If they collide at O and are deflected along the colored paths, determine their speeds just after impact. Assume that the icy surface over which they slide is smooth. Hint: Since the y' axis is not along the line of impact, apply the conservation of momentum along the x' and y' axes. Given: M 250 g= θ 1 30 deg= v 1 40 m s = θ 2 20 deg= v 2 60 m s = θ 3 45 deg= Solution: Initial Guess: v A2 5 m s = v B2 4 m s = Given Mv 2 cos θ 3 ()Mv 1 cos θ 1 ()+ Mv A2 cos θ 1 ()Mv B2 cos θ 2 ()+= M− v 2 sin θ 3 ()Mv 1 sin θ 1 ()+ Mv A2 sin θ 1 ()Mv B2 sin θ 2 ()−= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A2 v B2 ,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 6.90 75.66 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-83 Two smooth coins A and B, each having the same mass, slide on a smooth surface with the motion shown. Determine the speed of each coin after collision if they move off along the dashed paths. Hint: Since the line of impact has not been defined, apply the conservation of momentum along the x and y axes, respectively. 358 Problem 15-82 Engineering Mechanics - Dynamics Chapter 15 Given: v A1 0.5 ft s = v B1 0.8 ft s = α 30 deg= β 45 deg= γ 30 deg= c 4= d 3= Solution: Guesses v B2 0.25 ft s = v A2 0.5 ft s = Given v A1 − c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B1 sin γ()− v A2 − sin β() v B2 cos α()−= v A1 − d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B1 cos γ()+ v A2 cos β() v B2 sin α()−= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v A2 v B2 ,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.766 0.298 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ft s = *Problem 15-84 The two disks A and B have a mass M A and M B , respectively. If they collide with the initial velocities shown, determine their velocities just after impact. The coefficient of restitution is e. Given: M A 3kg= M B 5kg= θ 60 deg= v B1 7 m s = 359 Engineering Mechanics - Dynamics Chapter 15 v A1 6 m s = e 0.65= Solution: Guesses v A2 1 m s = v B2 1 m s = θ 2 20 deg= Given M A v A1 M B v B1 cos θ()− M A v A2 M B v B2 cos θ 2 ()+= ev A1 v B1 cos θ()+()v B2 cos θ 2 ()v A2 −= v B1 sin θ() v B2 sin θ 2 ()= v A2 v B2 θ 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , θ 2 ,()= θ 2 68.6 deg= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3.80− 6.51 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-85 Two smooth disks A and B each have mass M. If both disks are moving with the velocities shown when they collide, determine their final velocities just after collision. The coefficient of restitution is e. Given: M 0.5 kg= c 4= v A1 6 m s = e 0.75= d 3= v B1 4 m s = Solution: Guesses v A2 1 m s = v B2 1 m s = θ A 10 deg= θ B 10 deg= Given v A1 0() v A2 sin θ A ()= v B1 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 sin θ B ()= Mv B1 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Mv A1 − Mv A2 cos θ A ()Mv B2 cos θ B ()−= ev A1 v B1 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ v A2 cos θ A ()v B2 cos θ B ()+= 360 Engineering Mechanics - Dynamics Chapter 15 v A2 v B2 θ A θ B ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , θ A , θ B ,()= θ A θ B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0.00 32.88 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ deg= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.35 5.89 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = Problem 15-86 Two smooth disks A and B each have mass M. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision B travels along a line angle θ counterclockwise from the y axis. Given: M 0.5 kg= c 4= v A1 6 m s = θ B 30 deg= d 3= v B1 4 m s = Solution: Guesses v A2 2 m s = v B2 1 m s = θ A 10 deg= e 0.5= Given v A1 0 v A2 sin θ A ()= v B1 c c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B2 cos θ B ()= Mv B1 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Mv A1 − Mv A2 cos θ A ()Mv B2 sin θ B ()−= ev A1 v B1 d c 2 d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ v A2 cos θ A ()v B2 sin θ B ()+= v A2 v B2 θ A e ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A2 v B2 , θ A , e,()= v A2 v B2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.75− 3.70 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s = e 0.0113= Problem 15-87 Two smooth disks A and B have the initial velocities shown just before they collide at O. If they have masses m A and m B , determine their speeds just after impact. The coefficient of restitution is e. 361 Engineering Mechanics - Dynamics Chapter 15 Given: v A 7 m s = m A 8kg= c 12= e 0.5= v B 3 m s = m B 6kg= d 5= Solution: θ atan d c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 22.62 deg= Guesses v A2t 1 m s = v A2n 1 m s = v B2t 1 m s = v B2n 1 m s = Given v B cos θ() v B2t = v A − cos θ() v A2t = m B v B sin θ() m A v A sin θ()− m B v B2n m A v A2n += ev B v A +()sin θ() v A2n v B2n −= v A2t v A2n v B2t v B2n ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A2t v A2n , v B2t , v B2n ,()= v A2t v A2n v B2t v B2n ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 6.46− 0.22− 2.77 2.14− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ m s = v A2 v A2t 2 v A2n 2 += v A2 6.47 m s = v B2 v B2t 2 v B2n 2 += v B2 3.50 m s = *Problem 15-88 The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has weight W, and the coefficient of restitution between the “stones” is e, determine their speeds just after collision. Initially A has velocity v A1 and B is at rest. Neglect friction. Given: W 47 lb= v A1 8 ft s = e 0.8= θ 30 deg= 362 Engineering Mechanics - Dynamics Chapter 15 Solution: Guesses v A2t 1 ft s = v A2n 1 ft s = v B2t 1 ft s = v B2n 1 ft s = Given v A1 sin θ() v A2t = 0 v B2t = v A1 cos θ() v A2n v B2n += ev A1 cos θ() v B2n v A2n −= v A2t v A2n v B2t v B2n ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find v A2t v A2n , v B2t , v B2n ,()= v A2t v A2n v B2t v B2n ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 4.00 0.69 0.00 6.24 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ft s = v A2 v A2t 2 v A2n 2 += v A2 4.06 ft s = v B2 v B2t 2 v B2n 2 += v B2 6.24 ft s = Problem 15-89 The two billiard balls A and B are originally in contact with one another when a third ball C strikes each of them at the same time as shown. If ball C remains at rest after the collision, determine the coefficient of restitution. All the balls have the same mass. Neglect the size of each ball. Solution: Conservation of “x” momentum: mv 2mv'cos 30 deg()= v 2v' cos 30 deg()= 1() Coefficient of restitution: e v' v cos 30 deg() = 2() Substituiting Eq. (1) into Eq. (2) yields: 363 Engineering Mechanics - Dynamics Chapter 15 e v' 2v' cos 30 deg() 2 = e 2 3 = Problem 15-90 Determine the angular momentum of particle A of weight W about point O. Use a Cartesian vector solution. Given: W 2lb= a 3ft= b 2ft= v A 12 ft s = c 2ft= g 32.2 ft s 2 = d 4ft= Solution: r OA c− ab+ d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r v c b− d− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = v Av v A r v r v = H O r OA Wv Av ()×= H O 1.827− 0.000 0.914− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ slug ft 2 s ⋅= Problem 15-91 Determine the angular momentum H O of the particle about point O. Given: M 1.5 kg= v 6 m s = a 4m= b 3m= c 2m= d 4m= 364 Engineering Mechanics - Dynamics Chapter 15 Solution: r OA c− b− d ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB c a− d− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = v A v r AB r AB = H O r OA Mv A ()×= H O 42.0 0.0 21.0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kg m 2 ⋅ s = *Problem 15-92 Determine the angular momentum H O of each of the particles about point O. Given: θ 30 deg= φ 60 deg= m A 6kg= c 2m= m B 4kg= d 5m= m C 2kg= e 2m= v A 4 m s = f 1.5 m= v B 6 m s = g 6m= v C 2.6 m s = h 2m= a 8m= l 5= b 12 m= n 12= Solution: H AO am A v A sin φ() bm A v A cos φ()−= H AO 22.3 kg m 2 ⋅ s = H BO f− m B v B cos θ() em B v B sin θ()+= H BO 7.18− kg m 2 ⋅ s = H CO h− m C n l 2 n 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v C gm C l l 2 n 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v C −= H CO 21.60− kg m 2 ⋅ s = 365 Engineering Mechanics - Dynamics Chapter 15 Problem 15-93 Determine the angular momentum H P of each of the particles about point P. Given: θ 30 deg= φ 60 deg= a 8m= f 1.5 m= m A 6kg= b 12 m= g 6m= v A 4 m s = c 2m= h 2m= m B 4kg= v B 6 m s = d 5m= l 5= m C 2kg= v C 2.6 m s = e 2m= n 12= Solution: H AP m A v A sin φ()ad−()m A v A cos φ()bc−()−= H AP 57.6− kg m 2 ⋅ s = H BP m B v B cos θ()cf−()m B v B sin θ()de+()+= H BP 94.4 kg m 2 ⋅ s = H CP m C − n l 2 n 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v C ch+()m C l l 2 n 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v C dg+()−= H CP 41.2− kg m 2 ⋅ s = Problem 15-94 Determine the angular momentum H O of the particle about point O. Given: W 10 lb= d 9ft= v 14 ft s = e 8ft= a 5ft= f 4ft= b 2ft= g 5ft= c 3ft= h 6ft= 366 Engineering Mechanics - Dynamics Chapter 15 Solution: r OA f− g h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB fe+ dg− h− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = v A v r AB r AB = H O r OA Wv A ()×= H O 16.78− 14.92 23.62− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ slug ft 2 s ⋅= Problem 15-95 Determine the angular momentum H P of the particle about point P. Given: W 10 lb= d 9ft= v 14 ft s = e 8ft= a 5ft= f 4ft= b 2ft= g 5ft= c 3ft= h 6ft= Solution: r PA f− c− bg+ ha− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = r AB fe+ dg− h− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = v A v r AB r AB = H P r PA Wv A ()×= H P 14.30− 9.32− 34.81− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ slug ft 2 s ⋅= *Problem 15-96 Determine the total angular momentum H O for the system of three particles about point O. All the particles are moving in the x-y plane. Given: m A 1.5 kg= a 900 mm= 367 Engineering Mechanics - Dynamics Chapter 15 v A 4 m s = b 700 mm= m B 2.5 kg= c 600 mm= v B 2 m s = d 800 mm= m C 3kg= e 200 mm= v C 6 m s = Solution: H O a 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m A 0 v A − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ × c b 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m B v B − 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ×+ d− e− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ m C 0 v C − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ×+= H O 0.00 0.00 12.50 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kg m 2 ⋅ s = Problem 15-97 Determine the angular momentum H O of each of the two particles about point O. Use a scalar solution. Given: m A 2kg= c 1.5 m= m B 1.5 kg= d 2m= e 4m= v A 15 m s = f 1m= v B 10 m s = θ 30 deg= a 5m= l 3= b 4m= n 4= 368 Engineering Mechanics - Dynamics Chapter 15 Solution: H OA m A − n n 2 l 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A cm A l n 2 l 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A d−= H OA 72.0− kg m 2 ⋅ s = H OB m B − v B cos θ()em B v B sin θ()f−= H OB 59.5− kg m 2 ⋅ s = Problem 15-98 Determine the angular momentum H P of each of the two particles about point P. Use a scalar solution. Given: m A 2kg= c 1.5 m= d 2m= m B 1.5 kg= e 4m= v A 15 m s = f 1m= v B 10 m s = θ 30 deg= a 5m= l 3= b 4m= n 4= Solution: H PA m A n n 2 l 2 + v A bc−()m A l n 2 l 2 + v A ad+()−= H PA 66.0− kg m 2 ⋅ s = H PB m B − v B cos θ()be+()m B v B sin θ()af−()+= H PB 73.9− kg m 2 ⋅ s = Problem 15-99 The ball B has mass M and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque M = at 2 + bt + c, determine the speed of the ball when t = t 1 . The ball has a speed v = v 0 when t = 0. 369 Engineering Mechanics - Dynamics Chapter 15 Given: M 10 kg= a 3 Nm⋅ s 2 = b 5 Nm⋅ s = c 2Nm⋅= t 1 2s= v 0 2 m s = L 1.5 m= Solution: Principle of angular impulse momentum Mv 0 L 0 t 1 tat 2 bt+ c+ ⌠ ⎮ ⌡ d+ Mv 1 L= v 1 v 0 1 ML 0 t 1 tat 2 bt+ c+ ⌠ ⎮ ⌡ d+= v 1 3.47 m s = *Problem 15-100 The two blocks A and B each have a mass M 0 . The blocks are fixed to the horizontal rods, and their initial velocity is v' in the direction shown. If a couple moment of M is applied about shaft CD of the frame, determine the speed of the blocks at time t. The mass of the frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks. Given: M 0 0.4 kg= a 0.3 m= v' 2 m s = M 0.6 N m⋅= t 3s= Solution: 2aM 0 v' M t+ 2aM 0 v= 370 Engineering Mechanics - Dynamics Chapter 15 vv' Mt 2aM 0 += v 9.50 m s = Problem 15-101 The small cylinder C has mass m C and is attached to the end of a rod whose mass may be neglected. If the frame is subjected to a couple M = at 2 + b, and the cylinder is subjected to force F, which is always directed as shown, determine the speed of the cylinder when t = t 1 . The cylinder has a speed v 0 when t = 0. Given: m C 10 kg= t 1 2s= a 8N m s 2 = v 0 2 m s = d 0.75 m= b 5Nm⋅= e 4= F 60 N= f 3= Solution: m C v 0 d 0 t 1 tat 2 b+ ⌠ ⎮ ⌡ d+ f e 2 f 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Fdt 1 + m C v 1 d= v 1 v 0 1 m C d 0 t 1 tat 2 b+ ⌠ ⎮ ⌡ d f e 2 f 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Fdt 1 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ += v 1 13.38 m s = Problem 15-102 A box having a weight W is moving around in a circle of radius r A with a speed v A1 while connected to the end of a rope. If the rope is pulled inward with a constant speed v r , determine the speed of the box at the instant r = r B . How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box. Given: W 8lb= r A 2ft= 371 Engineering Mechanics - Dynamics Chapter 15 v A1 5 ft s = v r 4 ft s = r B 1ft= g 32.21 ft s 2 = Solution: W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r A v A1 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r B v Btangent = v Btangent r A v A1 r B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v Btangent 10.00 ft s = v B v Btangent 2 v r 2 += v B 10.8 ft s = U AB 1 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B 2 1 2 W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A1 2 −= U AB 11.3 ft lb⋅= Problem 15-103 An earth satellite of mass M is launched into a free-flight trajectory about the earth with initial speed v A when the distance from the center of the earth is r A . If the launch angle at this position is φ A determine the speed v B of the satellite and its closest distance r B from the center of the earth. The earth has a mass M e . Hint: Under these conditions, the satellite is subjected only to the earth’s gravitational force, F, Eq. 13-1. For part of the solution, use the conservation of energy. Units used: Mm 10 3 km= Given: φ A 70 deg= M 700 kg= v A 10 km s = G 6.673 10 11− × Nm 2 ⋅ kg 2 = r A 15 Mm= M e 5.976 10 24 × kg= 372 Engineering Mechanics - Dynamics Chapter 15 Solution: Guesses v B 10 km s = r B 10 Mm= Given Mv A sin φ A ()r A Mv B r B = 1 2 Mv A 2 GM e M r A − 1 2 Mv B 2 GM e M r B −= v B r B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v B r B ,()= v B 10.2 km s = r B 13.8 Mm= *Problem 15-104 The ball B has weight W and is originally rotating in a circle. As shown, the cord AB has a length of L and passes through the hole A, which is a distance h above the plane of motion. If L/2 of the cord is pulled through the hole, determine the speed of the ball when it moves in a circular path at C. Given: W 5lb= L 3ft= h 2ft= g 32.2 ft s 2 = Solution: θ B acos h L ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ B 48.19 deg= Guesses T B 1lb= T C 1lb= v B 1 ft s = v C 1 ft s = θ C 10 deg= Given T B cos θ B ()W− 0= T B sin θ B () W g v B 2 L sin θ B () ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = T C cos θ C ()W− 0= T C sin θ C () W g v C 2 L 2 sin θ C () ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v B L sin θ B () W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v C L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ C ()= 373 Engineering Mechanics - Dynamics Chapter 15 T B T C v B v C θ C ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find T B T C , v B , v C , θ C ,()= T B T C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 7.50 20.85 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lb= θ C 76.12 deg= v B 8.97 ft s = v C 13.78 ft s = Problem 15-105 The block of weight W rests on a surface for which the kinetic coefficient of friction is μ k . It is acted upon by a radial force F R and a horizontal force F H , always directed at angle θ from the tangent to the path as shown. If the block is initially moving in a circular path with a speed v 1 at the instant the forces are applied, determine the time required before the tension in cord AB becomes T. Neglect the size of the block for the calculation. Given: W 10 lb= μ k 0.5= F R 2lb= T 20 lb= F H 7lb= r 4ft= v 1 2 ft s = g 32.2 ft s 2 = θ 30 deg= Solution: Guesses t 1s= v 2 1 ft s = Given W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 rF H cos θ()rt+ μ k Wrt− W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2 r= F R F H sin θ()+ T− W g − v 2 2 r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = t v 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find tv 2 ,()= v 2 13.67 ft s = t 3.41 s= 374 Engineering Mechanics - Dynamics Chapter 15 Problem 15-106 The block of weight W is originally at rest on the smooth surface. It is acted upon by a radial force F R and a horizontal force F H , always directed at θ from the tangent to the path as shown. Determine the time required to break the cord, which requires a tension T. What is the speed of the block when this occurs? Neglect the size of the block for the calculation. Given: W 10 lb= θ 30 deg= F R 2lb= T 30 lb= F H 7lb= r 4ft= v 1 0 ft s = g 32.2 ft s 2 = Solution: Guesses t 1s= v 2 1 ft s = Given W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 1 rF H cos θ()rt+ W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v 2 r= F R F H sin θ()+ T− W g − v 2 2 r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = t v 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find tv 2 ,()= v 2 17.76 ft s = t 0.91 s= Problem 15-107 The roller-coaster car of weight W starts from rest on the track having the shape of a cylindrical helix. If the helix descends a distance h for every one revolution, determine the time required for the car to attain a speed v. Neglect friction and the size of the car. Given: W 800 lb= h 8ft= v 60 ft s = 375 Engineering Mechanics - Dynamics Chapter 15 r 8ft= Solution: θ atan h 2πr ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 9.04 deg= F N W cos θ()− 0= F N W cos θ()= F N 790.06 lb= v t v cos θ()= v t 59.25 ft s = H A tM ⌠ ⎮ ⌡ d+ H 2 = 0 t tF N sin θ()r ⌠ ⎮ ⌡ d W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ hv t = F N sin θ()rt W g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ hv t = tW v t h F N sin θ()gr ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = t 11.9 s= *Problem 15-108 A child having mass M holds her legs up as shown as she swings downward from rest at θ 1 . Her center of mass is located at point G 1 . When she is at the bottom position θ = 0°, she suddenly lets her legs come down, shifting her center of mass to position G 2 . Determine her speed in the upswing due to this sudden movement and the angle θ 2 to which she swings before momentarily coming to rest. Treat the child’s body as a particle. Given: M 50 kg= r 1 2.80 m= g 9.81 m s 2 = θ 1 30 deg= r 2 3m= Solution: v 2b 2gr 1 1 cos θ 1 ()−()= v 2b 2.71 m s = r 1 v 2b r 2 v 2a = v 2a r 1 r 2 v 2b = v 2a 2.53 m s = θ 2 acos 1 v 2a 2 2gr 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = θ 2 27.0 deg= 376 Engineering Mechanics - Dynamics Chapter 15 Problem 15-109 A small particle having a mass m is placed inside the semicircular tube. The particle is placed at the position shown and released. Apply the principle of angular momentum about point O (ΣM 0 = H 0 ), and show that the motion of the particle is governed by the differential equation θ'' + (g / R) sin θ = 0. Solution: ΣM 0 t H 0 d d = R− mgsin θ() t mvR() d d = g sin θ() t v d d −= 2 t s d d 2 −= But, sRθ= Thus, g sin θ() R− θ''= or, θ'' g R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ sin θ()+ 0= Problem 15-110 A toboggan and rider, having a total mass M, enter horizontally tangent to a circular curve (θ 1 ) with a velocity v A . If the track is flat and banked at angle θ 2 , determine the speed v B and the angle θ of “descent”, measured from the horizontal in a vertical x–z plane, at which the toboggan exists at B. Neglect friction in the calculation. Given: M 150 kg= θ 1 90 deg= v A 70 km hr = θ 2 60 deg= r A 60 m= r B 57 m= r 55 m= 377 Engineering Mechanics - Dynamics Chapter 15 Solution: hr A r B −()tan θ 2 ()= Guesses v B 10 m s = θ 1 deg= Given 1 2 Mv A 2 Mgh+ 1 2 Mv B 2 = Mv A r A Mv B cos θ()r B = v B θ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v B θ,()= v B 21.9 m s = θ 1.1− 10 3 × deg= Problem 15-111 Water is discharged at speed v against the fixed cone diffuser. If the opening diameter of the nozzle is d, determine the horizontal force exerted by the water on the diffuser. Units Used: Mg 10 3 kg= Given: v 16 m s = θ 30 deg= d 40 mm= ρ w 1 Mg m 3 = Solution: Q π 4 d 2 v= m' ρ w Q= F x m' v− cos θ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F x 11.0 N= *Problem 15-112 A jet of water having cross-sectional area A strikes the fixed blade with speed v. Determine the horizontal and vertical components of force which the blade exerts on the water. Given: A 4in 2 = 378 dhanesh_h Mathcad - CombinedMathcads

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