Apostila Pt1-2023-2

1 Dinâmica TMEC 019 Ementa do curso 1. Introdução 2. Cinemática da Partícula 3. Cinética da Partícula 4. Cinética de um Sistema de Partícula 5. Cinemática Plana de Corpo Rígido 6. Introdução à Dinâmica Tridimensional de Corpos Rígidos Bibliografia 1) Meriam, J.L. e Kraige, L.G., Mecânica – Dinâmica, LTC. 2) Hibbeler, R. C., Dinâmica, Pearson Education. Sistema de Avaliação 𝐌𝐅 = 𝐏𝟏 + 𝐏𝟐 𝟐 2 Parte 1 - Introdução - vide bibliografia Parte 2 - Cinemática da Partícula Conceito: Geometria apenas, sem considerar a causa do movimento. 1) Movimento Curvilíneo Plano Posição r: posição da partícula no ponto A 𝐫 + ∆𝐫: posição da partícula no ponto A’ ∆𝐫: deslocamento da partícula ∆𝑠: distância percorrida da partícula Velocidade Média 𝐯𝐦𝐞𝐝 = ∆𝐫 ∆𝑡 Velocidade Instantânea 3 𝐯 = lim ∆𝑡→0 ∆𝐫 ∆𝑡 = 𝐫̇ 𝑣 = 𝐯 = 𝑑𝑠 𝑑𝑡 = 𝑠̇ Conceito: A velocidade é tangente à trajetória!!! Vide Figura 2/5 Aceleração 𝐚 = lim ∆𝑡→0 ∆𝐯 ∆𝑡 = 𝐯̇ Conceito: A aceleração inclui os efeitos das variações do módulo e da direção de v!!! Vide Figura 2/5 e 2/6 Conceito: Agora será visto a REPRESENTAÇÃO do movimento plano (posição, velocidade e aceleração) da partícula em coordenadas retangulares, normal-tangencial e polares. 4 Coordenadas Retangulares (x-y) 𝐫 = 𝑥𝐢 + 𝑦𝐣 onde i e j são vetores unitários. 𝐯 = 𝑑𝐫 𝑑𝑡 = 𝑥̇𝐢 + 𝑥𝐢̇ + 𝑦̇𝐣 + 𝑦𝐣̇ onde 𝐢̇ = 𝐣̇ = 0 pois os vetores unitários i e j não variam nem em módulo nem em direção. Portanto, 𝐯 = 𝑑𝐫 𝑑𝑡 = 𝑥̇𝐢 + 𝑦̇𝐣 𝐚 = 𝑑𝐯 𝑑𝑡 = 𝑥̈𝐢 + 𝑦̈𝐣 Execício: Uma partícula possui movimento circular de raio R para uma posição angular genérica 𝜃, com velocidade angular 𝜃̇ = 𝜔 e aceleração angular 𝜃̈ = 𝛼 , todas no sentido anti horário. Determine os vetores de posição velocidade e aceleração da partícula representados em coordenadas retangulares. 5 Exemplo – Movimento de projéteis (resolvido no livro) Lista de Exercíos 1: Problema Resolvido 2/5 (resolvido no livro), Problema Resolvido 2/6, Problema 2/70 e Problema 2/85 2/112 Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when \theta = 60^0 if \dot{\theta} = 2 rad/sec and \ddot{\theta} = 4.025 rad/sec^2. Problem 2/112 2/114 Magnetic tape is being transferred from reel A to reel B and passes around idler pulleys C and D. At a certain instant, point P_1 on the tape is in contact with pulley C and point P_2 is in contact with pulley D. If the normal component of acceleration of P_1 is 40 m/s^2 and the tangential component of accelera- tion of P_2 is 30 m/s^2 at this instant, compute the corresponding speed v of the tape, the magnitude of the total acceleration of P_1, and the magnitude of the total acceleration of P_2. Problem 2/114 t=0 100 80 60 40 20 t=5 s y, m x, m 0 20 40 60 80 v_x=-30 m/s \phi=53.19^ v=50 m/s v_y=-40 m/s Path Path a_x=-16 m/s^2 a=17.89 m/s^2 a_y=-8 m/s^2 8 Coordenadas Normal e Tangencial (n-t) Conceito: As coordenadas normal-tangencial são consideradas como se MOVENDO ao longo da tajetória com a partícula. O sentido de n é sempre tomado para o centro de curvatura da trajetória e o sentido de t é sempre tangente a trajetória. Vide Figura 2/9. Velocidade 𝐯 = 𝑣 𝐞t 𝑣 = 𝑑𝑠 𝑑𝑡 = 𝜌 𝑑𝛽 𝑑𝑡 = 𝜌𝛽̇ onde 𝜌 é o raio de curvatura da trajetória e d𝛽 é um ângulo infinitesimal que ocorre durante um intervalo de tempo dt e et é o versor na direção tangencial e en é o vetor unitário na direção normal. Conceito: A velocidade é sempre tangente à trajetórtia!!! Vide Figura abaixo. 9 Aceleração 𝐚 = 𝑑𝐯 𝑑𝑡 = 𝑑 𝑣 𝐞𝐭 𝑑𝑡 = 𝑣𝐞̇ t + 𝑣̇𝐞t onde 𝐞̇ 𝑡 = 𝑑𝐞t 𝑑𝑡 = 𝐞𝐭 𝑑𝛽 𝑑𝑡 𝐞𝐧 = 𝛽̇ 𝐞𝐧 = 𝑣 𝜌 𝐞𝐧 portanto, 𝐚 = 𝑣2 𝜌 𝐞n + 𝑣̇𝐞t Conceito: A aceleração possui duas componentes, uma tangete à trajetória (relacionada com a mudança em módulo da velocidade) e outra normal apontando para o centro de curvatura da trajetória ( relacionada com a mudança da direção da velocidade)!!! Vide Figura abaixo. 10 Execício: Uma partícula possui movimento circular de raio R para uma posição angular genérica 𝜃, com velocidade angular 𝜃̇ = 𝜔 e aceleração angular 𝜃̈ = 𝛼 , todas no sentido anti horário. Determine os vetores de posição velocidade e aceleração da partícula representados em coordenadas normal-tangencial. 11 Lista de Exercícios 2: Problema resolvido 2/8, Problema 2/109, Problema 2/112, Problema 2/114 e Problema 2/129. 2/70 The center of mass G of a high jumper follows the trajectory shown. Determine the component v_0, measured in the vertical plane of the figure, of his take- off velocity and angle \theta if the apex of the trajectory just clears the bar at A. (In general, must the mass center G of the jumper clear the bar during a suc- cessful jump?) Problem 2/70 2/85 Ball bearings leave the horizontal trough with a ve- locity of magnitude u and fall through the 70-mm- diameter hole as shown. Calculate the permissible range of u which will enable the balls to enter the hole. Take the dashed positions to represent the lim- iting conditions. Problem 2/85 13 Coordenadas Polares (r-θ) Figura 2/13 - Direções radial e angular, 𝐞𝑟 e 𝐞𝜃 são vetores unitários. 14 Posição 𝐫 = 𝑟𝐞r Velocidade 𝐯 = 𝑑𝐫 𝑑𝑡 = 𝑑𝑟 𝑑𝑡 𝐞𝑟 + 𝑟 𝑑𝐞𝑟 𝑑𝑡 𝑑𝑟 𝑑𝑡 = 𝑟̇ 𝑑𝐞𝑟 𝑑𝑡 = 𝑑(1 𝑑𝜃) 𝑑𝑡 𝐞𝜽 = 𝜃̇𝐞𝜽 Vide Figura 2/13 𝐯 = 𝑟̇𝐞𝐫 + 𝑟𝜃̇𝐞𝛉 Vide Figura 2/14 15 Aceleração 𝐚 = 𝑑𝐯 𝑑𝑡 = 𝑑(𝑟̇) 𝑑𝑡 𝐞r + 𝑟̇ 𝑑𝐞r 𝑑𝑡 + 𝑑𝑟 𝑑𝑡 𝜃̇ 𝐞θ + 𝑟 𝑑(𝜃̇) 𝑑𝑡 𝐞θ + 𝑟𝜃̇ 𝑑𝐞θ 𝑑𝑡 onde 𝑑𝐞𝜃 𝑑𝑡 = 𝑑(1 𝑑𝜃) 𝑑𝑡 ( − 𝐞𝒓) =− 𝜃̇ 𝐞𝜽 (Vide Figura 2/13) 𝐚 = 𝑟̈ − 𝑟𝜃̇ 2 𝐞r + 𝑟𝜃̈ + 2𝜃̇𝑟̇ 𝐞θ Vide Figura 2/15 16 Execício (resolvido no quadro): Uma partícula possui movimento circular de raio R para uma posição angular genérica 𝜃, com velocidade angular 𝜃̇ = 𝜔 e aceleração angular 𝜃̈ = 𝛼, todas no sentido anti horário. Determine os vetores de posição velocidade e aceleração da partícula representados em coordenadas polares. Lista de Exercícios 3: Problema resolvido 2/9, Problema Resolvido 2/10, Problema 2/142, Problema 2/156 e Problema 2/157. SAMPLE PROBLEM 2/10 A tracking radar lies in the vertical plane of the path of a rocket which is coasting in unpowered flight above the atmosphere. For the instant when \( \theta = 30^{\circ} \), the tracking data give \( \dot{r} = 250\left(10^{3}\right) \mathrm{ft}\right) \), \( \dot{\theta} = 0.80 \) deg/sec. The acceleration of the rocket is due only to gravitational attraction and for its particular altitude is 31.4 ft/sec² vertically down. For these conditions determine the velocity \( v \) of the rocket and the values of \( \dot{r} \) and \( \ddot{\theta} \). Solution. The components of velocity from Eq. 2/13 are \[ [v_{r} = \dot{r}] [v_{\theta} = r \dot{\theta}] \] \[ v_{\theta} = 4000 \text{ ft/sec} \] \[ v_{\theta} = r \dot{\theta} = 3490 \text{ ft/sec} \] \[ \] \[ \] [\[ v = \sqrt{(4000)^2 + (3490)^2} = 5310 \text{ ft/sec} \] \[ \] (Ans.] Since the total acceleration of the rocket is \( g = 31.4 \text{ ft/sec}^2 \) down, we can easily find its \( r- \) and \( \theta- \)components for the given position. As shown in the figure, they are \[ a_{r} = -31.4 \cos 30^{\circ} = -27.2 \text{ ft/sec}^2 \] \[ a_{\theta} = 31.4 \sin 30^{\circ} = 15.70 \text{ ft/sec}^2 \] We now equate these values to the polar-coordinate expressions for \( a_{r} \) and \( a_{\theta} \) which contain the unknowns \( \dot{r} \) and \( \ddot{\theta} \). Thus, from Eq. 2/14 \[ a_{r} = \ddot{r} - r\dot{\theta}^2 \] \[-27.2 = \ddot{r} - 250\left(10^{3}\right)\left(0.80 \frac{\pi}{180}\right)^2 \] \[\ddot{r} = 21.5 \text{ ft/sec}^2 \] \[ a_{\theta} = r \ddot{\theta} + 2\dot{r}\dot{\theta} \] \[15.70 = 250\left(10^{3}\right)\ddot{\theta} + 24000\left(0.80 \frac{\pi}{180}\right) \] \[\ddot{\theta} = -3.84\left(10^{-3}\right) \text{ rad/sec}^2\] (Ans.) 2/142 The piston of the hydraulic cylinder gives pin A a constant velocity \( v = 3 \) ft/sec in the direction shown for an interval of its motion. For the instant when \( \theta = 60^{\circ} \), determine \( {\dot{r}}, \ddot{r}, \dot{\theta}, \) and \( \ddot{r} \) where \( r = OA \). Problem 2/142 2/156 The member OA of the industrial robot telescopes and pivots about the fixed axis at point O. At the instant shown, \( \theta = 60^{\circ} \), \( \dot{\theta} = 1.2 \) rad/s, \( \ddot{\theta} = 0.8 \) rad/s², \( OA = 0.9 \) m, \( \dot{OA} = 0.5 \) m/s, and \( \ddot{OA} = −6 \) m/s². Determine the magnitudes of the velocity and acceleration of joint A of the robot. Also, sketch the velocity and acceleration of A and determine the angles which these vectors make with the positive x-axis. The base of the robot does not revolve about a vertical axis. Problem 2/156 2/157 The robot arm is elevating and extending simultaneously. At a given instant, \( \theta = 30^{\circ} \), \( \dot{\theta} = 10 \) deg/s = constant, \( l = 0.5 \) m, \( \dot{l} = 0.2 \) m/s, and \( \ddot{l} = −0.3 \) m/s². Compute the magnitudes of the velocity \( v \) and acceleration \( a \) of the gripped part P. In addition, express \( v \) and \( a \) in terms of the unit vectors \( i \) and \( j \). Problem 2/157 20 Aceleração da partícula 𝐚 = 𝑥̈ 𝐢 + 𝑦̈ 𝐣 + 𝑧̈ 𝐤 Coordenadas Cilíndricas (r-θ-z) Figura 2/16 - vetores unitários 𝐞𝑟, 𝐞𝜃 e k. Posição da partícula 𝐑 = 𝑟 𝐞𝐫 + 𝑧 𝐤 Velocidade da partícula 𝐯 = 𝑑𝐑 𝑑𝑡 = 𝑟̇ 𝐞𝐫 + 𝑟 𝑑𝐞𝐫 𝑑𝑡 + 𝑧̇ 𝐤 + 𝑧 𝑑𝐤 𝑑𝑡 𝑑𝐞𝐫 𝑑𝑡 = 𝑑(1 𝑑𝜃) 𝑑𝑡 𝐞𝜃 = 𝜃̇ 𝐞𝜃 𝑑𝐳 𝑑𝑡 = 0 𝐯 = 𝑟̇ 𝐞𝐫 + 𝑟𝜃̇ 𝐞𝛉 + 𝑧̇ 𝐤 Aceleração da partícula 𝐚 = 𝑑𝐯 𝑑𝑡 = 𝑟̈ − 𝑟𝜃̇ 2 𝐞𝐫 + 𝑟𝜃̈ + 2𝑟̇𝜃̇ 𝐞𝛉 + 𝑧̈𝐤 Exercício: Demosntrar a equação acima. Coordenadas Esféricas (R-θ-Φ) Figura 2/16 - vetores unitários 𝐞𝑅, 𝐞𝜃 e 𝐞𝜙. 21 Velocidade Aceleração Exercício: Demosntrar a equação acima. Lista de Exercícios 4: Problema resolvido 2/11, Problema Resolvido 2/12, Problema 2/167, Problema 2/174 e Problema 2/177. SAMPLE PROBLEM 2/12 An aircraft P takes off at A with a velocity vo of 250 km/h and climbs in the vertical x-y plane at the constant 15° angle with an acceleration along its flight path of 0.8 m/s². Flight progress is monitored by radar at point O. (a) Resolve the velocity of P into cylindrical-coordinate components 60 seconds after takeoff and find r˙, θ˙, and z˙ for that instant. (b) Resolve the velocity of the aircraft P into spherical-coordinate components 60 seconds after takeoff and find R˙, θ˙, and φ˙ for that instant. Solution. (a) The accompanying figure shows the velocity and acceleration vectors in the x-y plane. The takeoff speed is vo = 250 × 1000/3600 = 69.4 m/s and the speed after 60 seconds is v = vo + at = 69.4 + 0.8(60) = 117.4 m/s The distance y traveled after takeoff is s = vot + 1/2 at² = 69.4(60) + (1/2)0.8(60)² = 5610 m The y-coordinate and associated angle θ are y = 5610 cos 15° = 5420 m θ = tan⁻¹(5420/3000) = 61.0° From the figure (b) of x-y projections, we have R = √3000² + 5420² = 6190 m vrθ = R˙ cos 15° = 117.4 cos 15° = 113.4 m/s vθ = Rθ˙sin 61.0° = 99.2 m/s Ans. vrz = vθ˙sin θ = ˙ sin 61.0° = 55.0 m/s So θ˙ = 55.0/88.9(10⁻²) rad/s Ans. Finally vz = vo sin 15° = 117.4 sin 15° = 30.4 m/s (b) Refer to the accompanying figure (c), which shows the xy plane and various velocity components projected into the vertical plane containing r and R. Note that z = y tan 15° = 5420 tan 15° = 1451 m φ = tan⁻¹(5420/1451) = 13.19° R = √x² + z² = √6190² + 1451² = 6360 m From the figure, vRg = R˙ = 99.2 cos 13.19° + 30.4 sin 13.19° = 103.6 m/s Ans. θ˙ = 8.88(10⁻²) rad/s, as in part (a) vRp = Rθ˙ = 30.4 cos 13.19° − 99.2 sin 13.19° = 6.96 m/s φ˙ = θ˙ = 1.093(10⁻²) rad/s Ans. 2/167 An amusement ride called the “corkscrew” takes the passengers through the upside-down curve of a horizontal cylindrical helix. The velocity of the cars as they pass position A is 15 m/s, and the component of their acceleration measured along the tangent to the path is g cos γ at this point. The effective radius of the cylindrical helix is 5 m, and the helix angle is γ = 40°. Compute the magnitude of the acceleration of the passengers as they pass position A. Problem 2/167 2/174 The rotating nozzle sprays a large circular area and turns with the constant angular rate θ˙ = K. Particles of water move along the tube at the constant rate l˙ = c relative to the tube. Write expressions for the magnitudes of the velocity and acceleration of a water particle P for a given position l in the rotating tube. Problem 2/174 2/177 The base structure of the firetruck ladder rotates about a vertical axis through O with a constant angular velocity Ω = 10 deg/s. At the same time, the ladder unit OE elevates at a constant rate φ˙ = 7 deg/s, and section AB of the ladder extends from within section OA at the constant rate of 0.5 m/s. At the instant under consideration, φ = 30°, OA = 9 m, and AB = 6 m. Determine the magnitudes of the velocity and acceleration of the end B of the ladder. Problem 2/177 Exercício Computacional *2/252 A projectile is launched from point A with speed v0 = 30 m/s. Determine the value of the launch angle α which maximizes the range R indicated in the figure. Determine the corresponding value R. Problem 2/252 26 Parte 3 - Cinética da Partícula Conceito: Em cinética da partícula é considerada a causa do movimento (forças que atuam sobre a partícula). 3.1 Segunda Lei de Newton Da verificação experimental, 𝐅 = m𝐚 Referencial inercial: sistema de eixos não girante que se translade com velocidade constante. A segunda lei de Newton se aplica em um sistema inercial. 3.2 Equação do Movimento 𝐅 = m𝐚 (chamada equação do movimento) A equação do movimento fornece o valor instantâneo da aceleração correpondente aos valores instantâneos das forças que estão atuando sobre uma partícula de massa. Conceito: Tem dois tipos de problemas de cinética da partícula: Problema 1) Dado variáveis cinéticas (forças externas que atuam sobre a partícula) e variáveis cinemáticas (posição, velocidade e aceleração da partícula), determine variáveis cinéticas (forças internas, reaçoes). Este problema recai em um sistema de equaçoes algébricas. Problema 2) Dado variáveis cinéticas (forças externas que atuam sobre a partícula), determine variáveis cinemáticas (posição, velocidade e aceleração da partícula). Este problema recai em um sistema de EDOs. 27 Exemplo: Pêndulo simples para pequenas oscilaçoes. (resolvido em sala de aula). Neste exemplo, resolvido no quadro, são mostrados os dois problemas de cinética da partícula. 3.3 Cinética da Partícula para o Movimento Retilíneo Para o caso geral, em três dimensões, a equação do movimento é, Lista de Exercícios 5: Problema resolvido 3/1, Problema Resolvido 3/2, Problema Resolvido 3/3, Problema Resolvido 3/4 e Problema Resolvido 3/5, Problema 3/2, Problema 3/3, Problema 3/15, Problema 3/32, Problema 3/37 e Problema 3/42. SAMPLE PROBLEM 3/1 A 75-kg man stands on a spring scale in an elevator. During the first 3 seconds of motion from rest, the tension T in the hoisting cable is 8300 N. Find the reading R of the scale in newtons during this interval and the upward velocity v of the elevator at the end of the 3 seconds. The total mass of the elevator, man, and scale is 750 kg. Solution. The force registered by the scale and the velocity both depend on the acceleration of the elevator, which is constant during the interval for which the forces are constant. From the free-body diagram of the elevator, scale, and man taken together, the acceleration is found to be ΣF_y = ma_y 8300 - 7360 = 750a_y a_y = 1.257 m/s^2 The scale reads the downward force exerted on it by the man’s feet. The equal and opposite reaction R to this action is shown on the free-body diagram of the man alone together with his weight, and the equation of motion for him gives ΣF_y = ma_y R - 736 = 75(1.257) R = 800 N The velocity reached at the end of the 3 seconds is Δv = ∫ a dt v = 0 ∫ 3 1.257 dt v = 3.77 m/s Ans. Helpful Hint If the scale were calibrated in kilograms, it would read 800/9.81 = 81.46 kg which, of course, is not his true mass since the measurement was made in a noninertial (accelerating) system. Suggestion: Rework this problem in U.S. customary units. SAMPLE PROBLEM 3/2 A small inspection car with a mass of 200 kg runs along the fixed overhead cable and is controlled by the attached cable at A. Determine the acceleration of the car when the control cable is horizontal and under a tension T = 2.4 kN. Also find the total force P exerted by the supporting cable on the wheels. Solution. The free-body diagram of the car and wheels taken together and treated as a particle discloses the 2.4-kN tension T, the weight W = mg = 200(9.81) = 1962 N, and the force P exerted on the wheel assembly by the cable. The car is in equilibrium in the y-direction since there is no acceleration in this direction. Thus, ΣF_y = 0 P - 2.4(5/13) - 1.962(12/13) = 0 P = 2.73 kN Ans. In the x-direction the equation of motion gives ΣF_x = ma_x 2400(12/13) - 1962(5/13) = 200a a = 7.30 m/s^2 Ans. Helpful Hint SAMPLE PROBLEM 3/3 The 250-lb concrete block A is released from rest in the position shown and pulls the 400-lb log up the 30° ramp. If the coefficient of kinetic friction between the log and the ramp is 0.5, determine the velocity of the block as it hits the ground at B. Solution. The motions of the log and the block A are clearly dependent. Although by now it should be evident that the acceleration of the log up the incline is half the downward acceleration of A, we may prove it formally. The constant total length of the cable is L = z_C + x_A + constant, where the constant accounts for the cable portions wrapped around the pulleys. Differentiating twice with respect to time gives 0 = 2z_C ̇ + x ̈A, or 0 = 2a_zC + a_A We assume here that the masses of the pulleys are negligible and that they turn with negligible friction. With these assumptions the free-body diagram of the pulley C discloses force and moment equilibriums. Thus, the tension in the cable attached to the log is twice that applied to the block. Note that the accelerations of the log and the center of pulley C are identical. The free-body diagram of the log shows the friction force µ_kN for motion up the plane. Equilibrium of the log in the y-direction gives ΣF_y = 0 N - 400 cos 30° = 0 N = 346 lb and its equation of motion in the x-direction gives ΣF_x = ma_x 0.5(346) = 2T + 400 sin 30° - 400/32.2 a_zC For the block in the positive downward direction, we have ΣF = ma 250 - T = 250/32.2 a_A Solving the three equations in a_zC, a_A, and T gives us _a_A = 5.83 ft/sec^2 _a_zC = -2.92 ft/sec^2 T = 205 lb For the 20-ft drop with constant acceleration, the block acquires a velocity [v_A^2 = 2ax] v_A = √[2(5.83)(20)] = 15.27 ft/sec Ans. Helpful Hints 1. The coordinates used in expressing the final kinematic relationship must be consistent with those used for the kinetic equations of motion. 2. We can verify that the log will indeed move up the ramp by calculating the force in the cable necessary to initiate motion from the equilibrium condition. This force is 2T = 0.5N + 400 sin 30° = 373 lb or 2T = 186.5 lb, which is less than the 250-lb weight of block A. Hence, the log will move up. 3. Note the serious error in assuming that T = 250 lb, in which case, block A would not accelerate. 4. Because the forces on this system remain constant, the resulting accelerations also remain constant. SAMPLE PROBLEM 3/4 The design model for a new ship has a mass of 10 kg and is tested in an experimental towing tank to determine its resistance to motion through the water at various speeds. The test results are plotted on the accompanying graph, and the resistance may be closely approximated by the dashed parabolic curve shown. If the model is released when it has a speed of 2 m/s, determine the time t required for it to reduce its speed to 1 m/s and the corresponding travel distance x. Solution. We approximate the resistance-velocity relation by R = ku^2 and find k by substituting R = 8 N and u = 2 m/s into the equation, which gives k = 8/2^2 = 2 N·s^2/m^2. Thus, R = 2u^2. The only horizontal force on the model is R, so that ΣF_x = ma_x -R = ma_x or -2u^2 = 10 du/dt We separate the variables and integrate to obtain ∫2 udt = -5 ∫du/u^2 Thus, when u = v/2 = 1 m/s, the time is t = 5(1.5) - (1 - 1/2) = 2.5 s Ans. The distance traveled during the 2.5 seconds is obtained by integrating v = dx/dt. Thus, x = 10(t^5 + 2t) so that dx = ∫0^2 15 10 dx/2 x = (10/2)ln(5 + 2u) | 1.5 = 34.7 m Ans. Helpful Hints ○ Be careful to observe the minus sign for R. ● Suggestion: Express the distance x after release in terms of the velocity x and see if you agree with the resulting relation x = 5ln(v/g). SAMPLE PROBLEM 3/5 The collar of mass m slides up the vertical shaft under the action of a force F of constant magnitude but variable direction. If θ = kt where k is a constant and if the collar starts from rest with θ = 0, determine the magnitude P of the force which will result in the collar coming to rest as θ reaches π/2. The coefficient of kinetic friction between the collar and shaft is μ. Solution. After drawing the free-body diagram, we apply the equation of motion in the y-direction to get ΣF_y = ma_y F cos θ = μ_k N - mg = m dv/dt where equilibrium in the horizontal direction requires N = F sin θ. Substituting θ = kt and integrating first between general limits give ∫12^0 F cos θk dt - μ_kF sin kt - mg kt = m 0 dv which becomes F/k ln kt + μ_k(cos kt - 1) - mgt = mv For θ = π/2 the time becomes t = π/2k, and v = 0 so that F/k[π/2 + μ_k(0 - 1)] - mgt = 0 and F = ng/2(1 - μ_k) Ans. Helpful Hints ○ If θ were expressed as a function of the vertical displacement y instead of the time t, the acceleration a would become a function of the displacement and we would use v dv = a dy. ● We see that the results do not depend on h, the rate at which the force changes direction. 3/2 The 50-kg crate of Prob. 3/1 is now projected down an incline as shown with an initial speed of 7 m/s. Investigate the time t required for the crate to come to rest and the corresponding distance x traveled if (a) θ = 15° and (b) θ = 30°. Problem 3/2 3/3 The 100-lb crate is carefully placed with zero velocity on the incline. Describe what happens if (a) θ = 15° and (b) θ = 20° Problem 3/3 3/15 A train consists of a 400,000-lb locomotive and one hundred 200,000-lb hopper cars. If the locomotive exerts a friction force of 40,000 lb on the rails in starting the train from rest, compute the forces in couplers 1 and 100. Assume no slack in the couplers and neglect friction associated with the hopper cars. Problem 3/15 3/32 The sliders A and B are connected by a light rigid bar of length l = 0.5 m and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position where xA = 0.4 m, the velocity of A is vA = 0.9 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant. Problem 3/32 3/37 Compute the acceleration of block A for the instant depicted. Neglect the masses of the pulleys. Problem 3/37 3/42 For what value(s) of the angle θ will the acceleration of the 80-lb block be 26 ft/s² to the right? Problem 3/42 SAMPLE PROBLEM 3/6 Determine the maximum speed v which the sliding block may have as it passes point A without losing contact with the surface. SOLUTION. The condition for loss of contact is that the normal force N which the surface exerts on the block goes to zero. Summing forces in the normal direction gives: [ΣFₙ = maₙ] mg = m\(v²/p) v = \sqrt{gp} Ans. If the speed at A were less than \sqrt{gp}, then an upward normal force exerted by the surface on the block would exist. In order for the block to have a speed at A which is greater than \sqrt{gp}, some type of constraint, such as a second curved surface above the block, would have to be introduced to provide additional downward force. SAMPLE PROBLEM 3/7 Small objects are released from rest at A and slide down the smooth circular surface of radius R to a conveyor B. Determine the expression for the normal contact force N between the guide and each object in terms of u and specify the correct angular velocity ω of the conveyor pulley of radius r to prevent any sliding on the belt as the objects transfer to the conveyor. SOLUTION. The free-body diagram of the object is shown together with the coordinate directions n and t. The normal force N depends on the n-component of the acceleration which, in turn, depends on the velocity. The velocity will be cumulative according to the tangential acceleration aₜ. Hence, we will find aₜ first for any general position. [ΣFₜ = maₜ] mg \cos θ = maₜ aₜ = g \cos θ Now we can find the velocity by integrating [v dv = aₜ ds] ∫₀ᵥ v dv = ∫₀^θ g \cos θ d(uθ) v² = 2gR \sin θ We obtain the normal force by summing forces in the positive n-direction, which is the direction of the n-component of acceleration. [ΣFₙ = maₙ] N - mg \sin θ = m v²/R N = 3mg \sin θ Ans. The conveyor pulley must turn at the rate \omega = ru for θ = π/2, so that ω = \sqrt{2gR/r} Ans. 140 Chapter 3 Kinetics of Particles SAMPLE PROBLEM 3/8 A 1500-kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h at A to a speed of 50 km/h as it passes C. The radius of curvature ρ of the road at A is 400 m and at C is 80 m. Determine the total horizontal force exerted by the road on the tires at positions A, B, and C. Point B is the inflection point where the curvature changes direction. Solution. The car will be treated as a particle so that the effect of all forces exerted by the road on the tires will be treated as a single force. Since the motion is described along the direction of the road, normal and tangential coordinates will be used to specify the acceleration of the car. We will then determine the forces from the accelerations. The constant tangential acceleration is in the negative x-direction, and its magnitude is given by |v_C^2 = v_A^2 + 2a_t s_1 a_t = ( (50(0.6))^2 - (100(0.6))^2 ) / 2(200) = -1.447 m/s^2 The normal components of acceleration at A, B, and C are [a_n = v^2/ρ] At A, a_n = (100(0.6))^2 / 400 = 1.929 m/s^2 At B, a_n = 0 At C, a_n = (50(0.6))^2 / 80 = 2.41 m/s^2 Application of Newton’s second law in both the n- and t-directions to the free-body diagrams of the car gives ΣF_t = ma_t F_t = 1500(1.447) = 2170 N ΣF_n = ma_n At A, F_n = 1500(1.929) = 2890 N At B, F_n = 0 At C, F_n = 1500(2.41) = 3620 N Thus, the total horizontal force acting on the tires becomes At A, F = √(F_n^2 + F_t^2) = √(2890)^2 + (2170)^2 = 3620 N Ans. F = F_t = 2170 N Ans. At C, F = √(F_n^2 + F_t^2) = √(3620)^2 + (2170)^2 = 4220 N Ans. Helpful Hints Note that Recognize the numerical value of the conversion factor from km/h to m/s as 1000/3600 or 1/3.6. Note that a_n is always directed toward the center of curvature. 141ARTICLE 3/3 CURVILINEAR MOTION 141 SAMPLE PROBLEM 3/9 Compute the magnitude v of the velocity required for the spacecraft S to maintain a circular orbit of altitude 200 mi above the surface of the earth. Solution. The only external force acting on the spacecraft is the force of gravitational attraction to the earth (i.e., its weight), as shown in the free-body diagram. Summing forces in the normal direction yields [ΣF_n = ma_n] GMm ——— = mv^2 (R + h)^2 (R + h) v^2 = GMm (R + h) = (R + h)^2 = g (R + h) where the substitution gR^2 = GMm has been made. Substitution of numbers gives v = (3969)/(5280) ≈ 25,326 ft/sec Helpful Hint Note that, for observations made within an inertial frame of reference, there is no such quantity as “centrifugal force” acting in the minus n-direction. Note also that neither the spacecraft nor its occupants are “weightless,” because the weight in each case is given by Newton’s law of gravitation. For this altitude, the weights are only about 10 percent less than the earth-surface values. Finally, the term “zero-g” is also misleading. It is only when we make our observations with respect to a coordinate system which has an acceleration equal to the gravitational acceleration (such as in an orbiting spacecraft) that we appear to be in a “zero-g” environment. The quantity which does go to zero around orbiting spacecraft is the familiar normal force associated with, for example, an object in contact with a horizontal surface within the spacecraft. SAMPLE PROBLEM 3/10 Tube A rotates about the vertical O-axis with a constant angular rate θ ̇ = ω and contains a small cylindrical plug B of mass m whose radial position is controlled by the cord which passes freely through the tube and shaft and is wound around the drum of radius b. Determine the tension T in the cord and the horizontal component F_R of force exerted by the tube on the plug when the constant angular rate of rotation of the drum is θ ̇ = ω, first in the direction for case (a) and second in the direction for case (b). Neglect friction. Solution. With r a variable, we use the polar-coordinate form of the equations of motion, Eqs. 3/8. The free-body diagram of B is shown in the horizontal plane and discloses only T and F_R. The equations of motion are ΣF_θ = ma_t -T = m (r ̇ - rθ ̇^2) ΣF_r = ma_r F_r = m (r ̇^2 + rθ ̇) Case (a). With r ̇ = + bθ ̇, r ̇ = 0, and θ ̇ = 0, the forces become T = mrω^2 F_r = 2mbθ ̇ω Ans. Case (b). With r ̇ = - bθ ̇, r ̇ = 0, and θ ̇ = 0, the forces become T = mrω^2 F_r = - 2mbθ ̇ω Ans. Helpful Hint The minus sign shows that F_R is in the direction opposite to that shown on the free-body diagram. 3/52 The slotted arm rotates about its center in a hori­zontal plane at the constant angular rate θ ̇ = 10 rad/sec and carries a 3.22-lb spring-mounted slider which oscillates freely in the slot. If the slider has a speed of 24 in./sec relative to the slot as it crosses the center, calculate the horizontal side thrust P ex­erted by the slotted arm on the slider at this in­stant. Determine which side, A or B, of the slot is in contact with the slider. 3/54 The member OA rotates about a horizontal axis through O with a constant counterclockwise angu­lar velocity ω = 3 rad/sec. As it passes the position θ = 0, a small block of mass m is placed on it at a radial distance r = 18 in. If the block is observed to slip at θ = 50°, determine the coefficient of static friction μ_s between the block and the member. 3/76 Determine the speed v at which the race car will have no tendency to slip sideways on the banked track, that is, the speed at which there is no reliance on friction. In addition, determine the minimum and maximum speeds, using the coefficient of static friction μ_s = 0.90. State any assumptions. Problem 3/76 3/84 At the instant when θ = 30°, the horizontal guide is given a constant upward velocity v_0 = 2 m/s. For this instant calculate the force N exerted by the fixed circular slot and the force P exerted by the horizon- tal slot on the 0.5-kg pin A. The width of the slots is slightly greater than the diameter of the pin, and friction is negligible. Problem 3/84 3/91 The slotted arm OA rotates about a horizontal axis through point O. The 0.2-kg slider P moves with negligible friction in the slot and is controlled by the inextensible cable BP. For the instant under consideration, θ = 30°, θ̇ = 5 rad/s, θ̈ = 0, and r = 0.6 m. Determine the corresponding values of the tension in cable BP and the force reaction R perpendicular to the slot. Which side of the slot contacts the slider? Problem 3/91 3/92 The small pendulum of mass m is suspended from a trolley which runs on a horizontal rail. The trolley and pendulum are initially at rest with θ = 0. If the trolley is given a constant acceleration a = g, deter- mine the maximum angle θ_max through which the pendulum swings. Also find the tension T in the cord in terms of θ. Problem 3/92 40 3.5 Trabalho e Energia - Não há a necessidade de se calcular a aceleração da partícula para incorpora la na equação do movimento. - Envolve apenas foças que realizam trabalho, isto é, as forças ativas. Não considera forças internas. - Trata se de uma abordagem baseada na energia (escalar). Não vetorial!!! Definição de Trabalho: Diferencial do trabalho exercido pela força F sobre a partícula que está posicionada a r da origem O de um referencial fixo. dr: tangente à trajetória Produto escalar 41 Cálculo do trabalho Exemplo 1 - Trabalho da força P exercido sobre o corpo. P e α são constantes. O trabalho da força P sobre o corpo é, 42 Exemplo 2 - Trabalho exercido pela força elástica (da mola) sobra o corpo. onde 𝑥1 e 𝑥2 são os comprimentos da mola a partir de sua posição indeformada. 43 Exemplo 3 - Trabalho exercido pela força peso sobre uma partícula de massa m. 𝑑𝐫 = 𝑑𝑥𝐢 + 𝑑𝑦𝐣 Trabalho e o Movimento curvilíeneo 44 F=ma Regra da cadeia: 𝑎𝑡 = 𝑑𝑣 𝑑𝑠 𝑑𝑠 𝑑𝑡 = 𝑑𝑣 𝑑𝑠 𝑣 ou 𝑎𝑡 𝑑𝑠 = 𝑣 𝑑𝑣 Conceito: O trabalho de uma força F exercido sobre uma partícula de massa m para movê la de uma posição 1 para a posição 2 é gual a variação da energia cinética dessa partícula para essas posições. Potência P=F.v 45 1 W = 1 J/s 1 hp = 746 W Lista de Exercícios 7: Problema resolvido 3/11, Problema Resolvido 3/12, Problema Resolvido 3/13 e Problema Resolvido 3/14, Problema 3/101, Problema 3/109, Problema 3/128 e Problema 3/134. SAMPLE PROBLEM 3/13 The 50-kg block at A is mounted on rollers so that it moves along the fixed horizontal rail with negligible friction under the action of the constant 300-N force in the cable. The block is released from rest at A, with the spring to which it is attached extended an initial amount x_1 = 0.233 m. The spring has a stiffness k = 80 N/m. Calculate the velocity v of the block as it reaches position B. Solution. It will be assumed initially that the stiffness of the spring is small enough to allow the block to reach position B. The active-force diagram for the system composed of both block and cable is shown for a general position. The spring force 80x and the 300-N tension are the only forces external to this sys- tem which do work on the system. The force exerted on the block by the rail, the weight of the block, and the reaction of the small pulley on the cable do no work on the system and are not included on the active-force diagram. As the block moves from x_1 = 0.233 m to x_2 = 0.233 + 1.2 = 1.433 m, the work done by the spring force acting on the block is (U_{1-2}= \frac{1}{2} k[x_1^2 - x_2^2] = \frac{80}{0.233^2 - (0.233 + 1.2)^2} = -800 J The work done on the system by the constant 300-N force in the cable is the force times the net horizontal movement of the cable over pulley C, which is \sqrt{(1.2)^2 + (0.9)^2} - 0.9 = 0.6 m. Thus, the work done is 300(0.6) = 180 J. We now apply the work-energy equation to the system and get [T_1 + U_{1-2} = T_2] 0 - 800 + 180 = \frac{1}{2} (50v^2) v = 2.00 m/s Ans. We take special note of the advantage to our choice of system. If the block alone had constituted the system, the horizontal component of the 300-N cable force on the block would have to be integrated over the 1.2-m displacement. This step would require considerably more effort than was needed in the solu- tion as presented. If there had been appreciable friction between the block and its guiding rail, we would have found it necessary to isolate the block alone in order to compute the variable normal force and, hence, the variable friction force. Integration of the friction force over the displacement would then be re- quired to evaluate the negative work which it would do. Helpful Hint Recall that this general formula is valid for any initial and final spring deflections x_1 and x_2, positive (spring in tension) or negative (spring in compression). In deriving the spring- work formula, we assumed the spring to be linear, which is the case here. SAMPLE PROBLEM 3/14 The power winch A hoists the 800-lb log up the 30° incline at a constant speed of 4 ft/sec. If the power output of the winch is 6 hp, compute the coefficient of ki- netic friction μk between the log and the incline. If the power is suddenly increased to 8 hp, what is the corresponding instantaneous acceleration a of the log? Solution. From the free-body diagram of the log, we get N = 800 cos 30° = 693 lb, and the kinetic friction force becomes 693μk. For constant speed, the forces are in equilibrium so that ∑Fx = 0 T = 800 sin 30° − 0 = T = 693μk + 400 The power output of the winch gives the tension in the cable P = T v] T = P/v = 6550/4 = 825 lb Substituting T gives 825 = 693μk + 400 μk = 0.613 When the power is increased, the tension momentarily becomes P = T v] T = P/v = 8(550)/4 = 1100 lb and the corresponding acceleration is given by ∑Fx = ma] 1100 − 693(0.613) − 800 sin 30° = 800/32.2 a a = 11.07 ft/sec^2 Ans. Helpful Hints Note the conversion from horse- power to ft-lb/sec. As the speed increases, the accelera- tion will drop until the speed stabi- lizes at a value higher than 4 ft/sec. 3/101 In the design of a spring bumper for a 3500-lb car, it is desired to bring the car to a stop from a speed of 5 mi/hr in a distance equal to 6 in. of spring de- formation. Specify the required stiffness k for each of the two springs behind the bumper. The springs are undeformed at the start of impact. 3/109 The 2-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of kinetic friction is 0.40. Calculate (a) the velocity v of the collar as it strikes the spring and (b) the maximum deflection x of the spring. Problem 3/109 3/128 Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its respective guide, with b being in the vertical direction. A 20-N horizontal force is applied to the midpoint of the connecting link of negligible mass, and the assem- bly is released from rest with θ = 0. Calculate the velocity vA with which A strikes the horizontal guide when θ = 90°. Problem 3/128 49 Potencial Gravitacional Vg é o potencial gravitacional ou energia potencial gravitacional da partícula de massa m. 50 Variação da energia potencial para elevar uma partícula de massa m de um nível h1 para h2 é definida por, ∆𝑉𝑔 = 𝑚𝑔 ℎ2 − ℎ1 = 𝑚𝑔∆ℎ O trabalho da força pelo exercido sobre a partícula de massa m para eleva la de um nível h1 para h2 é igual a, − mg∆h =− ∆Vg Potencial Elático Deformação de uma corpo elástico, tal como uma mola. Ve é o potencial elático Variação da energia potencial elástica para produzir uma deformação, seja de tração ou compressão, de uma mola de x1 para x2 é igual a, que representa o trabalho realizado de uma força externa sobre a mola. Portanto, o trabalho realizado da mola sobre a partícula é igula a −∆Ve. onde 𝑥1 e 𝑥2 são os comprimentos da mola a partir de sua posição indeformada. 51 Equação do trabalho - Energia 𝑈1−2 ' + ( − ∆𝑉𝑔) + ( − ∆𝑉𝑒) = ∆𝑇 ∆𝑉 = ∆𝑉𝑔 + ∆𝑉𝑒 𝑈1−2 ' + ( − ∆𝑉) = ∆𝑇 𝑈1−2 ' = ∆𝑇 + ∆𝑉 𝑈1−2 ' : trabalho realizado pela forças externas (que não seja peso e nem foeça elática) sobre a partícula me massa m, −∆𝑉𝑔: trabalho realizada pela força peso para elevar a partícula de massa m de uma elevação h, 52 −∆𝑉𝑒 : trabalho realizado pela mola sobre a a partícula de massa m para deformar um corpo elásticamente e ∆𝑇: variação da energia cinética da partícula de massa m. ou 𝑈1−2 ' = (𝑇2 − 𝑇1) + (𝑉𝑔2 − 𝑉𝑔1) + (𝑉𝑒2 − 𝑉𝑒1) Nada de novidade aqui. Apenas se olhou de uma maneira especial de como a força elática e a força peso podem entrar na expressão do trabalho. NADA DE ESPECIAL!!!! Energia Mecânica E = T + Vg + Ve 𝑈1−2 ' = ∆𝐸 Quando E é constante nota se que poder haver transferência de energia cinética para energia potencial. Nesse caso o sistema é conservativo. Lei da conservção de energia mecânica. Lista de Exercícios 8: Problema resolvido 3/16, Problema Resolvido 3/17, Problema Resolvido 3/18, Problema 3/143, Problema 3/155, Problema 3/165, Problema 3/166 e Problema 3/167. SAMPLE PROBLEM 3/16 The 6-lb slider is released from rest at position 1 and slides with negligible friction in a vertical plane along the circular rod. The attached spring has a stiff- ness of 12 lb/ft, and has an unstretched length of 24 in. Determine the velocity of the slider as it passes position 2. Solution. The work done by the weight and the spring force on the slider will be treated using potential-energy methods. The reaction of the rod on the slider is normal to the motion and does no work. Hence, U1→2 = 0. We define the datum to be at the level of position 1, so that the gravitational potential energies are V1 = 0 V2 = −mgh = −6(2) = −12 ft-lb The initial and final elastic (spring) potential energies are V1 = 1/2 k(x2)2 = 1/2(12)(24 12 )2 = 48 ft-lb V2 = 1/2 k(x2)2 = 1/2(12)(34 12 )2 = 8.24 ft-lb Substitution into the alternative work-energy equation yields [T1 + V1 + U1→2 = T2 + V2] 0 + 48 + 0 = 1/2(6/32 2 )v2^2 − 12 + 8.24 v2 = 23.6 ft/sec Ans. Helpful Hint Note that if we evaluated the work done by the spring force acting on the slider by means of the integral ∫F•dr, it would necessitate a lengthy computation to account for the change in the magnitude of the force, along with the change in the angle between the force and the tangent to the path. Note further that dr de- pends only on the direction of the motion and does not require knowledge of the shape of the path. SAMPLE PROBLEM 3/17 The 10-kg slider moves with negligible friction up the inclined guide. The attached spring has a stiffness of 60 N/m and is stretched 0.6 m in position A, where the slider is released from rest. The 250-N force is a constant and the pulley offers negligible resistance to the motion of the cord. Calculate the velocity vC of the slider as it passes point C. Solution. The slider and inextensible cord together with the attached spring will be analyzed as a system, which permits the use of Eq. 3/12a. The only non- potential force doing work on this system is the 250-N tension applied to the cord. While the slider moves from A to C, the point of application of the 250-N force moves a distance of AB = 0.9 m − 0.6 m. UAC = 250(0.6) = 150 J We define a datum at position A so that the initial and final gravitational poten- tial energies are VA = 0 VC = mgh = 10(9.81)(1.2 sin 30°) = 58.9 J The initial and final elastic potential energies are VA = 1/2 k(xA)2 = 1/2(60)(0.6)2 = 10.8 J VC = 1/2 k(xC)2 = 1/2(60)(0.6 − 1.2)2 = 9.72 J Substitution into the alternative work-energy equation 3/12a gives [T1 + VA + UAC = TC + VC] 0 + 0 + 10.8 + 150 = 1/2(10)vC^2 + 58.9 + 9.72 vC = 0.974 m/s Ans. Helpful Hints Do not hesitate to use subscripts tai- lored to the problem at hand. Here we use A and C rather than 1 and 2 The reactions of the guides on the slider are normal to the direction of motion and do no work. SAMPLE PROBLEM 3/18 The system shown is released from rest with the lightweight slender bar OA in the vertical position shown. The unstressed spring at B is undeflected in the initial position and exerts a restoring moment of magnitude k0 = 9 N·m on the bar, where ø is the counterclockwise angular deflection of the bar. The string S is attached to point C of the bar and slips without friction through a vertical hole in the support surface. For the values mA = 2 kg, mB = 4 kg, L = 0.5 m, and k0 = 13 N·m/rad: (a) Determine the speed øA of particle A when ø reaches 90°. (b) Plot øA as a function of ø over the range 0 ≤ ø ≤ 90°. Identify the maximum value of øA and the value of ø at which this maximum occurs. Solution (a). We begin by establishing a general relationship for the potential energy associated with the deflection of the torsional spring. Recalling that the change in potential energy is the work done on the spring to deform it, we write We also need to establish the relationship between øA and øB when ø = 90°. Not ing that the speed of point C is always øA/2, and further noting that the speed of cylinder B is one-half the speed of point C at ø = 90°, we conclude that at ø = 90°, Establishing datums at the initial altitudes of bodies A and B, and with state 1 at ø = 0 and state 2 at ø = 90°, we write With numbers: Solving, (b). We leave our definition of the initial state 1 as is, but now redefine state 2 to be associated with an arbitrary value of ø. From the accompanying diagram constructed for an arbitrary value of ø, we see that the speed of cylinder B can be written as Finally, because Upon substitution of the given quantities, we vary ø to produce the plot of øA versus ø. The maximum value of øA is seen to be Helpful Hints Note that mass B will move down ward by one-half of the length of string initially above the supporting surface. This downward distance is The absolute-value signs reflect the fact that øA is known to be positive. 3/143 Point P on the 2-kg cylinder has an initial velocity v0 = 0.8 m/s as it passes position A. Neglect the mass of the pulleys and cable and determine the distance y of point P below A when the 3-kg cylinder has acquired an upward velocity of 0.6 m/s. Problem 3/143 3/155 The spring has an unstretched length of 25 in. If the system is released from rest in the position shown, determine the speed u of the ball (a) when it has dropped a vertical distance of 10 in. and (b) when the rod has rotated 35°. Problem 3/155 3/166 Calculate the maximum velocity of slider B if the system is released from rest with x = y. Motion is in the vertical plane. Assume that friction is negligible. The sliders have equal masses, and the motion is restricted to y >= 0. Problem 3/166 3/167 The mechanism is released from rest with ø = 180°, where the uncompressed spring of stiffness k = 900 N/m is just touching the underside of the 4-kg collar. Determine the angle ø corresponding to the maximum compression of the spring. Motion is in the vertical plane, and the mass of the links may be neglected. Problem 3/167 57 3.6 Impulso e Quantidade de Movimento Conceito Na seção anterior tratou se das equações de trabalho e energia mecânica integrando a equação do movimento 𝐅 = 𝑚 𝐚 com relação ao deslocamento. Assim, encontrou se que as variações de velocidade podiam ser expressas em termos do trabalho realizado ou em termos da variação total da energia. Nesta seção, a equação do movimento será integrada com relação ao tempo. Essa abordagem permite levar às equações de impulso e quantidade de movimento. Dessa forma, será permitido resolver problemas nos quais as forças aplicadas atuam durante períodos de tempo especificados. 58 Impulso Linear e Quantidade de Movimento 𝐆 = 𝑚 𝐯 quantidade de movimento linear O Princípio do Impulso - Quantidade de Movimento Linear O produto da força pelo tempo é definido como impulso total sobre a partícula de massa m e corresponde à variação da quantidade de movimento linear de m. Figure 3/13 Impulso: área sobre a curva m(v1)x + ∫ΣFx dt = m(v2)x m(v1)y + ∫ΣFy dt = m(v2)y m(v1)z + ∫ΣFz dt = m(v2)z Figure 3/12 G1 = m v1 + ∫ΣF dt = m v2 60 Conservação da Quantidade de Movimento Linear Se a força resultante sobre a partícula é nula durante um intervalo de tempo, nota se que a quantidade de movimento G permaneça constante. Lista de Exercícios 9: Problema resolvido 3/19, Problema Resolvido 3/20, Problema Resolvido 3/21, Problema resolvido 3/22, Problema Resolvido 3/23, Problema 3/181, Problema 3/187, Problema 3/205. SAMPLE PROBLEM 3/20 A 2-lb particle moves in the vertical x-y plane (x up, y horizontal) under the action of its weight and a force F which varies with time. The linear momentum of the particle in pound-seconds is given by the expression G = 2/3(t2 + 3)t – 1/2(t3 – 4)k, where t is the time in seconds. Determine F and its magnitude for the instant when t = 2 sec. Solution. The weight expressed as a vector is –2k lb. Thus, the force-momentum equation becomes ∫F = G F = 2k – d/dt[ 2/3(t2 + 3)t – 1/2(t3 – 4)k ] = 3j – 2ft/ For t = 2 sec, F = (2k + 3(2))i = 2(-2)k = 6j – 6k lb Thus, F = √(32 + 62)1b Helpful Hint 1. Don’t forget that ∫F includes all external forces acting on the particle, including the weight. Ana. Ana. SAMPLE PROBLEM 3/21 A particle with a mass of 0.5 kg has a velocity of 10 m/s in the x-direction at time t = 0. Forces F1 and F2 act on the particle, and their magnitudes change with time according to the graphical schedule shown. Determine the velocity v2 of the particle at the end of the 3-s interval. The motion occurs in the horizontal x-y plane. Solution. First, we construct the impulse-momentum diagram as shown: m(v1)x + ∫ΣFx dt = m(v2)x 0.5(10) – [4(1) + 2(3 – 1)] = 0.5(6)(v2)x (v2)x = 6 m/s [m(v1)y + ∫ΣFy dt = m(v2)y ] 0.50 + [1(2) + 2(3 – 2)] = 0.5(6)(v2)y (v2)y = 8 m/s Thus, v2 = √(–62 + 82) m/s and v2 = 10 m/s θ2 = tan-1(8/–6) = 126.9° Ana. Although not called for, the path of the particle for the first 3 seconds is plotted in the figure. The velocity at t = 3 s is shown together with its components. Helpful Hint 1. The impulse in each direction is the corresponding area under the force-time graph. Note that F1 is in the negative x-direction, so its impulse is negative. SAMPLE PROBLEM 3/19 A tennis player strikes the tennis ball with her racket when the ball is at the uppermost point of its trajectory as shown. The horizontal velocity of the ball just before impact with the racket is v1 = 50 ft/sec, and just after impact its velocity is v2 = 70 ft/sec directed at the 15° angle as shown. If the 2-oz ball is in contact with the racket for 0.02 sec, determine the magnitude of the average force R exerted by the racket on the ball. Also determine the angle β made by R with the horizontal. Solution. We construct the impulse-momentum diagrams for the ball as follows: Forces diagram [ m(v1)x + ∫ΣFx dt = m(v2)x ] [ m(v1)y + ∫ΣFy dt = m(v2)y ] Component equations We can now solve for the impact forces as Rx = 22.8 lb Ry = 3.64 lb We note that the impact force Ry = 3.64 lb is considerably larger than the 0.125-lb weight of the ball. Thus, the weight mg, a nonimpulse force, could have been neglected as small in comparison with Ry. Had we neglected the weight, the computed value of Ry would have been 3.52 lb. We now determine the magnitude and direction of R as R = √(Rx^2 + Ry^2) = √(22.8^2 + 3.64^2) = 23.1 lb β = tan^(-1)(Ry/Rx) = tan^(-1)(3.64/22.8) = 9.06° Ans. Helpful Hints 1. Recall that for the impulse-momentum diagrams, initial linear momentum goes in the first diagram, all external linear impulses go in the second diagram, and final linear momentum goes in the third diagram. 2. For the linear impulse ∫Rx dt, the average impact force Rx is constant, so that it can be brought outside the integral sign, resulting in ∫Rx dt = Rx Δt. The linear impulse in the y-direction has been similarly treated. SAMPLE PROBLEM 3/22 The loaded 150-kg skip is rolling down the incline at 4 m/s when a force P is applied to the cable as shown at time t = 0. The force P is increased uniformly with time until it reaches 600 N at t = 4 s, after which time it remains constant at this value. Calculate (a) the time t' at which the skip reverses its direction and (b) the velocity v of the skip at t = 8 s. Treat the skip as a particle. Solution. The stated variation of P with time is plotted, and the impulse-momentum diagrams of the skip are drawn. Part (a). The skip reverses direction when its velocity becomes zero. We will assume that this condition occurs at t' = 4 + Δt'. The impulse-momentum equation applied consistently in the positive-x-direction gives m1v1x1 + ∫ΣF dt = m2v2x2 150(-4) + 1/2(4)(2)(600) + 2600Δt' - 150(9.81) sin 30°(4 + Δt') = 150(0) Δt' = 2.46 t' = 4 + 2.46 = 6.46 s Ans. Part (b). Applying the momentum equation to the entire 8-s interval gives m1v1x1 + ∫ΣF dt = m2v2x2 150(-4) + 1/2(4)(2)(600) + 42(600) - 150(9.81) sin 30°(8) = 150(v2)x2 (v2)x2 = 4.76 m/s Ans. The same result is obtained by analyzing the interval from t' to 8 s. SAMPLE PROBLEM 3/23 The 50-g bullet traveling at 600 m/s strikes the 4-kg block centrally and is embedded within it. If the block slides on a smooth horizontal plane with a velocity of 12 m/s in the direction shown prior to impact, determine the velocity v2 of the block and embedded bullet immediately after impact. Solution. Since the force of impact is internal to the system composed of the block and bullet and since there are no other external forces acting on the system in the plane of motion, it follows that the linear momentum of the system is conserved. Thus, | G1 = G2 | 0.050(600) + 412(cos 30°i + sin 30°j) = (4 + 0.050)v2 v2 = 10.261 + 13.839 m/s Ans. The final velocity and its direction are given by (v = √(v2x^2 + v2y^2)) vg = √(10.2632 + 13.8392) = 16.83 m/s Ans. [tan θ = (v2y/v2x)] tan θ = 13.83 / 10.263 = 1.299 θ = 52.4° Ans. Helpful Hint ➤ The impulse-momentum diagram keeps us from making the error of using the impulse of P rather than 2P or of forgetting the impulse of the component of the weight. The first term in the linear impulse is the triangular area of the P-time relation for the first 4 s, doubled for the force of 2P. Helpful Hint ➤ Working with the vector form of the principle of conservation of linear momentum is clearly equivalent to working with the component form. 3/181 Freight car A with a gross weight of 150,000 lb is moving along the horizontal track in a switching yard at 2 mi/hr. Freight car B with a gross weight of 120,000 lb and moving at 3 mi/hr overtakes car A and is coupled to it. Determine (a) the common velocity v of the two cars as they move together after being coupled and (b) the loss of energy |ΔE| due to the impact. Problem 3/181 3/187 The 20-lb block is moving to the right with a velocity of 2 ft/sec on a horizontal surface when a force P is applied to it at time t = 0. Calculate the velocity v of the block when t = 0.4 sec. The coefficient of kinetic friction is μk = 0.30. Problem 3/187 3/205 The force P, which is applied to the 10-kg block initially at rest, varies linearly with time as indicated. If the coefficients of static and kinetic friction between the block and the horizontal surface are 0.60 and 0.40, respectively, determine the velocity of the block when t = 4 s. Problem 3/205 66 Taxa de Variação Temporal da Quqntidade de Movimento Angular 67 Princípio do Impulso Angular - Quantidade de Movimento Angular Conservação da Quantidade de Movimento Angular Se o momento resultante em torno de um ponto fixo O de todas as forças atuando sobre uma partícula é nulo durante um intervalo de tempo, a quantidade de movimento angular HO em torno desse ponto permanece constante. Lista de Exercícios 10: Problema resolvido 3/24, Problema Resolvido 3/25, Problema Resolvido 3/26, Problema resolvido 3/27, Problema 3/226, Problema 3/235, Problema 3/236, Problema 3/238. SAMPLE PROBLEM 3/24 A small sphere has the position and velocity indicated in the figure and is acted upon by the force F. Determine the angular momentum HO about point O and the time derivative ḢO. Solution. We begin with the definition of angular momentum and write HO = r × mv = (3i + 6j + 4k) × 2(5j) = -40i + 30k N·m/s Ans. From Eq. 3/31, Ḣ0 = MO = r × F = (3i + 6j + 4k) × 10k = 66i - 30j N·m Ans. As with moments of forces, the position vector must run from the reference point (O in this case) to the line of action of the linear momentum mv. Here r runs directly to the particle. SAMPLE PROBLEM 3/25 A comet is in the highly eccentric orbit shown in the figure. Its speed at the most distant point A, which is at the outer edge of the solar system, is vA = 740 m/s. Determine its speed at the point B of closest approach to the sun. Solution. Because the only significant force acting on the comet, the gravitational force exerted on it by the sun, is central (points to the sun center O), angular momentum about O is conserved. (HO)A = (HO)B mAvArA = mBvBrB vB = (mA vA rA) / (mB rB) = 6000(10^9)(740) / 75(10^9) vB = 59 200 m/s Ans. SAMPLE PROBLEM 3/26 The assembly of the light rod and two end masses is at rest when it is struck by the falling wad of putty traveling with speed v1 as shown. The putty adheres to and travels with the right-hand end mass. Determine the angular velocity θż of the assembly just after impact. The pivot O is frictionless, and all three masses may be assumed to be particles. Solution. If we ignore the angular impulses associated with the weights during the collision process, then system angular momentum about O is conserved during the impact. (HO)1 = (HO)2 mvl₁ = (m + 2m)(vż₁)l + 4m(vż₂)y2l vż = \(\frac{v}{12}\) CW Ans. Note that each angular-momentum term is written in the form mvd, and the final transverse velocities are expressed as radial distances times the common final angular velocity θż. SAMPLE PROBLEM 3/27 A small mass particle is given an initial velocity vo tangent to the horizontal rim of a smooth hemispherical bowl at a radius ro from the vertical centerline, as shown at point A. As the particle slides past point B, a distance h below A and a distance r from the vertical centerline, its velocity v makes an angle θ with the horizontal tangent to the bowl through B. Determine θ. Solution. The forces on the particle are its weight and the normal reaction exerted by the smooth surface of the bowl. Neither force exerts a moment about the axis OO, so that angular momentum is conserved about that axis. Thus, (Ioω)1 = (Ioω)2 mυro = mυr cos θ Also, energy is conserved so that E1 = E2. Thus T1 + V1 = T2 + V2 12mυo² + mgh = 12mυ² + 0 υ² = υo² + 2gh Eliminating υ and substituting r² = ro² − h² give υoro = \(\sqrt{υo² + 2gh} ⋅ \frac{r}{ro} = \frac{υ^2}{ro²} ⋅ cos θ\) θ = cos−1 \(\frac{1}{\sqrt{1 + \frac{2gh}{υo²}\bigg)^2}}{\sqrt{1 − \frac{h²}{ro²}}\bigg)}}\) Ans. Helpful Hint * The angle θ is measured in the plane tangent to the hemispherical surface at B. 3/226 The small particle of mass m and its restraining cord are spinning with an angular velocity ω on the horizontal surface of a smooth disk, shown in section. As the force F is slightly relaxed, r increases and ω changes. Determine the rate of change of ω with respect to r and show that the work done by F during a movement dr equals the change in kinetic energy of the particle. 3/235 A pendulum consists of two 3.2-kg concentrated masses positioned as shown on a light but rigid bar. The pendulum is swinging through the vertical position with a clockwise angular velocity ω = 6 rad/s when a 50-g bullet traveling with velocity υ = 300 m/s in the direction shown strikes the lower mass and becomes embedded in it. Calculate the angular velocity ω' which the pendulum has immediately after impact and find the maximum angular deflection θ of the pendulum.