# Dynamics_Part3.pdf

## Civil Engineering 2120 with Bowman at Marquette University *

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Engineering Mechanics - Dynamics Chapter 12 Problem 12–99 The projectile is launched from a height h with a velocity v 0 . Determine the range R. Solution: a x 0= a y g−= v x v 0 cos θ()= v y g− tv 0 sin θ()+= s x v 0 cos θ()t= s y 1− 2 gt 2 v 0 sin θ()t+ h+= When it hits Rv 0 cos θ()t= t R v 0 cos θ() = 0 1− 2 gt 2 v 0 sin θ()t+ h+= g− 2 R v 0 cos θ() ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 v 0 sin θ() R v 0 cos θ() + h+= Solving for R we find R v 0 2 cos θ() 2 g tan θ() tan θ() 2 2gh v 0 2 cos θ() 2 ++ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = *Problem 12-100 A car is traveling along a circular curve that has radius ρ. If its speed is v and the speed is increasing uniformly at rate a t , determine the magnitude of its acceleration at this instant. Given: ρ 50 m= v 16 m s = a t 8 m s 2 = Solution: a n v 2 ρ = a n 5.12 m s 2 = aa n 2 a t 2 += a 9.498 m s 2 = Problem 12-101 A car moves along a circular track of radius ρ such that its speed for a short period of time 0 t≤ t 2 ≤ , is v = b t + c t 2 . Determine the magnitude of its acceleration when t = t 1 . How far has it traveled at time t 1 ? 75 Engineering Mechanics - Dynamics Chapter 12 Given: ρ 250 ft= t 2 4s= b 3 ft s 2 = c 3 ft s 3 = t 1 3s= Solution: vbtct 2 += a t b 2ct+= At t 1 v 1 bt 1 ct 1 2 += a t1 b 2ct 1 += a n1 v 1 2 ρ = a 1 a t1 2 a n1 2 += a 1 21.63 ft s 2 = Distance traveled d 1 b 2 t 1 2 c 3 t 1 3 += d 1 40.5 ft= Problem 12-102 At a given instant the jet plane has speed v and acceleration a acting in the directions shown. Determine the rate of increase in the plane’s speed and the radius of curvature ρ of the path. Given: v 400 ft s = a 70 ft s 2 = θ 60 deg= Solution: Rate of increase a t a( )cos θ()= a t 35 ft s 2 = Radius of curvature a n a( )sin θ()= v 2 ρ = ρ v 2 a( )sin θ() = ρ 2639 ft= Problem 12–103 A particle is moving along a curved path at a constant speed v. The radii of curvature of the path at points P and P' are ρ and ρ', respectively. If it takes the particle time t to go from P to P', determine the acceleration of the particle at P and P'. Given: v 60 ft s = ρ 20 ft= ρ' 50 ft= t 20 s= 76 Engineering Mechanics - Dynamics Chapter 12 Solution: a v 2 ρ = a 180 ft s 2 = a' v 2 ρ' = a' 72 ft s 2 = Note that the time doesn’t matter here because the speed is constant. *Problem 12-104 A boat is traveling along a circular path having radius ρ. Determine the magnitude of the boat’s acceleration when the speed is v and the rate of increase in the speed is a t . Given: ρ 20 m= v 5 m s = a t 2 m s 2 = Solution: a n v 2 ρ = a n 1.25 m s 2 = aa t 2 a n 2 += a 2.358 m s 2 = Problem 12-105 Starting from rest, a bicyclist travels around a horizontal circular path of radius ρ at a speed v = b t 2 + c t. Determine the magnitudes of his velocity and acceleration when he has traveled a distance s 1. Given: ρ 10 m= b 0.09 m s 3 = c 0.1 m s 2 = s 1 3m= Solution: Guess t 1 1s= Given s 1 b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 3 c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 2 += t 1 Find t 1 ()= t 1 4.147 s= v 1 bt 1 2 ct 1 += v 1 1.963 m s = a t1 2bt 1 c+= a t1 0.847 m s 2 = a n1 v 1 2 ρ = a n1 0.385 m s 2 = a 1 a t1 2 a n1 2 += a 1 0.93 m s 2 = 77 Engineering Mechanics - Dynamics Chapter 12 Problem 12-106 The jet plane travels along the vertical parabolic path. When it is at point A it has speed v which is increasing at the rate a t . Determine the magnitude of acceleration of the plane when it is at point A. Given: v 200 m s = a t 0.8 m s 2 = d 5km= h 10 km= Solution: yx() h x d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = y' x() x yx() d d = y'' x() x y' x() d d = ρ x() 1 y' x() 2 + () 3 y'' x() = a n v 2 ρ d() = aa t 2 a n 2 += a 0.921 m s 2 = Problem 12–107 The car travels along the curve having a radius of R. If its speed is uniformly increased from v 1 to v 2 in time t, determine the magnitude of its acceleration at the instant its speed is v 3 . Given: v 1 15 m s = t 3s= 78 Engineering Mechanics - Dynamics Chapter 12 v 2 27 m s = R 300 m= v 3 20 m s = Solution: a t v 2 v 1 − t = a n v 3 2 R = aa t 2 a n 2 += a 4.22 m s 2 = *Problem 12–108 The satellite S travels around the earth in a circular path with a constant speed v 1 . If the acceleration is a, determine the altitude h. Assume the earth’s diameter to be d. Units Used: Mm 10 3 km= Given: v 1 20 Mm hr = a 2.5 m s 2 = d 12713 km= Solution: Guess h 1Mm= Given a v 1 2 h d 2 + = h Find h()= h 5.99 Mm= Problem 12–109 A particle P moves along the curve y = b x 2 + c with a constant speed v. Determine the point on the curve where the maximum magnitude of acceleration occurs and compute its value. Given: b 1 1 m = c 4− m= v 5 m s = Solution: Maximum acceleration occurs where the radius of curvature is the smallest which occurs at x = 0. 79 Engineering Mechanics - Dynamics Chapter 12 yx() bx 2 c+= y' x() x yx() d d = y'' x() x y' x() d d = ρ x() 1 y' x() 2 + () 3 y'' x() = ρ min ρ 0m()= ρ min 0.5 m= a max v 2 ρ min = a max 50 m s 2 = Problem 12–110 The Ferris wheel turns such that the speed of the passengers is increased by a t = bt. If the wheel starts from rest when θ = 0°, determine the magnitudes of the velocity and acceleration of the passengers when the wheel turns θ = θ 1 . Given: b 4 ft s 3 = θ 1 30 deg= r 40 ft= Solution: Guesses t 1 1s= v 1 1 ft s = a t1 1 ft s 2 = Given a t1 bt 1 = v 1 b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 2 = rθ 1 b 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 3 = a t1 v 1 t 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find a t1 v 1 , t 1 ,()= t 1 3.16 s= v 1 19.91 ft s = a t1 12.62 ft s 2 = a 1 a t1 2 v 1 2 r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += v 1 19.91 ft s = a 1 16.05 ft s 2 = Problem 12-111 At a given instant the train engine at E has speed v and acceleration a acting in the direction shown. Determine the rate of increase in the train's speed and the radius of curvature ρ of the path. 80 Engineering Mechanics - Dynamics Chapter 12 Given: v 20 m s = a 14 m s 2 = θ 75 deg= Solution: a t a( )cos θ()= a t 3.62 m s 2 = a n a( )sin θ()= a n 13.523 m s 2 = ρ v 2 a n = ρ 29.579 m= *Problem 12–112 A package is dropped from the plane which is flying with a constant horizontal velocity v A . Determine the normal and tangential components of acceleration and the radius of curvature of the path of motion (a) at the moment the package is released at A, where it has a horizontal velocity v A , and (b) just before it strikes the ground at B. Given: v A 150 ft s = h 1500 ft= g 32.2 ft s 2 = Solution: At A: a An g= ρ A v A 2 a An = ρ A 699 ft= 81 Engineering Mechanics - Dynamics Chapter 12 At B: t 2h g = v x v A = v y gt= θ atan v y v x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = v B v x 2 v y 2 += a Bn g cos θ()= ρ B v B 2 a Bn = ρ B 8510 ft= Problem 12-113 The automobile is originally at rest at s = 0. If its speed is increased by dv/dt = bt 2 , determine the magnitudes of its velocity and acceleration when t = t 1 . Given: b 0.05 ft s 4 = t 1 18 s= ρ 240 ft= d 300 ft= Solution: a t1 bt 1 2 = a t1 16.2 ft s 2 = v 1 b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 3 = v 1 97.2 ft s = s 1 b 12 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 4 = s 1 437.4 ft= If s 1 437.4 ft= > d 300 ft= then we are on the curved part of the track. a n1 v 1 2 ρ = a n1 39.366 ft s 2 = aa n1 2 a t1 2 += a 42.569 ft s 2 = If s 1 437.4 ft= < d 300 ft= then we are on the straight part of the track. a n1 0 ft s 2 = a n1 0 ft s 2 = aa n1 2 a t1 2 += a 16.2 ft s 2 = 82 Engineering Mechanics - Dynamics Chapter 12 Problem 12-114 The automobile is originally at rest at s = 0. If it then starts to increase its speed at dv/dt = bt 2 , determine the magnitudes of its velocity and acceleration at s = s 1 . Given: d 300 ft= ρ 240 ft= b 0.05 ft s 4 = s 1 550 ft= Solution: a t bt 2 = v b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 3 = s b 12 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 4 = t 1 12s 1 b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 4 = t 1 19.061 s= v 1 b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 3 = v 1 115.4 ft s = If s 1 550 ft= > d 300 ft= the car is on the curved path a t bt 1 2 = v b 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 3 = a n v 2 ρ = aa t 2 a n 2 += a 58.404 ft s 2 = If s 1 550 ft= < d 300 ft= the car is on the straight path a t bt 1 2 = a n 0 ft s 2 = aa t 2 a n 2 += a 18.166 ft s 2 = Problem 12-115 The truck travels in a circular path having a radius ρ at a speed v 0 . For a short distance from s = 0, its speed is increased by a t = bs. Determine its speed and the magnitude of its acceleration when it has moved a distance s = s 1 . Given: ρ 50 m= s 1 10 m= v 0 4 m s = b 0.05 1 s 2 = 83 Engineering Mechanics - Dynamics Chapter 12 Solution: a t bs= v 0 v 1 vv ⌠ ⎮ ⌡ d 0 s 1 sbs ⌠ ⎮ ⌡ d= v 1 2 2 v 0 2 2 − b 2 s 1 2 = v 1 v 0 2 bs 1 2 += v 1 4.583 m s = a t1 bs 1 = a n1 v 1 2 ρ = a 1 a t1 2 a n1 2 += a 1 0.653 m s 2 = *Problem 12–116 The particle travels with a constant speed v along the curve. Determine the particle’s acceleration when it is located at point x = x 1 . Given: v 300 mm s = k 20 10 3 × mm 2 = x 1 200 mm= Solution: yx() k x = y' x() x yx() d d = y'' x() x y' x() d d = ρ x() 1 y' x() 2 + () 3 y'' x() = θ x( ) atan y' x()()= θ 1 θ x 1 ()= θ 1 26.6− deg= a v 2 ρ x 1 () sin θ 1 ()− cos θ 1 () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = a 144 288 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ mm s 2 = a 322 mm s 2 = 84 Engineering Mechanics - Dynamics Chapter 12 Problem 12–117 Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at v, determine the maximum acceleration experienced by the passengers. Given: v 60 km hr = a 60 m= b 40 m= Solution: Maximum acceleration occurs where the radius of curvature is the smallest. In this case that happens when y = 0. xy() a 1 y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= x' y() y xy() d d = x'' y() y x' y() d d = ρ y() 1 x' y() 2 + () 3 − x'' y() = ρ min ρ 0m()= ρ min 26.667 m= a max v 2 ρ min = a max 10.42 m s 2 = Problem 12–118 Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at v, determine the minimum acceleration experienced by the passengers. Given: v 60 km hr = a 60 m= b 40 m= 85 Engineering Mechanics - Dynamics Chapter 12 Solution: Minimum acceleration occurs where the radius of curvature is the largest. In this case that happens when x = 0. yx() b 1 x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= y' x() x yx() d d = y'' x() x y' x() d d = ρ x() 1 y' x() 2 + () 3 − y'' x() = ρ max ρ 0m()= ρ max 90 m= a min v 2 ρ max = a min 3.09 m s 2 = Problem 12-119 The car B turns such that its speed is increased by dv B /dt = be ct . If the car starts from rest when θ = 0, determine the magnitudes of its velocity and acceleration when the arm AB rotates to θ = θ 1 . Neglect the size of the car. Given: b 0.5 m s 2 = c 1s 1− = θ 1 30 deg= ρ 5m= Solution: a Bt be ct = v B b c e ct 1− () = ρθ b c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e ct b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t− b c 2 −= Guess t 1 1s= 86 Engineering Mechanics - Dynamics Chapter 12 Given ρθ 1 b c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e ct 1 b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 − b c 2 −= t 1 Find t 1 ()= t 1 2.123 s= v B1 b c e ct 1 1− () = v B1 3.68 m s = a Bt1 be ct 1 = a Bn1 v B1 2 ρ = a B1 a Bt1 2 a Bn1 2 += a Bt1 4.180 m s 2 = a Bn1 2.708 m s 2 = a B1 4.98 m s 2 = *Problem 12-120 The car B turns such that its speed is increased by dv B /dt = b e ct . If the car starts from rest when θ = 0, determine the magnitudes of its velocity and acceleration when t = t 1 . Neglect the size of the car. Also, through what angle θ has it traveled? Given: b 0.5 m s 2 = c 1s 1− = t 1 2s= ρ 5m= Solution: a Bt be ct = v B b c e ct 1− () = ρθ b c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e ct b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t− b c 2 −= v B1 b c e ct 1 1− () = v B1 3.19 m s = a Bt1 be ct 1 = a Bn1 v B1 2 ρ = a B1 a Bt1 2 a Bn1 2 += 87 Engineering Mechanics - Dynamics Chapter 12 a Bt1 3.695 m s 2 = a Bn1 2.041 m s 2 = a B1 4.22 m s 2 = θ 1 1 ρ b c 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e ct 1 b c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 1 − b c 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = θ 1 25.1 deg= Problem 12–121 The motorcycle is traveling at v 0 when it is at A. If the speed is then increased at dv/dt = a t , determine its speed and acceleration at the instant t = t 1 . Given: k 0.5 m 1− = a t 0.1 m s 2 = v 0 1 m s = t 1 5s= Solution: yx() kx 2 = y' x() 2kx= y'' x() 2k= ρ x() 1 y' x() 2 + () 3 y'' x() = v 1 v 0 a t t 1 += s 1 v 0 t 1 1 2 a t t 1 2 += v 1 1.5 m s = Guess x 1 1m= Given s 1 0 x 1 x1 y' x() 2 + ⌠ ⎮ ⌡ d= x 1 Find x 1 ()= a 1t a t = a 1n v 1 2 ρ x 1 () = a 1 a 1t 2 a 1n 2 += a 1 0.117 m s 2 = Problem 12-122 The ball is ejected horizontally from the tube with speed v A . Find the equation of the path y = f (x), and then find the ball’s velocity and the normal and tangential components of acceleration when t = t 1 . 88 Engineering Mechanics - Dynamics Chapter 12 Given: v A 8 m s = t 1 0.25 s= g 9.81 m s 2 = Solution: xv A t= t x v A = y g− 2 t 2 = y g− 2v A 2 x 2 = parabola when t = t 1 v x v A = v y g− t 1 = θ atan v y − v x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 17.044 deg= a n g cos θ()= a n 9.379 m s 2 = a t g sin θ()= a t 2.875 m s 2 = Problem 12–123 The car travels around the circular track having a radius r such that when it is at point A it has a velocity v 1 which is increasing at the rate dv/dt = kt. Determine the magnitudes of its velocity and acceleration when it has traveled one-third the way around the track. Given: k 0.06 m s 3 = r 300 m= v 1 5 m s = Solution: a t t() kt= vt() v 1 k 2 t 2 += s p t() v 1 t k 6 t 3 += 89 Engineering Mechanics - Dynamics Chapter 12 Guess t 1 1s= Given s p t 1 () 2πr 3 = t 1 Find t 1 ()= t 1 35.58 s= v 1 vt 1 ()= a t1 a t t 1 ()= a n1 v 1 2 r = a 1 a t1 2 a n1 2 += v 1 43.0 m s = a 1 6.52 m s 2 = *Problem 12–124 The car travels around the portion of a circular track having a radius r such that when it is at point A it has a velocity v 1 which is increasing at the rate of dv/dt = ks. Determine the magnitudes of its velocity and acceleration when it has traveled three-fourths the way around the track. Given: k 0.002 s 2− = r 500 ft= v 1 2 ft s = Solution: s p1 3 4 2πr= a t v s p v d d = ks p = Guess v 1 1 ft s = Given 0 v 1 vv ⌠ ⎮ ⌡ d 0 s p1 s p ks p ⌠ ⎮ ⌡ d= v 1 Find v 1 ()= a t1 ks p1 = a n1 v 1 2 r = a 1 a t1 2 a n1 2 += v 1 105.4 ft s = a 1 22.7 ft s 2 = Problem 12-125 The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds v A and v B respectively. Determine at t = t 1 , (a) the displacement along the path of each particle, (b) the position vector to each particle, and (c) the shortest distance between the particles. 90 Engineering Mechanics - Dynamics Chapter 12 Given: v A 0.7 m s = v B 1.5 m s = t 1 2s= ρ 5m= Solution: (a) The displacement along the path s A v A t 1 = s A 1.4 m= s B v B t 1 = s B 3m= (b) The position vector to each particle θ A s A ρ = r A ρ sin θ A () ρρcos θ A ()− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = r A 1.382 0.195 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= θ B s B ρ = r B ρ− sin θ B () ρρcos θ B ()− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = r B 2.823− 0.873 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m= (c) The shortest distance between the particles d r B r A −= d 4.26 m= Problem 12-126 The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds v A and v B respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens. Given: v A 0.7 m s = v B 1.5 m s = ρ 5m= 91 Engineering Mechanics - Dynamics Chapter 12 Solution: v A v B +()t 2πρ= t 2πρ v A v B + = t 14.28 s= a B v B 2 ρ = a B 0.45 m s 2 = Problem 12-127 The race car has an initial speed v A at A. If it increases its speed along the circular track at the rate a t = bs, determine the time needed for the car to travel distance s 1 . Given: v A 15 m s = b 0.4 s 2− = s 1 20 m= ρ 150 m= Solution: a t bs= v s v d d = v A v vv ⌠ ⎮ ⌡ d 0 s sbs ⌠ ⎮ ⌡ d= v 2 2 v A 2 2 − b s 2 2 = v t s d d = v A 2 bs 2 += 92 Engineering Mechanics - Dynamics Chapter 12 0 s s 1 v A 2 bs 2 + ⌠ ⎮ ⎮ ⎮ ⌡ d 0 t t1 ⌠ ⎮ ⌡ d= t 0 s 1 s 1 v A 2 bs 2 + ⌠ ⎮ ⎮ ⎮ ⌡ d= t 1.211 s= *Problem 12-128 A boy sits on a merry-go-round so that he is always located a distance r from the center of rotation. The merry-go-round is originally at rest, and then due to rotation the boy’s speed is increased at the rate a t . Determine the time needed for his acceleration to become a. Given: r 8ft= a t 2 ft s 2 = a 4 ft s 2 = Solution: a n a 2 a t 2 −= va n r= t v a t = t 2.63 s= Problem 12–129 A particle moves along the curve y = bsin(cx) with a constant speed v. Determine the normal and tangential components of its velocity and acceleration at any instant. Given: v 2 m s = b 1m= c 1 m = Solution: ybsin cx()= y' b c cos cx()= y'' b− c 2 sin cx()= ρ 1 y' 2 + () 3 y'' = 1 bccos cx()() 2 + ⎡ ⎣ ⎤ ⎦ 3 2 b− c 2 sin cx() = a n v 2 bcsin cx() 1 bccos cx()() 2 + ⎡ ⎣ ⎤ ⎦ 3 2 = a t 0= v t 0= v n 0= Problem 12–130 The motion of a particle along a fixed path is defined by the parametric equations r = b, θ = ct 93 Engineering Mechanics - Dynamics Chapter 12 pgp ypq and z = dt 2 . Determine the unit vector that specifies the direction of the binormal axis to the osculating plane with respect to a set of fixed x, y, z coordinate axes when t = t 1 . Hint: Formulate the particle’s velocity v p and acceleration a p in terms of their i, j, k components. Note that xrcos θ ()= and yrsin θ ()= . The binormal is parallel to v p × a p . Why? Given: b 8ft= c 4 rad s = d 6 ft s 2 = t 1 2s= Solution: r p1 b cos ct 1 () b sin ct 1 () dt 1 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = v p1 b− c sin ct 1 () bccos ct 1 () 2dt 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = a p1 b− c 2 cos ct 1 () b− c 2 sin ct 1 () 2d ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = Since v p and a p are in the normal plane and the binormal direction is perpendicular to this plane then we can use the cross product to define the binormal direction. u v p1 a p1 × v p1 a p1 × = u 0.581 0.161 0.798 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Problem 12-131 Particles A and B are traveling counter-clockwise around a circular track at constant speed v 0 . If at the instant shown the speed of A is increased by dv A /dt = bs A , determine the distance measured counterclockwise along the track from B to A when t = t 1 . What is the magnitude of the acceleration of each particle at this instant? Given: v 0 8 m s = b 4s 2− = t 1 1s= r 5m= θ 120 deg= Solution: Distance a At v A dv A ds A = bs A = v 0 v A v A v A ⌠ ⎮ ⌡ d 0 s A s A bs A ⌠ ⎮ ⌡ d= 94 Engineering Mechanics - Dynamics Chapter 12 v A 2 2 v 0 2 2 − b 2 s A 2 = v A v 0 2 bs A 2 += ds A dt = Guess s A1 1m= Given 0 t 1 t1 ⌠ ⎮ ⌡ d 0 s A1 s A 1 v 0 2 bs A 2 + ⌠ ⎮ ⎮ ⎮ ⌡ d= s A1 Find s A1 ()= s A1 14.507 m= s B1 v 0 t 1 = s B1 8m= s AB s A1 rθ+ s B1 −= s AB 16.979 m= a A bs A1 () 2 v 0 2 bs A1 2 + r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += a A 190.24 m s 2 = a B v 0 2 r = a B 12.8 m s 2 = Problem 12-132 Particles A and B are traveling around a circular track at speed v 0 at the instant shown. If the speed of B is increased by dv B /dt = a Bt , and at the same instant A has an increase in speed dv A /dt = bt, determine how long it takes for a collision to occur. What is the magnitude of the acceleration of each particle just before the collision occurs? Given: v 0 8 m s = r 5m= a Bt 4 m s 2 = θ 120 deg= b 0.8 m s 3 = Solution: v B a Bt tv 0 += s B a Bt 2 t 2 v 0 t+= a At bt= v A b 2 t 2 v 0 += s A b 6 t 3 v 0 t+= Assume that B catches A Guess t 1 1s= 95 Engineering Mechanics - Dynamics Chapter 12 Given a Bt 2 t 1 2 v 0 t 1 + b 6 t 1 3 v 0 t 1 + rθ+= t 1 Find t 1 ()= t 1 2.507 s= Assume that A catches B Guess t 2 13 s= Given a Bt 2 t 2 2 v 0 t 2 + r 2πθ−()+ b 6 t 2 3 v 0 t 2 += t 2 Find t 2 ()= t 2 15.642 s= Take the smaller time t min t 1 t 2 ,()= t 2.507 s= a A bt() 2 b 2 t 2 v 0 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 r ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 2 += a B a Bt 2 a Bt tv 0 +() 2 r ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 += a A a B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 22.2 65.14 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ m s 2 = Problem 12-133 The truck travels at speed v 0 along a circular road that has radius ρ. For a short distance from s = 0, its speed is then increased by dv/dt = bs. Determine its speed and the magnitude of its acceleration when it has moved a distance s 1 . Given: v 0 4 m s = ρ 50 m= b 0.05 s 2 = s 1 10 m= Solution: a t v s v d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = bs= v 0 v 1 vv ⌠ ⎮ ⌡ d 0 s 1 sbs ⌠ ⎮ ⌡ d= v 1 2 2 v 0 2 2 − b 2 s 1 2 = v 1 v 0 2 bs 1 2 += v 1 4.58 m s = 96 Engineering Mechanics - Dynamics Chapter 12 a t bs 1 = a n v 1 2 ρ = aa t 2 a n 2 += a 0.653 m s 2 = Problem 12-134 A go-cart moves along a circular track of radius ρ such that its speed for a short period of time, 0 < t < t 1 , is vb1 e ct 2 − ⎛ ⎝ ⎞ ⎠ = . Determine the magnitude of its acceleration when t = t 2 . How far has it traveled in t = t 2 ? Use Simpson’s rule with n steps to evaluate the integral. Given: ρ 100 ft= t 1 4s= b 60 ft s = c 1− s 2− = t 2 2s= n 50= Solution: tt 2 = vb1 e ct 2 − ⎛ ⎝ ⎞ ⎠ = a t 2− bcte ct 2 = a n v 2 ρ = aa t 2 a n 2 += a 35.0 ft s 2 = s 2 0 t 2 tb 1 e ct 2 − ⎛ ⎝ ⎞ ⎠ ⌠ ⎮ ⌡ d= s 2 67.1 ft= Problem 12-135 A particle P travels along an elliptical spiral path such that its position vector r is defined by r = (a cos bt i + c sin dt j + et k). When t = t 1 , determine the coordinate direction angles α, β, and γ, which the binormal axis to the osculating plane makes with the x, y, and z axes. Hint: Solve for the velocity v p and acceleration a p of the particle in terms of their i, j, k components. The binormal is parallel to v p × a p . Why? 97 Engineering Mechanics - Dynamics Chapter 12 Given: a 2m= d 0.1 s 1− = b 0.1 s 1− = e 2 m s = c 1.5 m= t 1 8s= Solution: tt 1 = r p a( )cos bt() c sin dt() et ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = v p a− b sin bt() cdcos dt() e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = a p a− b 2 cos bt() c− d 2 sin dt() 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = u b v p a p × v p a p × = u b 0.609 0.789− 0.085 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ acos u b ()= α β γ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 52.5 142.1 85.1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ deg= *Problem 12-136 The time rate of change of acceleration is referred to as the jerk, which is often used as a means of measuring passenger discomfort. Calculate this vector, a', in terms of its cylindrical components, using Eq. 12-32. Solution: a r'' rθ' 2 − () u r rθ'' 2r'θ'+()u θ + z''u z += a' r''' r'θ' 2 − 2rθ'θ''− () u r r'' rθ' 2 − () u' r + r'θ'' rθ'''+ 2r''θ'+ 2r'θ''+ u θ rθ'' 2r'θ''+()u' θ + z'''u z + z''u' z ++ ...= But u r θ'u θ = u' θ θ'− u r = u' z 0= Substituting and combining terms yields a' r''' 3r'θ' 2 − 3rθ'θ''− () u r rθ''' 3r'θ''+ 3r''θ'+ rθ' 3 − () u θ + z'''()u z += 98 Engineering Mechanics - Dynamics Chapter 12 If a particle’s position is described by the polar coordinates r = a(1 + sin bt) and θ = ce dt , determine the radial and tangential components of the particle’s velocity and acceleration when t = t 1 . Given: a 4m= b 1s 1− = c 2 rad= d 1− s 1− = t 1 2s= Solution: When tt 1 = ra1 sin bt()+()= r' a b cos bt()= r'' a− b 2 sin bt()= θ ce dt = θ'cde dt = θ'' c d 2 e dt = v r r'= v r 1.66− m s = v θ rθ'= v θ 2.07− m s = a r r'' rθ' 2 −= a r 4.20− m s 2 = a θ rθ'' 2r'θ'+= a θ 2.97 m s 2 = Problem 12–138 The slotted fork is rotating about O at a constant rate θ'. Determine the radial and transverse components of the velocity and acceleration of the pin A at the instant θ = θ 1 . The path is defined by the spiral groove r = b + cθ , where θ is in radians. Given: θ' 3 rad s = b 5in= c 1 π in= θ 1 2 π rad= Solution: θθ 1 = 99 Problem 12-137 Engineering Mechanics - Dynamics Chapter 12 rbcθ+= r' cθ'= r'' 0 in s 2 = θ'' 0 rad s 2 = v r r'= v θ rθ'= a r r'' rθ' 2 −= a θ rθ'' 2r'θ'+= v r 0.955 in s = v θ 21 in s = a r 63− in s 2 = a θ 5.73 in s 2 = Problem 12–139 The slotted fork is rotating about O at the rate θ ' which is increasing at θ '' when θ = θ 1 . Determine the radial and transverse components of the velocity and acceleration of the pin A at this instant. The path is defined by the spiral groove r = (5 + θ /π) in., where θ is in radians. Given: θ' 3 rad s = θ'' 2 rad s 2 = b 5in= c 1 π in= θ 1 2 π rad= Solution: θθ 1 = rbcθ+= r' cθ'= r'' cθ''= v r r'= v θ rθ'= a r r'' rθ' 2 −= a θ rθ'' 2r'θ'+= v r 0.955 in s = v θ 21 in s = a r 62.363− in s 2 = a θ 19.73 in s 2 = *Problem 12-140 If a particle moves along a path such that r = acos(bt) and θ = ct, plot the path r = f(θ) and determine the particle’s radial and transverse components of velocity and acceleration. 100 Engineering Mechanics - Dynamics Chapter 12 Given: a 2ft= b 1s 1− = c 0.5 rad s = The plot t θ c = ra( )cos b θ c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 0 0.01 2π(), 2π..= r θ() a( )cos b θ c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 ft = 01234567 2 0 2 Angle in radians Distance in ft r θ () θ ra( )cos bt()= r' a− b sin bt()= r'' a− b 2 cos bt()= θ ct= θ'c= θ'' 0= v r r'= a− b sin bt()= a r r'' rθ' 2 −= a− b 2 c 2 + () cos bt()= v θ rθ'= accos bt()= a θ rθ'' 2r'θ'+= 2− abcsin bt()= Problem 12-141 If a particle’s position is described by the polar coordinates r = asinbθ and θ = ct, determine the radial and tangential components of its velocity and acceleration when t = t 1 . Given: a 2m= b 2 rad= c 4 rad s = t 1 1s= Solution: tt 1 = ra( )sin bct()= r' a b c cos bct()= r'' a− b 2 c 2 sin bct()= θ ct= θ'c= θ'' 0 rad s 2 = v r r'= v r 2.328− m s = v θ rθ'= v θ 7.915 m s = 101 Engineering Mechanics - Dynamics Chapter 12 a r r'' rθ' 2 −= a r 158.3− m s 2 = a θ rθ'' 2r'θ'+= a θ 18.624− m s 2 = Problem 12-142 A particle is moving along a circular path having a radius r. Its position as a function of time is given by θ = bt 2 . Determine the magnitude of the particle’s acceleration when θ = θ 1 . The particle starts from rest when θ = 0°. Given: r 400 mm= b 2 rad s 2 = θ 1 30 deg= Solution: t θ 1 b = t 0.512 s= θ bt 2 = θ' 2bt= θ'' 2b= ar− θ' 2 () 2 rθ''() 2 += a 2.317 m s 2 = Problem 12-143 A particle moves in the x - y plane such that its position is defined by r = ati + bt 2 j. Determine the radial and tangential components of the particle’s velocity and acceleration when t = t 1 . Given: a 2 ft s = b 4 ft s 2 = t 1 2s= Solution: tt 1 = Rectangular xat= v x a= a x 0 ft s 2 = ybt 2 = v y 2bt= a y 2b= Polar θ atan y x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 75.964 deg= 102 Engineering Mechanics - Dynamics Chapter 12 v r v x cos θ() v y sin θ()+= v r 16.007 ft s = v θ v x − sin θ() v y cos θ()+= v θ 1.94 ft s = a r a x cos θ() a y sin θ()+= a r 7.761 ft s 2 = a θ a x − sin θ() a y cos θ()+= a θ 1.94 ft s 2 = *Problem 12-144 A truck is traveling along the horizontal circular curve of radius r with a constant speed v. Determine the angular rate of rotation θ' of the radial line r and the magnitude of the truck’s acceleration. Given: r 60 m= v 20 m s = Solution: θ' v r = θ' 0.333 rad s = ar− θ' 2 = a 6.667 m s 2 = Problem 12-145 A truck is traveling along the horizontal circular curve of radius r with speed v which is increasing at the rate v'. Determine the truck’s radial and transverse components of acceleration. 103 Engineering Mechanics - Dynamics Chapter 12 Given: r 60 m= v 20 m s = v' 3 m s 2 = Solution: a r v 2 − r = a r 6.667− m s 2 = a θ v'= a θ 3 m s 2 = Problem 12-146 A particle is moving along a circular path having radius r such that its position as a function of time is given by θ = c sin bt. Determine the acceleration of the particle at θ = θ 1 . The particle starts from rest at θ = 0°. Given: r 6in= c 1 rad= b 3s 1− = θ 1 30 deg= Solution: t 1 b asin θ 1 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = t 0.184 s= θ c sin bt()= θ'cbcos bt()= θ'' c b 2 sin bt()= ar− θ' 2 () 2 rθ''() 2 += a 48.329 in s 2 = Problem 12-147 The slotted link is pinned at O, and as a result of the constant angular velocity θ' it drives the peg P for a short distance along the spiral guide r = aθ. Determine the radial and transverse components of the velocity and acceleration of P at the instant θ = θ 1 . 104 Engineering Mechanics - Dynamics Chapter 12 Given: θ' 3 rad s = θ 1 π 3 rad= a 0.4 m= b 0.5 m= Solution: θθ 1 = raθ= r' aθ'= r'' 0 m s 2 = v r r'= v r 1.2 m s = v θ rθ'= v θ 1.257 m s = a r r'' rθ' 2 −= a r 3.77− m s 2 = a θ 2r'θ'= a θ 7.2 m s 2 = *Problem 12-148 The slotted link is pinned at O, and as a result of the angular velocity θ' and the angular acceleration θ'' it drives the peg P for a short distance along the spiral guide r = aθ. Determine the radial and transverse components of the velocity and acceleration of P at the instant θ = θ 1 . Given: θ' 3 rad s = θ 1 π 3 rad= a 0.4 m= θ'' 8 rad s 2 = b 0.5 m= Solution: θθ 1 = 105 Engineering Mechanics - Dynamics Chapter 12 raθ= r' aθ'= r'' aθ''= v r r'= v r 1.2 m s = v θ rθ'= v θ 1.257 m s = a r r'' rθ' 2 −= a r 0.57− m s 2 = a θ rθ'' 2r'θ'+= a θ 10.551 m s 2 = Problem 12-149 The slotted link is pinned at O, and as a result of the constant angular velocity θ' it drives the peg P for a short distance along the spiral guide r = aθ where θ is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = b. Given: θ' 3 rad s = a 0.4 m= b 0.5 m= Solution: θ b a = raθ= r' aθ'= r'' 0 m s 2 = v r r'= v θ rθ'= a r r'' rθ' 2 −= a θ 2r'θ'= vv r 2 v θ 2 += aa r 2 a θ 2 += v 1.921 m s = a 8.491 m s 2 = Problem 12–150 A train is traveling along the circular curve of radius r. At the instant shown, its angular rate of rotation is θ', which is decreasing at θ''. Determine the magnitudes of the train’s velocity and acceleration at this instant. 106 Engineering Mechanics - Dynamics Chapter 12 r 600 ft= θ' 0.02 rad s = θ'' 0.001− rad s 2 = Solution: vrθ'= v 12 ft s = ar− θ' 2 () 2 rθ''() 2 += a 0.646 ft s 2 = Problem 12–151 A particle travels along a portion of the “four-leaf rose” defined by the equation r = a cos(bθ). If the angular velocity of the radial coordinate line is θ' = ct 2 , determine the radial and transverse components of the particle’s velocity and acceleration at the instant θ = θ 1 . When t = 0, θ = 0°. Given: a 5m= b 2= c 3 rad s 3 = θ 1 30 deg= Solution: θ t() c 3 t 3 = θ't() ct 2 = θ'' t() 2ct= rt() a( )cos bθ t()()= r' t() t rt() d d = r'' t() t r' t() d d = When θ = θ 1 t 1 3θ 1 c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 = v r r' t 1 ()= v r 16.88− m s = 107 Given: Engineering Mechanics - Dynamics Chapter 12 v θ rt 1 ()θ't 1 ()= v θ 4.87 m s = a r r'' t 1 () rt 1 ()θ't 1 () 2 −= a r 89.4− m s 2 = a θ rt 1 ()θ'' t 1 () 2 r' t 1 ()θ't 1 ()+= a θ 53.7− m s 2 = *Problem 12-152 At the instant shown, the watersprinkler is rotating with an angular speed θ' and an angular acceleration θ''. If the nozzle lies in the vertical plane and water is flowing through it at a constant rate r', determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r. Given: θ' 2 rad s = θ'' 3 rad s 2 = r' 3 m s = r 0.2 m= Solution: vr' 2 rθ'() 2 += v 3.027 m s = ar− θ' 2 () 2 rθ'' 2r'θ'+() 2 += a 12.625 m s 2 = Problem 12–153 The boy slides down the slide at a constant speed v. If the slide is in the form of a helix, defined by the equations r = constant and z = −(hθ )/(2π), determine the boy’s angular velocity about the z axis, θ' and the magnitude of his acceleration. Given: v 2 m s = r 1.5 m= h 2m= 108 Engineering Mechanics - Dynamics Chapter 12 Solution: z h 2π θ= z' h 2π θ'= vz' 2 rθ'() 2 += h 2π ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 r 2 + θ'= θ' v h 2π ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 r 2 + = θ' 1.304 rad s = ar− θ' 2 = a 2.55 m s 2 = Problem 12–154 A cameraman standing at A is following the movement of a race car, B, which is traveling along a straight track at a constant speed v. Determine the angular rate at which he must turn in order to keep the camera directed on the car at the instant θ = θ 1 . Given: v 80 ft s = θ 1 60 deg= a 100 ft= Solution: θθ 1 = arsin θ()= 0 r' sin θ() rθ' cos θ()+= xrcos θ()= v x v−= r' cos θ() rθ' sin θ()−= Guess r 1ft= r' 1 ft s = θ' 1 rad s = Given arsin θ()= 109 Engineering Mechanics - Dynamics Chapter 12 0 r' sin θ() rθ' cos θ()+= v− r' cos θ() rθ' sin θ()−= r r' θ' ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find rr', θ',()= r 115.47 ft= r' 40− ft s = θ' 0.6 rad s = Problem 12-155 For a short distance the train travels along a track having the shape of a spiral, r = a/θ. If it maintains a constant speed v, determine the radial and transverse components of its velocity when θ = θ 1 . Given: a 1000 m= v 20 m s = θ 1 9 π 4 rad= Solution: θθ 1 = r a θ = r' a− θ 2 θ'= v 2 r' 2 r 2 θ' 2 += a 2 θ 4 a 2 θ 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ θ' 2 = θ' vθ 2 a 1 θ 2 + = r a θ = r' a− θ 2 θ'= v r r'= v r 2.802− m s = v θ rθ'= v θ 19.803 m s = *Problem 12-156 For a short distance the train travels along a track having the shape of a spiral, r = a / θ. If the angular rate θ' is constant, determine the radial and transverse components of its velocity and acceleration when θ = θ 1 . Given: a 1000 m= θ' 0.2 rad s = θ 1 9 π 4 = Solution: θθ 1 = r a θ = r' a− θ 2 θ'= r'' 2a θ 3 θ' 2 = 110 Engineering Mechanics - Dynamics Chapter 12 v r r'= v r 4.003− m s = v θ rθ'= v θ 28.3 m s = a r r'' rθ' 2 −= a r 5.432− m s 2 = a θ 2r'θ'= a θ 1.601− m s 2 = Problem 12-157 The arm of the robot has a variable length so that r remains constant and its grip. A moves along the path z = a sinbθ. If θ = ct, determine the magnitudes of the grip’s velocity and acceleration when t = t 1 . Given: r 3ft= c 0.5 rad s = a 3ft= t 1 3s= b 4= Solution: tt 1 = θ ct= rr= zasin bct()= θ'c= r' 0 ft s = z' a b c cos bct()= θ'' 0 rad s 2 = r'' 0 ft s 2 = z'' a− b 2 c 2 sin bct()= vr' 2 rθ'() 2 + z' 2 += v 5.953 ft s = a r'' rθ' 2 − () 2 rθ'' 2r'θ'+() 2 + z'' 2 += a 3.436 ft s 2 = Problem 12-158 For a short time the arm of the robot is extending so that r' remains constant, z = bt 2 and θ = ct. Determine the magnitudes of the velocity and acceleration of the grip A when t = t 1 and r = r 1 . 111 Engineering Mechanics - Dynamics Chapter 12 Given: r' 1.5 ft s = b 4 ft s 2 = c 0.5 rad s = t 1 3s= r 1 3ft= Solution: tt 1 = rr 1 = θ ct= zbt 2 = θ'c= z' 2bt= z'' 2b= vr' 2 rθ'() 2 + z' 2 += v 24.1 ft s = ar− θ' 2 () 2 2r'θ'() 2 + z'' 2 += a 8.174 ft s 2 = Problem 12–159 The rod OA rotates counterclockwise with a constant angular velocity of θ'. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = b(c − cos(θ)). Determine the speed of the slider blocks at the instant θ = θ 1 . Given: θ' 5 rad s = b 100 mm= c 2= θ 1 120 deg= Solution: θθ 1 = rbccos θ()−()= r' b sin θ()θ'= 112 dhanesh_h Mathcad - CombinedMathcads

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