# Dynamics_Part11.pdf

## Civil Engineering 2120 with Bowman at Marquette University *

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Engineering Mechanics - Dynamics Chapter 15 v 25 ft s = θ 130 deg= γ w 62.4 lb ft 3 = Solution: QAv= Q 0.69 ft 3 s = t m d d m'= ρQ= m' γ w Q= m' 1.3468 slug s = v Ax v= v Ay 0 ft s = v Bx v cos θ()= v By v sin θ()= F x m'− g v Bx v Ax −()= F x 55.3 lb= F y m' g v By v Ay −()= F y 25.8 lb= Problem 15-113 Water is flowing from the fire hydrant opening of diameter d B with velocity v B . Determine the horizontal and vertical components of force and the moment developed at the base joint A, if the static (gauge) pressure at A is P A . The diameter of the fire hydrant at A is d A . Units Used: kPa 10 3 Pa= Mg 10 3 kg= kN 10 3 N= Given: d B 150 mm= h 500 mm= v B 15 m s = d A 200 mm= ρ w 1 Mg m 3 = P A 50 kPa= Solution: A B π d B 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = A A π d A 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = m' ρ w v B π d B 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 = v A m' ρ w A A = 379 Engineering Mechanics - Dynamics Chapter 15 A x m' v B = A x 3.98 kN= A y − 50π d A 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + m' 0 v A −()= A y m' v A P A π d A 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 += A y 3.81 kN= Mm'hv B = M 1.99 kN m⋅= Problem 15-114 The chute is used to divert the flow of water Q. If the water has a cross-sectional area A, determine the force components at the pin A and roller B necessary for equilibrium. Neglect both the weight of the chute and the weight of the water on the chute. Units Used: Mg 10 3 kg= kN 10 3 N= Given: Q 0.6 m 3 s = ρ w 1 Mg m 3 = A 0.05 m 2 = h 2m= a 1.5 m= b 0.12 m= Solution: t m d d m'= m' ρ w Q= v A Q A = v B v A = ΣF x m' v Ax v Bx −()= B x A x − m' v Ax v Bx −()= ΣF y m' v Ay v By −()= A y m' 0 v B −()− ⎡ ⎣ ⎤ ⎦ = A y 7.20 kN= ΣM A m' d 0A v A d 0B v B −()= B x 1 h m' b v A ab−()v A + ⎡ ⎣ ⎤ ⎦ = B x 5.40 kN= A x B x m' v A −= A x 1.80− kN= 380 Engineering Mechanics - Dynamics Chapter 15 Problem 15-115 The fan draws air through a vent with speed v. If the cross-sectional area of the vent is A, determine the horizontal thrust on the blade. The specific weight of the air is γ a . Given: v 12 ft s = A 2ft 2 = γ a 0.076 lb ft 3 = g 32.20 ft s 2 = Solution: m' t m d d = m' γ a vA= m' 0.05669 slug s = T m' v 0−() g = T 0.68 lb= *Problem 15-116 The buckets on the Pelton wheel are subjected to a jet of water of diameter d, which has velocity v w . If each bucket is traveling at speed v b when the water strikes it, determine the power developed by the wheel. The density of water is γ w . Given: d 2in= θ 20 deg= v w 150 ft s = g 32.2 ft s 2 = v b 95 ft s = γ w 62.4 lbf ft 3 = 381 Engineering Mechanics - Dynamics Chapter 15 Solution: v A v w v b −= v A 55 ft s = v Bx v A − cos θ() v b += v Bx 43.317 ft s = ΣF x m' v Bx v Ax −()= F x γ w g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ π d 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ v A v Bx − v A −()− ⎡ ⎣ ⎤ ⎦ = F x 266.41 m s 2 lb⋅= PF x v b = P 4.69 hp= Problem 15-117 The boat of mass M is powered by a fan F which develops a slipstream having a diameter d. If the fan ejects air with a speed v, measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest. Assume that air has a constant density ρ a and that the entering air is essentially at rest. Neglect the drag resistance of the water. Given: M 200 kg= h 0.375 m= d 0.75 m= v 14 m s = ρ a 1.22 kg m 3 = Solution: QAv= Q π 4 d 2 v= Q 6.1850 m 3 s = t m d d m'= m' ρ a Q= m' 7.5457 kg s = ΣF x m' v Bx v Ax −()= F ρ a Qv= F 105.64 N= ΣF x Ma x = FMa= 382 Engineering Mechanics - Dynamics Chapter 15 a F M = a 0.53 m s 2 = Problem 15-118 The rocket car has a mass M C (empty) and carries fuel of mass M F . If the fuel is consumed at a constant rate c and ejected from the car with a relative velocity v DR , determine the maximum speed attained by the car starting from rest. The drag resistance due to the atmosphere is F D = kv 2 and the speed is measured in m/s. Units Used: Mg 10 3 kg= Given: M C 3Mg= M F 150 kg= v DR 250 m s = c 4 kg s = k 60 N s 2 m 2 ⋅= Solution: m 0 M C M F += At time t the mass of the car is m 0 ct− Set Fkv 2 = , then k− v 2 m 0 ct−() t v d d v DR c−= Maximum speed occurs at the instant the fuel runs out. t M F c = t 37.50 s= Thus, Initial Guess: v 4 m s = Given 0 v v 1 cv DR kv 2 − ⌠ ⎮ ⎮ ⎮ ⌡ d 0 t t 1 m 0 ct− ⌠ ⎮ ⎮ ⌡ d= v Find v()= v 4.06 m s = 383 Engineering Mechanics - Dynamics Chapter 15 Problem 15-119 A power lawn mower hovers very close over the ground. This is done by drawing air in at speed v A through an intake unit A, which has cross-sectional area A A and then discharging it at the ground, B, where the cross-sectional area is A B . If air at A is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has mass M with center of mass at G. Assume that air has a constant density of ρ a . Given: v A 6 m s = A A 0.25 m 2 = A B 0.35 m 2 = M 15 kg= ρ a 1.22 kg m 3 = Solution: m' ρ a A A v A = m' 1.83 kg s = + ↑ ΣF y m' v By v Ay −()= PA B Mg− m' 0 v A −()− ⎡ ⎣ ⎤ ⎦ = P 1 A B m' v A Mg+()= P 452 Pa= *Problem 15-120 The elbow for a buried pipe of diameter d is subjected to static pressure P. The speed of the water passing through it is v. Assuming the pipe connection at A and B do not offer any vertical force resistance on the elbow, determine the resultant vertical force F that the soil must then exert on the elbow in order to hold it in equilibrium. Neglect the weight of the elbow and the water within it. The density of water is γ w . Given: d 5in= θ 45 deg= 384 Engineering Mechanics - Dynamics Chapter 15 P 10 lb in 2 = γ w 62.4 lb ft 3 = v 8 ft s = Solution: Qv π 4 d 2⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = m' γ w g Q= Also, the force induced by the water pressure at A is A π 4 d 2 = FPA= F 196.35 lb= 2F cos θ() F 1 − m' v− cos θ() v cos θ()−()= F 1 2 F cos θ() m' v cos θ()+= F 1 302 lb= Problem 15-121 The car is used to scoop up water that is lying in a trough at the tracks. Determine the force needed to pull the car forward at constant velocity v for each of the three cases. The scoop has a cross-sectional area A and the density of water is ρ w . 385 Engineering Mechanics - Dynamics Chapter 15 Solution: The system consists of the car and the scoop. In all cases ΣF s m t v d d V De t m e d d −= F 0 VρAV−= FV 2 ρA= Problem 15-122 A rocket has an empty weight W 1 and carries fuel of weight W 2 . If the fuel is burned at the rate c and ejected with a relative velocity v DR , determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket. Given: W 1 500 lb= W 2 300 lb= c 15 lb s = v DR 4400 ft s = g 32.2 ft s 2 = Solution: m 0 W 1 W 2 + g = The maximum speed occurs when all the fuel is consumed, that is, where t W 2 c = t 20.00 s= ΣF x m t v d d v DR t m e d d −= At a time t, Mm 0 c g t−= , where c g t m e d d = . In space the weight of the rocket is zero. 0 m 0 ct−() t v d d v DR c−= Guess v max 1 ft s = Given 0 v max v1 ⌠ ⎮ ⌡ d 0 t t c g v DR m 0 c g t− ⌠ ⎮ ⎮ ⎮ ⎮ ⎮ ⌡ d= v max Find v max ()= v max 2068 ft s = 386 Engineering Mechanics - Dynamics Chapter 15 Problem 15-123 The boat has mass M and is traveling forward on a river with constant velocity v b , measured relative to the river. The river is flowing in the opposite direction at speed v R . If a tube is placed in the water, as shown, and it collects water of mass M w in the boat in time t, determine the horizontal thrust T on the tube that is required to overcome the resistance to the water collection. Units Used: Mg 10 3 kg= Given: M 180 kg= M w 40 kg= v b 70 km hr = t 80 s= ρw 1 Mg m 3 = v R 5 km hr = Solution: m' M w t = m' 0.50 kg s = v di v b = v di 19.44 m s = ΣF i m t v d d v di m'+= Tv di m'= T 9.72 N= *Problem 15-124 The second stage of a two-stage rocket has weight W 2 and is launched from the first stage with velocity v. The fuel in the second stage has weight W f . If it is consumed at rate r and ejected with relative velocity v r , determine the acceleration of the second stage just after the engine is fired. What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of gravitation. Given: W 2 2000 lb= W f 1000 lb= r 50 lb s = v 3000 mi hr = v r 8000 ft s = g 32.2 ft s 2 = 387 Engineering Mechanics - Dynamics Chapter 15 Solution: Initially, ΣF s m t v d d v di t m e d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= 0 W 2 W f + g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ av r r g −= av r r W 2 W f + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = a 133 ft s 2 = Finally, a 1 v r r W 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = a 1 200 ft s 2 = 0 W 2 g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 1 v r r g ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ −= Problem 15-125 The earthmover initially carries volume V of sand having a density ρ. The sand is unloaded horizontally through A dumping port P at a rate m' measured relative to the port. If the earthmover maintains a constant resultant tractive force F at its front wheels to provide forward motion, determine its acceleration when half the sand is dumped. When empty, the earthmover has a mass M. Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll. Units Used: Mg 10 3 kg= kN 10 3 N= Given: A 2.5 m 2 = ρ 1520 kg m 3 = m' 900 kg s = V 10 m 3 = F 4kN= M 30 Mg= Solution: When half the sand remains, M 1 M 1 2 Vρ+= M 1 37600 kg= 388 Engineering Mechanics - Dynamics Chapter 15 t m d d m'= ρvA= v m' ρA = v 0.24 m s = Σ Fm t v d d t m d d v DR −= FM 1 am'v−= a Fm'v+ M 1 = a 0.11 m s 2 = a 112 mm s 2 = Problem 15-126 The earthmover initially carries sand of volume V having density ρ. The sand is unloaded horizontally through a dumping port P of area A at rate of r measured relative to the port. Determine the resultant tractive force F at its front wheels if the acceleration of the earthmover is a when half the sand is dumped. When empty, the earthmover has mass M. Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll. Units Used: kN 10 3 N= Mg 1000 kg= Given: V 10 m 3 = r 900 kg s = ρ 1520 kg m 3 = a 0.1 m s 2 = A 2.5 m 2 = M 30 Mg= Solution: When half the sand remains, M 1 M 1 2 Vρ+= M 1 37600 kg= t m d d r= r ρvA= v r ρA = v 0.237 m s = Fm t v d d t m d d v−= FM 1 arv−= F 3.55 kN= 389 Engineering Mechanics - Dynamics Chapter 15 Problem 15-127 If the chain is lowered at a constant speed v, determine the normal reaction exerted on the floor as a function of time. The chain has a weight W and a total length l. Given: W 5 lb ft = l 20 ft= v 4 ft s = Solution: At time t, the weight of the chain on the floor is WMgvt()= t v d d 0= M t Mvt()= t M t d d Mv= Σ F s M t v d d v Dt t M t d d += RMgvt()− 0 vMv()+= RMgvtv 2 + () = R W g gvt v 2 + () = *Problem 15-128 The rocket has mass M including the fuel. Determine the constant rate at which the fuel must be burned so that its thrust gives the rocket a speed v in time t starting from rest. The fuel is expelled from the rocket at relative speed v r . Neglect the effects of air resistance and assume that g is constant. Given: M 65000 lb= v r 3000 ft s = v 200 ft s = g 32.2 ft s 2 = t 10 s= 390 Engineering Mechanics - Dynamics Chapter 15 Solution: A System That Losses Mass: Here, Wm 0 t m e d d t− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ g= Applying Eq. 15-29, we have + ↑ Σ F z m t v d d v DE t m e d d −= integrating we find vv DE ln m o m 0 t m e d d t− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ gt−= with m 0 M= v DE v r = vv r ln m 0 m 0 t m e d d t− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ gt()−= t m e d d A= m 0 − e vgt+ v r m 0 + ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 t = A m 0 − e vgt+ v r m 0 + ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 t = A 43.3 slug s = Problem 15-129 The rocket has an initial mass m 0 , including the fuel. For practical reasons desired for the crew, it is required that it maintain a constant upward acceleration a 0 . If the fuel is expelled from the rocket at a relative speed v er , determine the rate at which the fuel should be consumed to maintain the motion. Neglect air resistance, and assume that the gravitational acceleration is constant. Solution: a 0 t v d d = + ↑ ΣF s = m t v d d v er t m e d d − mg− ma o v er t m d d −= v er dm m a 0 g+()dt= 391 Engineering Mechanics - Dynamics Chapter 15 Since v er is constant, integrating, with t = 0 when m = m 0 yields v er ln m m 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 0 g+()t= m m 0 e a 0 g+ v er ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t = The time rate fuel consumption is determined from Eq.[1] t m d d m a 0 g+ v er = t m d d m 0 a 0 g+ v er ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ e a 0 g+ v er ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t = Note : v er must be considered a negative quantity. Problem 15-130 The jet airplane of mass M has constant speed v j when it is flying along a horizontal straight line. Air enters the intake scoops S at rate r 1 . If the engine burns fuel at the rate r 2 and the gas (air and fuel) is exhausted relative to the plane with speed v e , determine the resultant drag force exerted on the plane by air resistance. Assume that air has a constant density ρ. Hint: Since mass both enters and exits the plane, Eqs. 15-29 and 15-30 must be combined. Units Used: Mg 1000 kg= kN 10 3 N= Given: M 12 Mg= r 2 0.4 kg s = v j 950 km hr = v e 450 m s = r 1 50 m 3 s = ρ 1.22 kg m 3 = Solution: ΣF s m t v d d t m e d d v DE ()− t m i d d v Di ()+= t v d d 0= v DE V e = v Di v j = t m i d d r 1 ρ= 392 Engineering Mechanics - Dynamics Chapter 15 Ar 1 ρ= t m e d d r 2 A+= Br 2 A+= Forces T and F D are incorporated as the last two terms in the equation, F D v e Bv j A−= F D 11.5 kN= Problem 15-131 The jet is traveling at speed v, angle θ with the horizontal. If the fuel is being spent at rate r 1 and the engine takes in air at r 2 whereas the exhaust gas (air and fuel) has relative speed v e , determine the acceleration of the plane at this instant. The drag resistance of the air is F D = kv 2 The jet has weight W. Hint: See Prob. 15-130. Given: v 500 mi hr = v e 32800 ft s = θ 30 deg= k 1 0.7 lb s 2 ft 2 = r 1 3 lb s = W 15000 lb= r 2 400 lb s = g 32.2 ft s 2 = Solution: t m i d d r 2 g 1 = A 1 r 2 = t m e d d r 1 r 2 + g 1 = Br 1 r 2 += v 1 v= ΣF s m t v d d v De t m e d d − v Di t m i d d += W− sin θ() k 1 v 1 2 − Wa v e B− v 1 A 1 += a W− sin θ() k 1 v 1 2 − v e B g + v 1 A 1 g − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ g W = a 37.5 ft s 2 = 393 Engineering Mechanics - Dynamics Chapter 15 *Problem 15-132 The rope has a mass m' per unit length. If the end length y = h is draped off the edge of the table, and released, determine the velocity of its end A for any position y, as the rope uncoils and begins to fall. Solution: F s m t v d d v Di t m i d d += At a time t, mm'y= and t m i d d m' t y d d = m' v= . Here, v Di v= , t v d d g= . m' g y m' y t v d d vm'v()+= gy y t v d d v 2 += Since v t y d d = , then dt dy v = gy vy y v d d v 2 += Multiply both sides by 2ydy 2gy 2 dy 2vy 2 dv 2yv 2 dy+= y2gy 2 ⌠ ⎮ ⎮ ⌡ d v 2 y 2 1 ⌠ ⎮ ⌡ d= 2 3 gy 3 C+ v 2 y 2 = C 2− 3 gh 3 = v 0= at yh= 2 3 gh 3 C+ 0= 2 3 gy 3 2 3 gh 3 − v 2 y 2 = v 2 3 g y 3 h 3 − y 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Problem 15-133 The car has a mass m 0 and is used to tow the smooth chain having a total length l and a mass per unit of length m'. If the chain is originally piled up, determine the tractive force F that must be supplied by the rear wheels of the car, necessary to maintain a constant speed v while the chain is being drawn out. 394 Engineering Mechanics - Dynamics Chapter 15 Solution: + ⎯ ⎯ → ΣF s = m t v d d v Di t m i d d + At a time t, mm 0 ct+= Where, c t m i d d = m' t x d d = m'v= Here, v Di v= t v d d 0= Fm 0 m'vt−()0() vm'v()+= m'v 2 = F m'v 2 = Problem 15-134 Determine the magnitude of force F as a function of time, which must be applied to the end of the cord at A to raise the hook H with a constant speed v. Initially the chain is at rest on the ground. Neglect the mass of the cord and the hook. The chain has a mass density ρ. Given: v 0.4 m s = ρ 2 kg m = g 9.81 m s 2 = Solution: t v d d 0= yvt= m i my= mvt= t m i d d mv= + ↑ ΣF s m t v d d v Di t m i d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ += Fmgvt− 0 vmv+= Fmgvtvmv+= F ρgvt v 2 += f 1 ρgv= f 1 7.85 N s = f 2 ρv 2 = f 2 0.320 N= Ff 1 tf 2 += 395 Engineering Mechanics - Dynamics Chapter 16 Problem 16-1 A wheel has an initial clockwise angular velocity ω and a constant angular acceleration α. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity ω f . What time is required? Units Used: rev 2π rad= Given: ω 10 rad s = α 3 rad s 2 = ω f 15 rad s = Solution: ω f 2 ω 2 2αθ+= θ ω f 2 ω 2 − 2α = θ 3.32 rev= ω f ωαt+= t ω f ω− α = t 1.67 s= Problem 16-2 A flywheel has its angular speed increased uniformly from ω 1 to ω 2 in time t. If the diameter of the wheel is D, determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel at time t, and the total distance the point travels during the time period. Given: ω 1 15 rad s = ω 2 60 rad s = t 80 s= D 2ft= Solution: r D 2 = ω 2 ω 1 αt+= α ω 2 ω 1 − t = α 0.56 rad s 2 = a t αr= a t 0.563 ft s 2 = a n ω 2 2 r= a n 3600 ft s 2 = θ ω 2 2 ω 1 2 − 2α = θ 3000 rad= d θr= d 3000 ft= Problem 16-3 The angular velocity of the disk is defined by ω = at 2 + b. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = t 1 . 396 Engineering Mechanics - Dynamics Chapter 16 Given: a 5 rad s 3 = b 2 rad s = r 0.8 m= t 1 0.5 s= Solution: tt 1 = ω at 2 b+= ω 3.25 rad s = α 2at= α 5.00 rad s 2 = v ωr= v 2.60 m s = a αr() 2 ω 2 r () 2 += a 9.35 m s 2 = *Problem 16-4 The figure shows the internal gearing of a “spinner” used for drilling wells. With constant angular acceleration, the motor M rotates the shaft S to angular velocity ω M in time t starting from rest. Determine the angular acceleration of the drill-pipe connection D and the number of revolutions it makes during the start up at t. Units Used: rev 2π= Given: ω M 100 rev min = r M 60 mm= r D 150 mm= t 2s= Solution: ω M α M t= α M ω M t = α M 5.24 rad s 2 = α M r M α D r D = 397 Engineering Mechanics - Dynamics Chapter 16 α D α M r M r D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α D 2.09 rad s 2 = θ 1 2 α D t 2 = θ 0.67 rev= Problem 16-5 If gear A starts from rest and has a constant angular acceleration α A , determine the time needed for gear B to attain an angular velocity ω B . Given: α A 2 rad s 2 = r B 0.5 ft= r A 0.2 ft= ω B 50 rad s = Solution: The point in contact with both gears has a speed of v p ω B r B = v p 25.00 ft s = Thus, ω A v p r A = ω A 125.00 rad s = So that ωα c t= t ω A α A = t 62.50 s= Problem 16-6 If the armature A of the electric motor in the drill has a constant angular acceleration α A , determine its angular velocity and angular displacement at time t. The motor starts from rest. Given: α A 20 rad s 2 = t 3s= Solution: ωα c t= ωα A t= ω 60.00 rad s = 398 Engineering Mechanics - Dynamics Chapter 16 θ 1 2 α A t 2 = θ 90.00 rad= Problem 16-7 The mechanism for a car window winder is shown in the figure. Here the handle turns the small cog C, which rotates the spur gear S, thereby rotating the fixed-connected lever AB which raises track D in which the window rests. The window is free to slide on the track. If the handle is wound with angular velocity ω c , determine the speed of points A and E and the speed v w of the window at the instant θ. Given: ω c 0.5 rad s = r C 20 mm= θ 30 deg= r s 50 mm= r A 200 mm= Solution: v C ω c r C = v C 0.01 m s = ω s v C r s = ω s 0.20 rad s = v A v E = ω s r A = v A ω s r A = v A v E = 40.00 mm s = Points A and E move along circular paths. The vertical component closes the window. v w v A cos θ()= v w 34.6 mm s = *Problem 16-8 The pinion gear A on the motor shaft is given a constant angular acceleration α. If the gears A and B have the dimensions shown, determine the angular velocity and angular displacement of the output shaft C, when t = t 1 starting from rest. The shaft is fixed to B and turns with it. Given: α 3 rad s 2 = 399 Engineering Mechanics - Dynamics Chapter 16 t 1 2s= r 1 35 mm= r 2 125 mm= Solution: α A α= r 1 α A r 2 α C = α C r 1 r 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ α A = ω C α C t 1 = ω C 1.68 rad s = θ C 1 2 α C t 1 2 = θ C 1.68 rad= Problem 16-9 The motor M begins rotating at an angular rate ω = a(1 − e bt ). If the pulleys and fan have the radii shown, determine the magnitudes of the velocity and acceleration of point P on the fan blade when t = t 1 . Also, what is the maximum speed of this point? Given: a 4 rad s = r 1 1in= b 1− 1 s = r 2 4in= t 1 0.5 s= r 3 16 in= Solution: tt 1 = r 1 ω 1 r 2 ω 2 = ω 1 a 1 e bt − () = ω 2 r 1 r 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ω 1 = v P r 3 ω 2 = v P 6.30 in s = 400 Engineering Mechanics - Dynamics Chapter 16 α 1 a− be bt = α 2 r 1 r 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ α 1 = a P α 2 r 3 () 2 ω 2 2 r 3 () 2 += a P 10.02 in s 2 = As t approaches ∞ ω 1 a= ω f r 1 r 2 ω 1 = v f r 3 ω f = v f 16.00 in s = Problem 16-10 The disk is originally rotating at angular velocity ω 0 . If it is subjected to a constant angular acceleration α, determine the magnitudes of the velocity and the n and t components of acceleration of point A at the instant t. Given: ω 0 8 rad s = α 6 rad s 2 = t 0.5 s= r 2ft= Solution: ωω 0 αt+= v A rω= v A 22.00 ft s = a t rα= a t 12.00 ft s 2 = a n rω 2 = a n 242.00 ft s 2 = 401 Engineering Mechanics - Dynamics Chapter 16 Problem 16-11 The disk is originally rotating at angular velocity ω 0 . If it is subjected to a constant angular acceleration α, determine the magnitudes of the velocity and the n and t components of acceleration of point B just after the wheel undergoes a rotation θ. Given: rev 2π rad= α 6 rad s 2 = r 1.5 ft= ω 0 8 rad s = θ 2 rev= Solution: ωω 0 2 2αθ+= ω 14.66 rad s = v B rω= v B 22 ft s = a Bt rα= a Bt 9 ft s 2 = a Bn rω 2 = a Bn 322 ft s 2 = *Problem 16-12 The anemometer measures the speed of the wind due to the rotation of the three cups. If during a time period t 1 a wind gust causes the cups to have an angular velocity ω = (At 2 + B), determine (a) the speed of the cups when t = t 2 , (b) the total distance traveled by each cup during the time period t 1 , and (c) the angular acceleration of the cups when t = t 2 . Neglect the size of the cups for the calculation. Given: t 1 3s= t 2 2s= r 1.5 ft= A 2 1 s 3 = B 3 1 s = 402 Engineering Mechanics - Dynamics Chapter 16 Solution: ω 2 At 2 2 B+= v 2 rω 2 = v 2 16.50 ft s = dr 0 t 1 tAt 2 B+ ⌠ ⎮ ⌡ d= d 40.50 ft= α dω 2 dt = α 2At 2 = α 8.00 rad s 2 = Problem 16-13 A motor gives disk A a clockwise angular acceleration α A = at 2 + b. If the initial angular velocity of the disk is ω 0 , determine the magnitudes of the velocity and acceleration of block B when t = t 1 . Given: a 0.6 rad s 4 = ω 0 6 rad s = r 0.15 m= b 0.75 rad s 2 = t 1 2s= Solution: α A at 1 2 b+= ω A a 3 t 1 3 bt 1 + ω 0 += v B ω A r= v B 1.365 m s = a B α A r= a B 0.472 m s 2 = Problem 16-14 The turntable T is driven by the frictional idler wheel A, which simultaneously bears against the inner rim of the turntable and the motor-shaft spindle B. Determine the required diameter d of the spindle if the motor turns it with angular velocity ω B and it is required that the turntable rotate with angular velocity ω T . 403 Engineering Mechanics - Dynamics Chapter 16 ω B 25 rad s = ω T 2 rad s = a 9in= Solution: ω B d 2 ω A a d 2 − 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = ω A ω B d a d 2 − = ω A a d 2 − 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ω T a= ω B d 2 ω T a= d 2ω T a ω B = d 1.44 in= Problem 16-15 Gear A is in mesh with gear B as shown. If A starts from rest and has constant angular acceleration α Α , determine the time needed for B to attain an angular velocity ω B . Given: α A 2 rad s 2 = r A 25 mm= ω B 50 rad s = r B 100 mm= Solution: α A r A α B r B = α B r A r B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ α A = ω B α B t= t ω B α B = t 100.0 s= 404 Given: Engineering Mechanics - Dynamics Chapter 16 *Problem 16-16 The blade on the horizontal-axis windmill is turning with an angular velocity ω 0 . Determine the distance point P on the tip of the blade has traveled if the blade attains an angular velocity ω in time t. The angular acceleration is constant. Also, what is the magnitude of the acceleration of this point at time t? Given: ω 0 2 rad s = ω 5 rad s = t 3s= r p 15 ft= Solution: α ωω 0 − t = d p r p 0 t tω 0 αt+ ⌠ ⎮ ⌡ d= d p 157.50 ft= a n r p ω 2 = a t r p α= a p a n a t ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = a p 375.30 ft s 2 = Problem 16-17 The blade on the horizontal-axis windmill is turning with an angular velocity ω 0 . If it is given an angular acceleration α, determine the angular velocity and the magnitude of acceleration of point P on the tip of the blade at time t. 405 Engineering Mechanics - Dynamics Chapter 16 Given: ω 0 2 rad s = α 0.6 rad s 2 = r 15 ft= t 3s= Solution: ωω 0 αt+= ω 3.80 rad s = a pt αr= a pt 9.00 ft s 2 = a pn ω 2 r= a pn 216.60 ft s 2 = a p a pt 2 a pn 2 += a p 217 ft s 2 = Problem 16-18 Starting from rest when s = 0, pulley A is given an angular acceleration α Α = kθ. Determine the speed of block B when it has risen to s = s 1 . The pulley has an inner hub D which is fixed to C and turns with it. Given: k 6s 2− = r C 150 mm= s 1 6m= r D 75 mm= r A 50 mm= Solution: θ A r A θ C r C = θ C r D s 1 = θ A r C r A ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ s 1 r D = α A kθ= ω A 2 2 k θ A 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = ω A kθ A = ω A r A ω C r C = ω C r A r C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ω A = v B ω C r D = v B 14.70 m s = 406 Engineering Mechanics - Dynamics Chapter 16 Problem 16-19 Starting from rest when s = 0, pulley A is given a constant angular acceleration α Α . Determine the speed of block B when it has risen to s = s 1 .The pulley has an inner hub D which is fixed to C and turns with it. Given: α A 6 rad s 2 = s 1 6m= r A 50 mm= r C 150 mm= r D 75 mm= Solution: θ A r A θ C r C = θ C r D s 1 = θ A r C r A ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ s 1 r D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = ω A 2 2 α A θ A = ω A 2α A θ A = ω A r A ω C r C = ω C r A r C ω A = v B ω C r D = v B 1.34 m s = *Problem 16-20 Initially the motor on the circular saw turns its drive shaft at ω = kt 2/3 . If the radii of gears A and B are r A and r B respectively, determine the magnitudes of the velocity and acceleration of a tooth C on the saw blade after the drive shaft rotates through angle θ = θ 1 starting from rest. 407 Engineering Mechanics - Dynamics Chapter 16 Given: r A 0.25 in= r B 1in= r C 2.5 in= θ 1 5 rad= k 20 rad s 5 3 = Solution: ω A kt 2 3 = θ A 3 5 kt 5 3 = t 1 5θ 1 3k ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 5 = t 1 0.59 s= ω A kt 1 2 3 = ω A 14.09 rad s = ω B r A r B ω A = ω B 3.52 rad s = α A 2 3 kt 1 1− 3 = α A 15.88 rad s 2 = α B r A r B α A = α B 3.97 rad s 2 = v C r C ω B = v C 8.81 in s = a C r C α B () 2 r C ω B 2 () 2 += a C 32.6 in s 2 = Problem 16-21 Due to the screw at E, the actuator provides linear motion to the arm at F when the motor turns the gear at A. If the gears have the radii listed, and the screw at E has pitch p, determine the speed at F when the motor turns A with angular velocity ω A . Hint: The screw pitch indicates the amount of advance of the screw for each full revolution. 408 Engineering Mechanics - Dynamics Chapter 16 Given: rev 2π rad= p 2 mm rev = ω A 20 rad s = r A 10 mm= r B 50 mm= r C 15 mm= r D 60 mm= Solution: ω A r A ω B r B = ω B r C ω D r D = ω D r A r B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r C r D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ω A = ω D 1 rad s = v F ω D p= v F 0.318 mm s = Problem 16-22 A motor gives gear A angular acceleration α A = aθ 3 + b. If this gear is initially turning with angular velocity ω A0 , determine the angular velocity of gear B after A undergoes an angular displacement θ 1 . Given: rev 2π rad= a 0.25 rad s 2 = b 0.5 rad s 2 = ω A0 20 rad s = 409 Engineering Mechanics - Dynamics Chapter 16 r A 0.05 m= r B 0.15 m= θ 1 10 rev= Solution: α A aθ 3 b+= ω A 2 ω A0 2 2 0 θ 1 θaθ 3 b+ () ⌠ ⎮ ⌡ d+= ω A ω A0 2 2 0 θ 1 θaθ 3 b+ ⌠ ⎮ ⌡ d+= ω A 1395.94 rad s = ω B r A r B ω A = ω B 465 rad s = Problem 16-23 A motor gives gear A angular acceleration α A = kt 3 . If this gear is initially turning with angular velocity ω A0 , determine the angular velocity of gear B when t = t 1 . Given: t 1 2s= k 4 rad s 5 = r A 0.05 m= ω A0 20 rad s = r B 0.15 m= Solution: tt 1 = α A kt 3 = ω A k 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ t 4 ω A0 += ω A 36.00 rad s = ω B r A r B ω A = ω B 12.00 rad s = *Problem 16-24 For a short time a motor of the random-orbit sander drives the gear A with an angular velocity ω A = A(t 3 + Bt). This gear is connected to gear B, which is fixed connected to the shaft CD. The end of this shaft is connected to the eccentric spindle EF and pad P, which causes the pad to orbit around shaft CD at a radius r E . Determine the magnitudes of the velocity and the 410 Engineering Mechanics - Dynamics Chapter 16 tangential and normal components of acceleration of the spindle EF at time t after starting from rest. Given: r A 10 mm= r B 40 mm= r E 15 mm= A 40 rad s 4 = B 6s 2 = t 2s= Solution: ω A At 3 Bt+ () = ω B r A r B ω A = α A A 3t 2 B+ () = α B r A r B α A = v ω B r E = v 3.00 m s = a t α B r E = a t 2.70 m s 2 = a n ω B 2 r E = a n 600.00 m s 2 = Problem 16-25 For a short time the motor of the random-orbit sander drives the gear A with an angular velocity ω A = kθ 2 . This gear is connected to gear B, which is fixed connected to the shaft CD. The end of this shaft is connected to the eccentric spindle EF and pad P, which causes the pad to orbit around shaft CD at a radius r E . Determine the magnitudes of the velocity and the tangential and normal components of acceleration of the spindle EF when θ = θ 1 starting from rest. Units Used: rev 2π rad= 411 Engineering Mechanics - Dynamics Chapter 16 Given: r A 10 mm= k 5 rad s = r B 40 mm= θ 1 0.5 rev= r E 15 mm= Solution: ω A kθ 1 2 = α A kθ 1 2 ()2kθ 1 ()= ω B r A r B ω A = α B r A r B α A = v ω B r E = v 0.19 m s = a t α B r E = a t 5.81 m s 2 = a n ω B 2 r E = a n 2.28 m s 2 = Problem 16-26 The engine shaft S on the lawnmower rotates at a constant angular rate ω A . Determine the magnitudes of the velocity and acceleration of point P on the blade and the distance P travels in time t. The shaft S is connected to the driver pulley A, and the motion is transmitted to the belt that passes over the idler pulleys at B and C and to the pulley at D. This pulley is connected to the blade and to another belt that drives the other blade. Given: ω A 40 rad s = r P 200 mm= r A 75 mm= α A 0= r D 50 mm= t 3s= 412 Engineering Mechanics - Dynamics Chapter 16 Solution: ω D r A r D ω A = v P ω D r P = v P 12.00 m s = a P ω D 2 r P = a P 720.00 m s 2 = s P r P ω A tr A r D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = s P 36.00 m= Problem 16-27 The operation of “reverse” for a three-speed automotive transmission is illustrated schematically in the figure. If the crank shaft G is turning with angular speed ω G , determine the angular speed of the drive shaft H. Each of the gears rotates about a fixed axis. Note that gears A and B, C and D, and E and F are in mesh. The radii of each of these gears are listed. Given: ω G 60 rad s = r A 90 mm= r B 30 mm= r C 30 mm= r D 50 mm= r E 70 mm= r F 60 mm= Solution: ω B r A r B ω G = ω B 180.00 rad s = ω D r C r D ω B = ω D 108.00 rad s = ω H r E r F ω D = ω H 126.00 rad s = 413 Engineering Mechanics - Dynamics Chapter 16 *Problem 16-28 Morse Industrial manufactures the speed reducer shown. If a motor drives the gear shaft S with an angular acceleration α = ke bt , determine the angular velocity of shaft E at time t after starting from rest. The radius of each gear is listed. Note that gears B and C are fixed connected to the same shaft. Given: r A 20 mm= r B 80 mm= r C 30 mm= r D 120 mm= k 0.4 rad s 2 = b 1s 1− = t 2s= Solution: ω 0 t tke bt ⌠ ⎮ ⌡ d= ω 2.56 rad s = ω E r A r B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r C r D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ω= ω E 0.160 rad s = Problem 16-29 Morse Industrial manufactures the speed reducer shown. If a motor drives the gear shaft S with an angular acceleration α = kω -3 , determine the angular velocity of shaft E at time t 1 after gear S starts from an angular velocity ω 0 when t = 0. The radius of each gear is listed. Note that gears B and C are fixed connected to the same shaft. Given: r A 20 mm= r B 80 mm= 414 Engineering Mechanics - Dynamics Chapter 16 r C 30 mm= r D 120 mm= ω 0 1 rad s = k 4 rad s 5 = t 1 2s= Solution: Guess ω 1 1 rad s = Given 0 t 1 tk ⌠ ⎮ ⌡ d ω 0 ω 1 ωω 3 ⌠ ⎮ ⌡ d= ω 1 Find ω 1 ()= ω 1 2.40 rad s = ω E r A r B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r C r D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ω 1 = ω E 0.150 rad s = Problem 16-30 A tape having a thickness s wraps around the wheel which is turning at a constant rate ω. Assuming the unwrapped portion of tape remains horizontal, determine the acceleration of point P of the unwrapped tape when the radius of the wrapped tape is r. Hint: Since v P = ω r, take the time derivative and note that dr/dt = ω (s/2π). Solution: v P ωr= a p dv p dt = dω dt r ω dr dt += since dω dt 0= , a p ω dr dt ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = In one revolution r is increased by s, so that 2π Δθ s Δr = 415 Engineering Mechanics - Dynamics Chapter 16 Hence, Δr s 2π Δθ= dr dt s 2π ω= a p s 2π ω 2 = Problem 16-31 The sphere starts from rest at θ = 0° and rotates with an angular acceleration α = kθ. Determine the magnitudes of the velocity and acceleration of point P on the sphere at the instant θ = θ 1 . Given: θ 1 6 rad= r 8in= φ 30 deg= k 4 rad s 2 = Solution: α kθ 1 = ω 2 2 k θ 1 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = ω kθ 1 = v P ωrcos φ()= v P 6.93 ft s = a P αrcos φ()() 2 ω 2 rcos φ() () 2 += a P 84.3 ft s 2 = *Problem 16-32 The rod assembly is supported by ball-and-socket joints at A and B. At the instant shown it is rotating about the y axis with angular velocity ω and has angular acceleration α. Determine the magnitudes of the velocity and acceleration of point C at this instant. Solve the problem using Cartesian vectors and Eqs. 16-9 and 16-13. 416 dhanesh_h Mathcad - CombinedMathcads

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