Dinâmica TMEC 019 - Resumo Cinemática da Partícula e Coordenadas

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 JL e Kraige LG Mecânica Dinâmica LTC 2 Hibbeler R C Dinâmica Pearson Education 3 Beer Johnston Cornwell Self e Sanshi Mecânica Vetorial para Engenheiros Dinâmica 11a edição 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 Notação em negrito F r𝛚 etc são vetores sem negrito s F r t etc são escalares 1 Movimento Curvilíneo Plano Posição r posição da partícula no ponto A um vetor 𝐫 𝐫 posição da partícula no ponto A um vetor 𝐫 deslocamento da partícula um vetor e 𝑠 distância percorrida da partícula um escalar 3 Velocidade Média 𝐯𝐦𝐞𝐝 𝐫 𝑡 Velocidade Instantânea 𝐯 lim 𝑡0 𝐫 𝑡 d𝐫 dt 𝐫 v 𝐯 lim t0 s t ds dt s Conceito A velocidade é tangente à trajetória Vide Figura 25 Aceleração 𝐚 lim 𝑡0 𝐯 𝑡 𝐯 Conceito A aceleração inclui os efeitos das variações do módulo e da direção de v Vide Figura 25 e 26 Isso ficará mais claro nas próximas seções dessas notas 4 Conceito Agora será visto a REPRESENTAÇÃO do movimento plano posição velocidade e aceleração da partícula em coordenadas retangulares normaltangencial e polares Coordenadas Retangulares xy 𝐫 x𝐢 y𝐣 onde i e j são vetores unitários 𝐯 d𝐫 dt x𝐢 x𝐢 y𝐣 y𝐣 onde 𝐢 𝐣 0 pois os vetores unitários i e j não variam nem em módulo nem em direção Portanto 𝐯 d𝐫 dt x𝐢 y𝐣 𝐚 d𝐯 dt x𝐢 y𝐣 5 Exercício Movimento de projéteis resolvido no livro do Merian Exercício Uma partícula possui movimento circular de raio R para uma posição angular genérica 𝜃 com velocidade angular 𝜃 𝜔 e aceleração angular 𝜃 𝛼 todas no sentido antihorário Determine os vetores de posição velocidade e aceleração da partícula representados em coordenadas retangulares Uma aplicação muito importante da representação do movimento plano da partícula em coordenadas retangulares consiste no movimento de lançamento de partículas Lista de Exercíos 1 Problema Resolvido 25 resolvido no livro Problema 270 e Problema 285 270 The center of mass G of a high jumper follows the trajectory shown Determine the component v0 measured in the vertical plane of the figure of his takeoff velocity and angle θ 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 successful jump Problem 270 285 Ball bearings leave the horizontal trough with a velocity of magnitude u and fall through the 70mmdiameter 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 limiting conditions Problem 285 8 Coordenadas Normal e Tangencial nt Conceito As coordenadas normaltangencial são tomadas como se estivessem MOVENDO sobre a trajetória junto com a partícula são não estacionárias Conceiro O sentido de n é sempre tomado para o centro de curvatura da trajetória e a componente tangencial t é perpendicular a componente normal como mostra a Figura 29 Velocidade 𝐯 v 𝐞t v ds dt ρ dβ dt ρβ 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 9 Conceito A velocidade é sempre tangente à trajetórtia Vide Figura abaixo Aceleração 𝐚 d𝐯 dt d v 𝐞𝐭 dt v𝐞 t v𝐞t onde 𝐞 t d𝐞t dt 𝐞𝐭 dβ dt 𝐞𝐧 1 dβ dt 𝐞𝐧 β𝐞𝐧 v ρ 𝐞𝐧 portanto 𝐚 v2 ρ 𝐞n v𝐞t an v2 ρ e at v 10 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 Execício Uma partícula possui movimento circular de raio R para uma posição angular genérica 𝜃 com velocidade angular 𝜃 𝜔 e aceleração angular 𝜃 𝛼 todas no sentido antihorário Determine os vetores de posição velocidade e aceleração da partícula representados em coordenadas normaltangencial 11 Aplicações importantes da representação do movimento plano da partícula em coordenadas normaltangencial consistem no movimento de pêndulos e movimentos envolvendo o raio de curvatura de trajetórias curvilíneo de partículas 12 Lista de Exercícios 2 Problema resolvido 28 Problema 2109 Problema 2112 Problema 2114 e Problema 2129 2112 Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both nt and xy coordinates for the instant when θ 60 if θ 2 radsec and θ 4025 radsec² Problem 2112 2114 Magnetic tape is being transferred from reel A to reel B and passes around idler pulleys C and D At a certain instant point P1 on the tape is in contact with pulley C and point P2 is in contact with pulley D If the normal component of acceleration of P1 is 40 ms² and the tangential component of acceleration of P2 is 30 ms² at this instant compute the corresponding speed v of the tape the magnitude of the total acceleration of P1 and the magnitude of the total acceleration of P2 Problem 2114 2129 The pin P is constrained to move in the slotted guides which move at right angles to one another At the instant represented A has a velocity to the right of 02 ms which is decreasing at the rate of 075 ms each second At the same time B is moving down with a velocity of 015 ms which is decreasing at the rate of 05 ms each second For this instant determine the radius of curvature p of the path followed by P Is it possible to also determine the time rate of change of p Problem 2129 15 Coordenadas Polares rθ r medida radial θ medida angular Conceito As coordenadas polares são tomadas como se estivessem MOVENDO sobre a tajetória junto com a partícula não estacionárias Conceiro O sentido de r e θ são mostrados na Figura 213 a Figura 213 Direções radial e angular 𝐞𝑟 e 𝐞𝜃 são vetores unitários Posição 𝐫 r 𝐞r 16 Velocidade 𝐯 d𝐫 dt dr dt 𝐞r r d𝐞r dt onde dr dt r escalar d𝐞r dt d1 dθ dt 𝐞𝛉 θ𝐞𝛉 vetor Vide Figura 213 Portanto a equação de velocidade acima pode ser reescrita como 𝐯 r𝐞𝐫 rθ𝐞𝛉 vr r e vθ rθ Vide Figura abaixo 17 Aceleração 𝐚 d𝐯 dt dr dt 𝐞r r d𝐞r dt dr dt θ 𝐞θ r dθ dt 𝐞θ rθ d𝐞θ dt onde d𝐞θ dt d1 dθ dt 𝐞𝐫 θ 𝐞𝛉 Vide Figura 213 b 𝐚 r rθ 2 𝐞r rθ 2θr 𝐞θ 𝑎𝑟 r rθ 2 e 𝑎𝜃 rθ 2θr Vide Figura 215 18 Execício Uma partícula possui movimento circular de raio R para uma posição angular genérica 𝜃 com velocidade angular 𝜃 𝜔 e aceleração angular 𝜃 𝛼 todas no sentido anti horário Determine os vetores de posição velocidade e aceleração da partícula representados em coordenadas polares Aplicações importantes da representação do movimento plano da partícula em coordenadas polares consistem no movimento de lançamento de projeteis e principalmente na análise cinemática de mecanismos de cadeias aberta e fechada 19 Lista de Exercícios 3 Problema resolvido 29 Problema Resolvido 210 Problema 2142 Problema 2150 Problema 2156 e Problema 2157 2142 The piston of the hydraulic cylinder gives pin A a constant velocity v 3 ftsec in the direction shown for an interval of its motion For the instant when θ 60 determine ṙ r θ and θ where rOA Problem 2142 2150 The slotted arm OA forces the small pin to move in the fixed spiral guide defined by r Kθ Arm OA starts from rest at θ π4 and has a constant counterclockwise angular acceleration θ α Determine the magnitude of the acceleration of the pin P when θ 3π4 Problem 2150 2156 The member OA of the industrial robot telescopes and pivots about the fixed axis at point O At the instant shown θ 60 θ 12 rads θ 08 rads² OA 09 m OA 05 ms and OA 6 ms² 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 xaxis The base of the robot does not revolve about a vertical axis Problem 2156 2157 The robot arm is elevating and extending simultaneously At a given instant θ 30 θ 10 degs constant l 05 m l 02 ms and l 03 ms² 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 2157 22 2 Movimento Curvilíneo Espacial Coordenadas Retangulares xyz Figura 216 vetores unitários i j e k Posição da partícula 𝐑 𝑥 𝐢 𝑦 𝐣 𝑧 𝐤 Velocidade da Partícula 𝐯 𝑑𝐑 𝑑𝑡 23 Como o vetores unitários i j e k não variam nem em módulo nem em direção com o tempo implica que 𝑑𝐢 𝑑𝑡 𝑑𝐣 𝑑𝑡 𝑑𝐤 𝑑𝑡 𝟎 Portanto 𝐯 x 𝐢 y 𝐣 z 𝐤 Aceleração da partícula 𝐚 d𝐯 dt 𝐚 x 𝐢 y 𝐣 z 𝐤 24 Coordenadas Cilíndricas rθz Figura 216 vetores unitários 𝐞𝑟 𝐞𝜃 e k Nesse tipo de representação incluise uma coordenada z sobre as coordenadas polares Posição da partícula 𝐑 r 𝐞𝐫 z 𝐤 Velocidade da partícula 𝐯 d𝐑 dt r 𝐞𝐫 r d𝐞𝐫 dt z 𝐤 z d𝐤 dt d𝐞𝐫 dt d1 dθ dt 𝐞θ θ 𝐞θ d𝐳 dt 0 𝐯 r 𝐞𝐫 rθ 𝐞𝛉 z 𝐤 Aceleração da partícula 𝐚 d𝐯 dt r rθ 2 𝐞𝐫 rθ 2rθ 𝐞𝛉 z𝐤 Exercício Demosntrar a equação acima 25 Coordenadas Esféricas RθΦ Figura 216 vetores unitários 𝐞R 𝐞θ e 𝐞ϕ Velocidade 𝐯 vR𝐞R vθ𝐞θ vϕ𝐞ϕ vR R vθ Rθcosϕ vϕ Rϕ Aceleração 𝐚 aR𝐞R aθ𝐞θ aϕ𝐞ϕ aR R Rϕ 2 Rθ 2cos2ϕ aθ cosϕ R d dt R2θ 2Rθϕ senϕ aϕ 1 R d dt R2ϕ Rθ 2senϕcosϕ 26 Lista de Exercícios 4 Problema resolvido 211 Problema Resolvido 212 Problema 2167 Problema 2182 e Problema 2177 SAMPLE PROBLEM 212 An aircraft P takes off at A with a velocity v0 of 250 kmh and climbs in the vertical yz plane at the constant 15 angle with an acceleration along its flight path of 08 ms2 Flight progress is monitored by radar at point O a Resolve the velocity of P into cylindricalcoordinate components 60 seconds after takeoff and find ṙ θ and ż for that instant b Resolve the velocity of the aircraft P into sphericalcoordinate components 60 seconds after takeoff and find Ṙ θ and ϕ for that instant Solution a The accompanying figure shows the velocity and acceleration vectors in the yz plane The takeoff speed is v0 25036 694 ms and the speed after 60 seconds is υ υ0 at 694 0860 1174 ms The distance s traveled after takeoff is s s0 υ0t 12 at2 0 69460 12 08602 5610 m The ycoordinate and associated angle θ are y 5610 cos 15 5420 m θ tan¹54203000 610 From the figure b of xy projections we have r 30002 54202 6190 m uxy υ cos 15 1174 cos 15 1134 ms uθ rθ uxy sin θ 1134 sin 610 992 ms uθ ṙθ uxy cos θ 1134 cos 610 550 ms So θ 5506190 888103 rads Finally ż uż υ sin 15 1174 sin 15 304 ms 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 tan¹zr tan¹14516190 1319 R r² z² 6190² 1451² 6360 m From the figure uR Ṙ 992 cos 1319 304 sin 1319 1036 ms θ 88810³ rads as in part a uϕ Rϕ 304 cos 1319 992 sin 1319 696 ms ϕ 6956360 109310³ rads 2167 An amusement ride called the corkscrew takes the passengers through the upsidedown curve of a horizontal cylindrical helix The velocity of the cars as they pass position A is 15 ms 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 2167 2182 The particle P moves down the spiral path which is wrapped around the surface of a right circular cone of base radius b and altitude h The angle γ between the tangent to the curve at any point and a horizontal tangent to the cone at this point is constant Also the motion of the particle is controlled so that θ is constant Determine the expression for the radial acceleration αr of the particle for any value of θ Problem 2182 2177 The base structure of the fire truck ladder rotates about a vertical axis through O with a constant angular velocity Ω 10 degs At the same time the ladder unit OB elevates at a constant rate ϕ 7 degs and section AB of the ladder extends from within section OA at the constant rate of 05 ms 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 2177 30 Parte 3 Cinemática Plana de Corpos Rígidos Conceito Geometria apenas sem considerar a causa do movimento Corpo Rígido é um sistema de partículas para o qual as distâncias entre elas permanecem inalteradas Tipos de Movimento Plano Translação rotação em torno de um eixo fixo e movimento plano geral Vide Figura 51 abaixo 31 31 Rotação em Torno de Um Eixo Fixo e Método do Movimento Absoluto Relações algébricas não vetoriais para o movimento circular de um ponto A de um corpo rígido em torno de um eixo fixo Figura 53 v ωr ω rotação do corpo rígido ou velocidade angular do corpo rígido v velocidade do ponto A do corpo rígido an ω2r v2 r vω at αr α aceleração angular do corpo rígido an e at são as componentes normal e tangencial da aceleração do ponto A do corpo rígido 32 Relações vetoriais para o movimento circular de um ponto A de um corpo rígido em torno de um eixo fixo Figura 54 𝐯 d𝐫 dt 𝐫 𝛚 𝐫 𝐫 posição do ponto A do corpo rígido 𝛚 rotação ou velocidade angular do corpo rígido produto vetorial 𝐯 velocidade do ponto A do corpo rígido 𝐚 d𝐯 dt 𝐯 𝛚 𝐫 𝛚 𝐫 𝐚 𝛚 𝛚 𝐫 𝛚 𝐫 𝐚 𝛚 𝛚 𝐫 𝛂 𝐫 𝛂 aceleração angular do corpo rígido 𝐚 aceleração do ponto A do corpo rígido 𝐚𝐧 𝛚 𝛚 𝐫 𝐚𝐭 𝛂 𝐫 33 Lista de Exercícios 11 Problema resolvido 52 Problema Resolvido 53 Problema 517 Problema 528 e Problema 57 517 The beltdriven pulley and attached disk are rotating with increasing angular velocity At a certain instant the speed v of the belt is 15 ms and the total acceleration of point A is 75 ms2 For this instant determine a the angular acceleration α of the pulley and disk b the total acceleration of point B and c the acceleration of point C on the belt Problem 517 528 The design characteristics of a gearreduction unit are under review Gear B is rotating clockwise with a speed of 300 revmin when a torque is applied to gear A at time t 2 s to give gear A a counterclockwise acceleration α which varies with time for a duration of 4 seconds as shown Determine the speed NB of gear B when t 6 s Problem 528 57 The rectangular plate is rotating about its corner axis through O with a constant angular velocity ω 10 rads Determine the magnitudes of the velocity v and acceleration a of the corner A by a using the scalar relations and b using the vector relations Problem 57 36 32 Método do Movimento Relativo para Eixos Transladados Análise de Velocidades Conceito Movimento Plano Geral Translação Pura Rotação em Torno de Um Eixo Fixo Figura 55 a XY sistema inercial xy sistema movél que translada não gira translada apenas com origem no ponto B A Figura 55 a mostra que tomando dois pontos A e B de um corpo rígido pode se escrever a seguinte equação 𝐫A 𝐫B 𝐫A B 𝐫A deslocamento absoluto do ponto A 𝐫B deslocamento absoluto do ponto B e 𝐫A B deslocamento relativo do ponto A com relação ao ponto B 37 Dividindo a equação acima por t e fazendo o limite quando t tende a zero resulta lim 𝑡0 𝐫𝐴 𝑡 lim 𝑡0 𝐫𝐵 𝑡 lim 𝑡0 𝐫𝐴 𝐵 𝑡 ou 𝐯A 𝐯B 𝐯A B 𝐯A velocidade absoluta do ponto A 𝐯B velocidade absoluta do ponto B e 𝐯A B velocidade relativa do ponto A com relação ao ponto B Conceito O movimento relativo do ponto A com relação ao ponto B é a rotação do ponto A com relação a um eixo fixo em B Vide Figura 55 b Portanto 𝐯A B 𝛚 𝐫 𝛚 vetor rotação absoluta do corpo rígido que contém os dois pontos A e B e 𝐫 vetor posição relativa do ponto A com relação ao ponto B Portanto a equação de velocidades fica 38 𝐯A 𝐯B 𝐯A B 𝐯B 𝛚 𝐫 𝐯A velocidade absoluta do ponto A Tangente à trajetória absoluta do ponto A 𝐯B velocidade absoluta do ponto B Tangente à trajetória absoluta do ponto B 𝐯A B velocidade realativa do ponto A com relação ao ponto B Trata se da rotação do ponto A em torno de B Tangente à trajetória relativa do ponto A em relação a B 𝛚 rotação ou velocidade angular absoluta da peça que contém os ponto A e B Vide Figura 56 abaixo 39 Lista de Exercícios 12 Problema resolvido 57 Problema Resolvido 58 Problema Resolvido 59 Problema 564 Problema 5119 e Problema 5120 SAMPLE PROBLEM 58 Crank CB oscillates about C through a limited arc causing crank OA to oscillate about O When the linkage passes the position shown with CB horizontal and OA vertical the angular velocity of CB is 2 rads counterclockwise For this instant determine the angular velocities of OA and AB Solution I Vector The relativevelocity equation vA vB vAB is rewritten as ωOA rA ωCB rB ωAB rAB where ωOA ωOA k ωCB 2k rads ωAB ωAB k rA 100j mm rB 75i mm rAB 175i 50j mm Substitution gives ωOAk 100j 2k 75i ωABk 175i 50j 100ωOAj 150j 175ωABi 50ωABj Matching coefficients of the respective i and jterms gives 100ωOA 50ωAB 0 256 7ωAB 0 the solutions of which are ωAB 67 rads and ωOA 37 rads Ans Solution II ScalarGeometric Solution by the scalar geometry of the vector triangle is particularly simple here since vA and vB are at right angles for this special position of the linkages First we compute vB which is v rω vB 00752 0150 ms and represent it in its correct direction as shown The vector vAB must be perpendicular to AB and the angle θ between vAB and vB is also the angle made by AB with the horizontal direction This angle is given by tan θ 100 50250 75 27 The horizontal vector vA completes the triangle for which we have vAB vBcos θ 0150cos θ vA vB tan θ 015027 0307 ms The angular velocities become ω vr ωAB vABAB 0150cos θ cos θ 0250 0075 67 rads CW Ans ω vr ωOA vAOA 0307 10100 37 rads CW Ans Helpful Hints 1 We are using here the first of Eqs 53 and Eq 56 2 The minus signs in the answers indicate that the vectors ωAB and ωOA are in the negative kdirection Hence the angular velocities are clockwise 3 Always make certain that the sequence of vectors in the vector polygon agrees with the equality of vectors specified by the vector equation SAMPLE PROBLEM 59 The common configuration of a reciprocating engine is that of the slidercrank mechanism shown If the crank OB has a clockwise rotational speed of 1500 revmin determine for the position where θ 60 the velocity of the piston A the velocity of point G on the connecting rod and the angular velocity of the connecting rod Solution The velocity of the crank pin B as a point on AB is easily found so that B will be used as the reference point for determining the velocity of A The relativevelocity equation may now be written vA vB vAB The crankpin velocity is 1 v rω vB 512 1500 2π60 654 ftsec and is normal to OB The direction of vA is of course along the horizontal cylinder axis The direction of vAB must be perpendicular to the line AB as explained in the present article and as indicated in the lower diagram where the reference point B is shown as fixed We obtain this direction by computing angle β from the law of sines which gives 5sin β 14sin 60º β sin¹ 0309 1802º We now complete the sketch of the velocity triangle where the angle between vAB and vA is 90 1802 720 and the third angle is 180 30 720 780 Vectors vA and vAB are shown with their proper sense such that the headtotail sum of vB and vAB equals vA The magnitudes of the unknowns are now calculated from the trigonometry of the vector triangle or are scaled from the diagram if a graphical solution is used Solving for vA and vAB by the law of sines gives 2 vAsin 780 654sin 720 vA 673 ftsec Ans vABsin 30 654sin 720 vAB 344 ftsec The angular velocity of AB is counterclockwise as revealed by the sense of vAB and is ω ur ωAB vABAB 3441412 295 radsec Ans We now determine the velocity of G by writing vG vB vGB where vGB GBeωAB GBAB vAB 414 344 983 ftsec As seen from the diagram vGB has the same direction as vAB The vector sum is shown on the last diagram We can calculate vG with some geometric labor or simply measure its magnitude and direction from the velocity diagram drawn to scale For simplicity we adopt the latter procedure here and obtain vG 641 ftsec Ans As seen the diagram may be superposed directly on the first velocity diagram Helpful Hints 1 Remember always to convert ω to radians per unit time when using v rω 2 A graphical solution to this problem is the quickest to achieve although its accuracy is limited Solution by vector algebra can of course be used but would involve somewhat more labor in this problem 564 The circular disk of radius 02 m is released very near the horizontal surface with a velocity of its center v0 07 ms to the right and a clockwise angular velocity ω 2 rads Determine the velocities of points A and P of the disk Describe the motion upon contact with the ground Problem 564 5119 The large roller bearing rolls to the left on its outer race with a velocity of its center O of 09 ms At the same time the central shaft and inner race rotate counterclockwise with an angular speed of 240 revmin Determine the angular velocity ω of each of the rollers Problem 5119 5120 The shaft at O drives the arm OA at a clockwise speed of 90 revmin about the fixed bearing at O Use the method of the instantaneous center of zero velocity to determine the rotational speed of gear B gear teeth not shown if a ring gear D is fixed and b ring gear D rotates counterclockwise about O with an angular speed of 80 revmin Problem 5120 43 33 Método do Movimento Relativo para Eixos Transladados Análise de Aceleração Derivando a equação de velocidade 𝐯A 𝐯B 𝐯A B 𝐯B 𝛚 𝐫 com relação ao tempo dá 𝐚A 𝐚B 𝛚 𝐫 𝛚 𝐫 𝐚B 𝛚 𝐫 𝛚 𝛚 𝐫 𝐚B 𝐚AB 𝐚A aceleração absoluta do ponto A 𝐚B aceleração absoluta do ponto B e 𝐚AB aceleração relativa do ponto A com relação ao ponto B 𝛚 𝛂 𝛂 vetor aceleração angular absoluta do corpo rígido Vide Figura abaixo 44 𝐚A 𝐚B 𝐚AB n 𝐚AB t 𝐚AB n 𝛚 𝛚 𝐫 𝐚AB t 𝛂 𝐫 𝐚AB n componente normal do vetor aceleração relativa 𝐚AB t componente tangencial do vetor aceleração relativa SAMPLE PROBLEM 513 The wheel of radius r rolls to the left without slipping and at the instant considered the center O has a velocity v0 and an acceleration a0 to the left Determine the acceleration of points A and C on the wheel for the instant considered Solution From our previous analysis of Sample Problem 54 we know that the angular velocity and angular acceleration of the wheel are ω v0r and α a0r The acceleration of A is written in terms of the given acceleration of O Thus aA a0 aNO a0 aNωt aNωn The relativeacceleration terms are viewed as though O were fixed and for this relative circular motion they have the magnitudes aAOt r0 ω² r0 v0r² aAOn r0 α r0 a0r and the directions shown Adding the vectors headtotail gives aA as shown In a numerical problem we may obtain the combination algebraically or graphically The algebraic expression for the magnitude of aA is found from the square root of the sum of the squares of its components If we use n and tdirections we have aA aAn² aAt² a0 cos θ aNωn² a0 sin θ aNωt² ra cos θ r0 ω²² ra sin θ r0 α² Ans The direction of aA can be computed if desired The acceleration of the instantaneous center C of zero velocity considered a point on the wheel is obtained from the expression aC a0 aCO where the components of the relativeacceleration term are aCOn r0² directed from C to O and aCOt ra directed to the right because of the counterclockwise angular acceleration of line CO about O The terms are added together in the lower diagram and it is seen that aC rω² Ans Helpful Hints 1 The counterclockwise angular acceleration α of OA determines the positive direction of aNωt The normal component aNOn is of course directed toward the reference center O 2 If the wheel were rolling to the right with the same velocity v0 but still had an acceleration a0 to the left note that the solution for aA would be unchanged 3 We note that the acceleration of the instantaneous center of zero velocity is independent of α and is directed toward the center of the SAMPLE PROBLEM 514 The linkage of Sample Problem 58 is repeated here Crank CB has a constant counterclockwise angular velocity of 2 rads in the position shown during a short interval of its motion Determine the angular acceleration of links AB and OA for this position Solve by using vector algebra Solution We first solve for the velocities which were obtained in Sample Problem 58 They are ωAB 67 rads and ωOA 37 rads where the counterclockwise direction kdirection is taken as positive The acceleration equation is aA aB aABn aABt where from Eqs 53 and 59a we may write aA αOA rA ωOA ωOA rA αOAk 100j 37 k 37 k 100j 100αOA i 100372 j mms2 aB αCB rB ωCB ωCB rB 0 2k 2k 75i 300i mms2 aABn ωAB k ωAB rAB 67 k 67 k 175i 50j 672 175i 50j mms2 aABt αAB rAB αABk 175i 50j 50αAB i 175αAB j mms2 We now substitute these results into the relativeacceleration equation and equate separately the coefficients of the iterms and the coefficients of the jterms to give 100αOA 429 50αAB 1837 367 175αAB The solutions are αAB 01050 rads2 and αOA 434 rads2 Ans Since the unit vector k points out from the paper in the positive zdirection we see that the angular accelerations of AB and OA are both clockwise negative It is recommended that the student sketch each of the acceleration vectors in its proper geometric relationship according to the relativeacceleration equation to help clarify the meaning of the solution Helpful Hints 1 Remember to preserve the order of the factors in the cross products 2 In expressing the term aAB be certain that rAB is written as the vector from B to A and not the reverse SAMPLE PROBLEM 515 The slidercrank mechanism of Sample Problem 59 is repeated here The crank OB has a constant clockwise angular speed of 1500 revmin For the instant when the crank angle θ is 60 determine the acceleration of the piston A and the angular acceleration of the connecting rod AB Solution The acceleration of A may be expressed in terms of the acceleration of the crank pin B Thus aA aB aABn aABt Point B moves in a circle of 5in radius with a constant speed so that it has only a normal component of acceleration directed from B to O aB rω2 aB 512 15002π602 10280 ftsec2 The relativeacceleration terms are visualized with A rotating in a circle relative to B which is considered fixed as shown From Sample Problem 59 the angular velocity of AB for these same conditions is ωAB 295 radsec so that aB rω2 aABn 1412 2952 1015 ftsec2 directed from A to B The tangential component aABt is known in direction only since its magnitude depends on the unknown angular acceleration of AB We also know the direction of aA since the piston is confined to move along the horizontal axis of the cylinder There are now only two scalar unknowns left in the equation namely the magnitudes of aA and aABt so the solution can be carried out If we adopt an algebraic solution using the geometry of the acceleration polygon we first compute the angle between AB and the horizontal With the law of sines this angle becomes 1802 Equating separately the horizontal components and the vertical components of the terms in the acceleration equation as seen from the acceleration polygon gives aA 10280 cos 60 1015 cos 1802 aABt sin 1802 0 10280 sin 60 1015 sin 1802 aABt cos 1802 The solution to these equations gives the magnitudes aABt 9030 ftsec2 and aA 3310 ftsec2 Ans With the sense of aABt also determined from the diagram the angular acceleration of AB is seen from the figure representing rotation relative to B to be α ar αAB 90301412 7740 radsec2 clockwise Ans If we adopt a graphical solution we begin with the known vectors aB and aABn and add them headtotail using a convenient scale Next we construct the direction of aABt through the head of the last vector The solution of the equation is obtained by the intersection P of this last line with a horizontal line through the starting point representing the known direction of the vector sum aA Scaling the magnitudes from the diagram gives values which agree with the calculated results aA 3310 ftsec2 and aABt 9030 ftsec2 Ans Helpful Hints 1 If the crank OB had an angular acceleration aB would also have a tangential component of acceleration 2 Alternatively the relation aB v2r may be used for calculating aABn provided the relative velocity vAB is used for v The equivalence is easily seen when it is recalled that ωAB vABrab 3 Except where extreme accuracy is required do not hesitate to use a graphical solution as it is quick and reveals the physical relationships among the vectors The known vectors of course may be added in any order as long as the governing equation is satisfied 5121 The center O of the wheel is mounted on the sliding block which has an acceleration aO 8 ms2 to the right At the instant when θ 45 θ 3 rads and θ 8 rads2 For this instant determine the magnitudes of the accelerations of points A and B Problem 5121 5146 The mechanism of Prob 575 is repeated here Each of the sliding bars A and B engages its respective rim of the two riveted wheels without slipping If in addition to the information shown bar A has an acceleration of 2 ms2 to the right and there is no acceleration of bar B calculate the magnitude of the acceleration of P for the instant depicted Problem 5146 5147 The fourbar linkage of Prob 588 is repeated here If the angular velocity and angular acceleration of drive link OA are 10 rads and 5 rads² respectively both counterclockwise determine the angular accelerations of bars AB and BC for the instant represented Problem 5147 5153 The elements of a power hacksaw are shown in the figure The saw blade is mounted in a frame which slides along the horizontal guide If the motor turns the flywheel at a constant counterclockwise speed of 60 revmin determine the acceleration of the blade for the position where θ 90 and find the corresponding angular acceleration of the link AB Problem 5153 5155 An oil pumping rig is shown in the figure The flexible pump rod D is fastened to the sector at E and is always vertical as it enters the fitting below D The link AB causes the beam BCE to oscillate as the weighted crank OA revolves If OA has a constant clockwise speed of 1 rev every 3 s determine the acceleration of the pump rod D when the beam and the crank OA are both in the horizontal position shown 50 Dados rotação da manivela OB é 1500 rpm no sentido horário Determine a velocidade e a aceleração do ponto G da biela do mecanismo bielemanivela abaixo para θ 60o 51 34 Método do Movimento Relativo para Eixos Girantes Análises de Velocidade e Aceleração A figura 511 mostra uma peça ranhurada que para essa fase do movimento gira com rotação 𝛚 O ponto A realiza movimento sobre a trajetória mstrada na peça ranhurada Para observar o movimento do ponto A sobre a peça ranhurada adotase um referencial girante xy que possui a mesma rotação 𝛚 da peça ranhurada É como se o referencial xy estivesse fixado sobre a peça ranhurada Por isso esse referencial recebe o nome de referencial girante Daí o termo eixos girantes 52 Eixos Girantes Posição 𝐫A 𝐫B 𝐫 𝐫B x𝐢 y𝐣 𝐫A posição absoluta do ponto A 𝐫B posição absoluta do ponto B 𝐫 x𝐢 y𝐣 posição relativa do ponto A com relação ao ponto B x e y são as coordenadas de r no sistema girante xy Velocidade 𝐫A 𝐫B d dt x𝐢 y𝐣 ou 𝐯A 𝐯B x𝐢 y𝐣 x𝐢 y𝐣 𝐢 d𝐢 dt 𝐢dθ dt 𝐣 1 dθ dt 𝐣 ω𝐣 𝐣 d𝐣 dt 𝐣dθ dt 𝐢 1 dθ dt 𝐢 ω𝐢 53 Com base nas duas últimas equações acima observe que 𝐢 𝛚 𝐢 𝐣 𝛚 𝐣 onde 𝛚 ω𝐤 é a rotação do sistema móvel Vide Figura 510 Agora x𝐢 y𝐣 x𝛚 𝐢 y𝛚 𝐢 𝛚 x𝐢 y𝐣 𝛚 𝐫 e x𝐢 y𝐣 𝐯rel 𝐯A 𝐯B 𝛚 𝐫 𝐯rel 𝐯A velocidade absoluta do ponto A Tangente à trajetória absoluta do ponto A 𝐯B velocidade absoluta do ponto B Velocidade absoluta da origem do sistema móvel girante Tangente à trajetória absoluta do ponto B O ponto B é a origem do sistema de eixos móveis Ele é adotado arbitrariamente pelo analista 54 𝛚 rotação do sistema móvel 𝐯rel velocidade relativa observada pelo sistema móvel Tangente à trajetória relativa Para visualização dos vetores acimavide Figura 511 desconsiderando o ponto P Aceleração Considere agora que além do referencial móvel e girante possuir a rotação 𝛚 igual da peça ranhurada ele possua aceleração angular 𝛂 𝛚 igual da peça ranhurada Derivando a equação de velocidade com relação ao tempo resulta 55 𝐚A 𝐚B 𝛚 𝐫 𝛚 𝐫 𝐯 rel onde 𝛚 𝛂 𝛂 𝛚 aceleração angular do sistema de eixos girantes que é igual a aceleração angular da peça ranhurada 𝐫 d dt x𝐢 y𝐣 x𝐢 y𝐣 x𝐢 y𝐣 𝛚 𝐫 𝐯rel 𝛚 𝐫 𝛚 𝛚 𝐫 𝐯rel 𝛚 𝛚 𝐫 𝛚 𝐯𝐫𝐞l 𝐯 rel d dt x𝐢 y𝐣 x𝐢 y𝐣 x𝐢 y𝐣 𝐯 rel x 𝛚 𝐢 y 𝛚 𝐣 x𝐢 y𝐣 𝛚 x𝐢 y𝐣 x𝐢 y𝐣 𝐯 rel 𝛚 vrel arel Portanto substituindo 𝐚A 𝐚B 𝛚 𝐫 𝛚 𝛚 𝐫 𝛚 𝐯𝐫𝐞l 𝛚 vrel arel 56 𝐚A 𝐚B 𝛚 𝐫 𝛚 𝛚 𝐫 2𝛚 𝐯rel 𝐚rel 𝐚A aceleração absoluta do ponto A Possui componetes tangencial tangente á trajetória absoluta de A e normal perpendicular a componente tangencial apondando para o centro de curvatura da trajetória absoluta de A 𝐚B aceleração absoluta do ponto B Possui componetes tangencial tangente á trajetória absoluta de B e normal perpendicular a componente tangencial apondando para o centro de curvatura da trajetória absoluta de B O ponto B é a origem do sistema de eixos móveis Ele é adotado arbitrariamente pelo analista 𝛂 𝛚 aceleração angular do sistema de eixos girantes que é igual a aceleração angular da peça ranhurada 𝛚 𝐫 componente tangencial do movimento de rotação do ponto A em relação à B 𝛚 𝛚 𝐫 componente normal do movimento de rotação do ponto A em relação à B 2𝛚 𝐯rel aceleração de Coriolis 57 𝐚rel aceleração relativa observada pelo sistema móvel Tem duas componentes uma tangencial à trajetória relativa e outra normal perpendicular a componente tangencial apontando para o centro de curvatura da trajerória relativa Para visualização dos vetores acimavide Figura 513 desconsiderando o ponto P SAMPLE PROBLEM 516 At the instant represented the disk with the radial slot is rotating about O with a counterclockwise angular velocity of 4 radsec which is decreasing at the rate of 10 radsec² The motion of slider A is separately controlled and at this instant r 6 in r 5 insec and r 81 insec² Determine the absolute velocity and acceleration of A for this position Solution We have motion relative to a rotating path so that a rotating coordinate system with origin at O is indicated We attach xy axes to the disk and use the unit vectors i and j Velocity With the origin at O the term vB of Eq 512 disappears and we have vA ω r vrel 1 The angular velocity as a vector is ω 4k radsec where k is the unit vector normal to the xy plane in the zdirection Our relativevelocity equation becomes vA 4k 6i 5i 24j 5i insec Ans in the direction indicated and has the magnitude vA 24² 5² 245 insec Ans Acceleration Equation 514 written for zero acceleration of the origin of the rotating coordinate system is aA ω ω r ω r 2ω vrel arel The terms become 3 ω ω r 4k 4k 6i 4k 24j 96i insec² ω r 10k 6i 60j insec² 2ω vrel 24k 5i 40j insec² arel 81i insec² The total acceleration is therefore aA 81 96i 40 60j 15i 20j insec² Ans in the direction indicated and has the magnitude aA 15² 20² 25 insec² Ans Vector notation is certainly not essential to the solution of this problem The student should be able to work out the steps with scalar notation just as easily The correct direction of the Coriolisacceleration term can always be found by the direction in which the head of the vrel vector would move if rotated about its tail in the sense of ω as shown Helpful Hints 1 This equation is the same as vA vP vAP where P is a point attached to the disk coincident with A at this instant 2 Note that the xyz axes chosen constitute a righthanded system 3 Be sure to recognize that ω ω r and ω r represent the normal and tangential components of acceleration of a point P on the disk coincident with A This description becomes that of Eq 514b SAMPLE PROBLEM 518 For the conditions of Sample Problem 517 determine the angular acceleration of AC and the acceleration of A relative to the rotating slot in arm OD Solution We attach the rotating coordinate system xy to arm OD and use Eq 514 With the origin at the fixed point O the term aB becomes zero so that aA ω r ω ω r 2ω vrel arel From the solution to Sample Problem 517 we make use of the values ω 2k rads ωCA 4k rads and vrel 450 2i mms and write aA ωCA rCA ωCA ωCA rCA ωCAk 2252 i j 4k 4k 2252 i j ω r 0 since ω constant ω ω r 2k 2k 2252i 900 2i mms² 2ω vrel 22k 450 2i 1800 2j mms² arel xi Substitution into the relativeacceleration equation yields 12 225ωCA 3600i 12 225ωCA 3600j 900 2i 1800 2j xi Equating separately the i and j terms gives 225ωCA 36002 900 2 x and 225ωCA 36002 18002 Solving for the two unknowns gives ωCA 32 rads² and x arel 8910 mms² Ans If desired the acceleration of A may also be written as aA 225232i j 36002i j 7640i 2550j mms² We make use here of the geometric representation of the relativeacceleration equation to further clarify the problem The geometric approach may be used as an alternative solution Again we introduce point P on OD coincident with A The equivalent scalar terms are aAt ωCA rCA riωCA raCA normal to CA sense unknown aAn ωCA ωCA rCA ravCA² from A to C aPn ω ω r 0Pω² from P to O aPt ω r riω 0 since ω constant 2ω vrel 20ωvrel directed as shown arel x along OD sense unknown Helpful Hints 1 If the slot had been curved with a radius of curvature ρs the term arel would have had a component vrel²ρs normal to the slot and directed toward the center of curvature in addition to its component along the slot 5160 The disk rotates about a fixed axis through O with angular velocity ω 5 radsec and angular acceleration α 3 radsec2 in the directions shown at a certain instant The small sphere A moves in the circular slot and at the same instant β 30 β 2 radsec and β 4 radsec2 Determine the absolute velocity and acceleration of A at this instant Problem 5160 5163 An experimental vehicle A travels with constant speed v relative to the earth along a northsouth track Determine the Coriolis acceleration aCor as a function of the latitude θ Assume an earthfixed rotating frame Bxyz and a spherical earth If the vehicle speed is v 500 kmh determine the magnitude of the Coriolis acceleration at a the equator and b the north pole Problem 5163 5176 For the instant represented link CB is rotating counterclockwise at a constant rate N 4 rads and its pin A causes a clockwise rotation of the slotted member ODE Determine the angular velocity ω and angular acceleration α of ODE for this instant Problem 5176 5182 The crank OA revolves clockwise with a constant angular velocity of 10 rads within a limited arc of its motion For the position θ 30 determine the angular velocity of the slotted link CB and the acceleration of A as measured relative to the slot in CB Problem 5182 5183 The Geneva wheel of Prob 556 is shown again here Determine the angular acceleration α2 of wheel C for the instant when θ 20 Wheel A has a constant clockwise angular velocity of 2 rads Problem 5183 5191 The pin A in the bell crank AOD is guided by the flanges of the collar B which slides with a constant velocity vB of 3 ftsec along the fixed shaft for an interval of motion For the position θ 30 determine the acceleration of the plunger CE whose upper end is positioned by the radial slot in the bell crank Problem 5191 Problem Answers When a problem asks for both a general and a specific result only the specific result might be listed below Chapter 1 11 180lb person m 559 slugs 816 kg W 801 N 12 W 14720 N 310 lb m 1028 slugs 13 m 001294 slugs 01888 kg W 1853 N 14 27 1392i 18j 1939j 6j 1787k 215 15 a 12551010j N b 3141011 j N 16 F 573i 331j104 N 17 h 0414R Chapter 2 21 v 75 ms 22 t 211 sec 789 sec 23 s 72 m v 42 ms2 24 a 150 mms2 25 Δs 27 mm D 45 mm a constant 26 Δs 24 m 27 a 361g 28 Δs 1248 m D 1419 m 29 v 3 30t 2t2 ms s 5 3t 15t2 23 t3 m 210 s 213 ft 211 va 1390 ftsec 212 h 2040 m t 408 s 213 h 494 ft t 424 sec vB 564 ftsec down 214 tAC 239 sec 215 v 25 ftsec a 312 ftsec2 tACC 08 sec 216 vxx 075 ms v 125 ms 217 a v 219 ms b v 256 ms 218 t 655 s s 1819 km 219 Δa 05 ms2 Δs 64 m 220 v 08 ms 221 s 326 m t 326 s 222 a 1168 ftsec2 v 998 mihr 223 s 713 m 224 s 330 m 225 s 2250 m 226 a t 00370 sec b t 00555 sec 701 227 vm 120 ms h 1934 km 228 s 9720 ft 229 t 0917 s 230 v 1587 insec 231 v 389 kmh 232 vmax 359 ftsec 233 a 667 ms2 t 234 t 1667 s 235 a 872 ftsec2 t 274 sec 236 v 1789 ftsec 237 s 5810 ft 238 D 3710 ft 239 c v02 2gym ym2 240 Particle 1 s v0k1 ekt v v0 ekt Particle 2 s vt 16 kt3 v v0 12 kt2 Particle 3 s v0k sin kt v v02 k s2 241 v 2 KL D2 LD 242 K 1073103 ft1 t 254 sec 243 t 508 s 244 D 653 ft Δs 667 ft s5 367 ft 245 e 23 s3 660 m 2ym 246 vmax 18 ms 247 D 12C2 ln 1 C2 C1 v02 248 Δ 10 s s 416 m 249 a v 13040 ftsec b v 12290 ftsec 250 a v 6490 ftsec b v 4990 ftsec 251 v v03 x cv0 1x13 ms 252 v v0k x v0k 1 ekt v v0 kx 253 a s 1206 m b s 1268 m 254 D 0693k t 1 kv0 255 x 0831 ft 256 h 1208 ft v 785 ftsec 257 v 6e ect 1 kc ebt ect c b 258 t 105 sec amax 1173 ftsec2 259 vav 206 ms θ 760 260 aav 5 ms2 θ 531 261 v 620i 336j ms θ 279 262 v 894 mms2 θx 634 263 v 447 mms2 θx 266 264 v 242 ftsec a 253 ftsec2 265 t 247 sec h 1786 mi 266 v 92 144r2 t 30 ftsec 267 v0 367 ms d 1340 m 268 Rmax v02 g 269 m 343 ms 270 v0 1633 ftsec θ 668 271 v0 771 ftsec θ 311 272 257 ft above B 273 θ 506 250 ft right of B 274 θ 1491 275 θ 217 276 θ 557 277 v 700 ftsec s 1185 ft 278 θ 487 or θ 536 279 R 2970 m 280 u 1441 ms 281 θ1 261 θ2 806 282 R 464 m θ 233 283 s 455 m t 1335 s 284 206 v0 224 ftsec 285 vmax 1135 ms umin 0744 ms 286 310 θ 343 or 531 θ 547 287 s 776 ft 288 a 0445 h 500 ft 289 h 1227 m 290 s 1046 km t 1775 s 291 Ri 0667 v02 gi 1155 v02 g 292 f2 1 f1 f2 1 2 293 vx v0 cos θ ekt x v0 cos θ k 1 ekt vy v0 sin θ g k ekt g k y 1k v0 sin θ g k 1 ekt g t k vx 0 vy g k 294 h 583 ft tf 1259 sec d 746 ft 295 θ 90 α θ 45 60 675 2 296 x 1242 ft y 627 ft 297 vA 1175 ms vag 1346 ms 298 ax 839 ftsec2 2100 v 713 kmh 2101 v 530 ftsec am 250 ftsec2 2102 a 0269 ms2 2103 p 266 m 2104 tA 897 s tB 992 s 239 m 2105 tA 897 s tB 889 s 250 m 2106 am 367 ftsec2 at 20 ftsec2 2107 pB 1630 m 2108 uA 258 ms uB 396 ms 2109 u 356 ms a 00260 ms2 2110 u 278103 kmh 2111 a a 7 ftsec2 b a 1797 ftsec2 c a 882 ftsec2 2112 a 16e 1610e ftsec2 2113 N 336 revmin 2114 P1 u 2 ms a1 50 ms2 P2 a 80 ms2 a2 854 ms2 2115 u 72 kmh 2116 a p 243 ft u 1847 ftsec2 b p 1334 ft u 0 2117 a p 1422 ft a 658 ftsec2 b p 1497 ft a 875 ftsec2 2118 p 417 in 2119 ap 338 ms2 ap 15 ms2 2120 au 1888 ftsec2 2121 p 19070 km a 1265 ms2 2122 a a 2g right a 0 b a 389 ms2 597 c a 973 ms2 e 1684 2123 e 08s a 731 ms2 a 1281 12s e 1962 ms2 e 180 2124 tA 1052 s tB 1086 s 2125 a 939 ftsec2 2126 p 18 480 km 2127 p 437 km a 874 mms2 a 363 mms2 2128 L 461 m 2129 p 125 m 2130 xC 225 m yc 229 m 2131 a 1280 ms2 a 880 ms2 2133 r 477 ftsec 6 410 degsec 2134 r 931 ms 0568 rads 2135 r 207 ms2 e 1653 rads2 2136 u 545 mms a 632 mms2 2137 r 425 ms e 01403 rads 2138 r 1152 ms2 00813 rads2 2139 i 328 mms 2140 a 2K2 R2 120 2141 u 0377 ms at a 260 a 0272 ms2 at a 1944 2142 r 15 ftsec e 450 radsec 2143 v 1169 ftsec2 6 234 radsec2 2144 r 1200 ftsec e 670 ftsec2 2145 r 1512 ms 6 00495 rads 2146 r d r v0 cos a r v2 sin2 a d a 0 6 U0 sin a d r 2u0 2 cos a sin a g 2147 u 1617 ftsec 6 00808 radsec 2148 a 862 ftsec2 0 001832 radsec2 2149 a o24ce2 4bc cos 9 62 2150 c 1075Ka 2151 u 529 ms E 489 a 976 ms2 2152 v b02 2153 u 699 mihr srin 0 h 22 cot 2 2154 i 1732 ms r 333 ms2 0 385 rads2 2155 r 0256 m r 472 ms 8 387 g 646 rads 2156 uA 1190 ms θ1 1252 αA 754 ms2 θ1 225 2157 u 0296 ms a 0345 ms2 v 00641 0289j ms a 0328j 01086j ms2 2158 u 962 ms ur 556 ms ar 1029 ms2 6 00390 rads2 2159 r 2b sin β r act cos 273 0 radiation r 0 2160 r 21 900 m r 730 ms r 207 ms2 g 432 t 000312 rads θ 901103 rads2 2161 i 358 ms a 1786 rads r 315 ms 6 1510 rads2 2162 r 224 m r 671 ms 6 459 ms2 θ 266 0 006 rads 0 00518 rads2 2163 r 8910 lsec r 1790 1sec θ 348104 radsec 0 1398107 radsec2 2164 r 510 ft r 914 ftsec r 1135 ftsec2 e 319 6 0334 radsec θ 0660 radsec2 2165 6 746 a 1571 ms2 p 859 m 2166 u 4700 ft y 1710 ft v 2220 ft ux 235 ftsec uy 855 ftsec uz 211 ftsec αx ay 0 ay 322 ftsec2 2167 α 275 ms2 2168 uB u sin 6 uR u cos 0 cos 0 ua u cos 0 sin e 2169 cmax V 2 161 4 4 2 2170 up V2 eo l2 12 2171 a 1982 ftsec2 a0 291 ftsec2 a 0386 ftsec2 2172 R 920 kmh θ 01988 rads 6 00731 rads 2173 R 201 ms2 a 0 3 5 00238rads2 2174 u 2 K2 r sin2 B a K sin B K2T2 4c2 2175 x y r sin z cos 3 2 3 2 2 176 uR bu sin B u0 a sin B 2 42 sin2 B h2 2177 u 296 ms a 0672 ms2 2178 uA 1347 ms aA 841 ms2 2179 uP 285 ms ap 580 ms2 2180 ap 510 ms2 a 764 ms2 a 03 ms2 2181 uR 0 U0 Rew1h2R2 u0 ho x1 h2R2 2182 a2 b0 e tan27 sin2 β 1etanT sin3 where B tan16b 2183 uAB 151 225j ms aA 2184 uAB 1086 ftsec2 2185 uAB 1442 kmh 13 337 west of south 2186 uWP 1229j 1860j mihr 223 mihr 334 west of south uWP 1229j 1396j mihr 1237 mihr 648 north of west 2187 u 718 u 794 mihr 2188 uR 3001 1999j ms 2189 uAB 363j 0628j ms2 2189 uR 1383 knots B 231 2190 a 238 2191 uB 643 ms 2192 e 287 below normal 2193 a 1887 2194 vPW 273i 85j mihr 893 mihr 1782 east of north 2195 vB 523i 1667j ftsec 2196 uB 206 kmh ab 0457 ms2 2197 a 1389 ms2 2198 B 556 2199 r r0 2 6 0 2200 uR 924 kmh uR 354 kmh 2201 uA 0787 ms2 B 935 2202 uAwA 0733i 292j ftsec2 2203 a 333 vA B 731i 731j ftsec 2204 r 0637 ms 0 1660106 rads2 2205 a vA B 501 50j ms auB 1251 ms2 b u 0884 ms2 p 5660 m 2206 vAB 715i 474j ftsec 2207 uA 04 ms down 2208 uA 18 ms up a 3 ms2 down 2209 aA 2 ftsec2 up 2210 rA aB 0 2211 h 400 mm 2212 t 200 s 2213 u 15 ms up 2214 uA uB 0 one 2215 uA uB WC 0 two 2216 uB 3yA 2y2 b2 2217 uA 2xy182 2218 uHA 1 ftsec aHA 2 ftsec2 uC 4 ftsec 2219 uA 276 ms 2220 u 838 mms L2bλ 2221 a L2y232 2222 uB Zx x V2 U1 2223 uA sin2 02 2224 uB 629 ms up 2225 ap 1193 mms2 up 2226 u 2x2 h2 r x2 h2 2227 u 727 ms 2228 u 15 ftsec uB 618 ftsec 2229 311 20 ft 2230 t1 227 sec t2 848 sec 2231 t 208 s h 418 km 2232 r r0 1 r22 2233 6 1386 radsec 6 215 radsec2 2234 343 ft 2235 p 953 km 2236 uP 272 ms 2237 t 535 sec 2238 a 5 ftsec2 a 5i ftsec2 a 0 ap 10 ftsec2 p 6 2239 B 1209 2240 1255 1193 0 ft 2241 uAB 458 ms2 p 206 west of north 2242 p 1499 m a 5 ms2 2243 i 15 ms r 444 ms2 6 0325 rads 0 00352 rads2 a 693 ms2 a 4 ms2 p 1299 m 2244 a a bV 4Q cos O b a bVK2 4cuoa2 2245 u 6 468 mms up 67 2246 a 786 mms2 up 2247 t 1473 s x 01178 m 2248 x 00023 fts u 998 ftsec u 1135 ftsec 2249 a a 303 ms2 b a 0 c a 303 ms2 2250 amax 1104 at t 0802 sec θmax 379 rads at t 0324 sec θ 90 at t 0526 sec 2251 uMBmax 70 ms at t 471 s and sB 1264 m uMBmin 10 ms at t 236 s and sB 557 m aMBmax 612 ms2 at t 0 and sB 0 aMBmin 252 ms2 at t 10 s and sB 150 m 2252 a 422 R 1013 m 2253 a 903 ms2 at 0 443 and r 0237 s amax 1076 ms2 at 0 0 and t 0 2254 a 10 insec at t 0330 sec x 245 in Chapter 3 31 t 1784 s x 624 m 32 a t 559 s x 1958 m b Crate does not stop 33 a no motion b a 345 ftsec2 down incline 34 R 846 N L 1104 N R L 0 35 F 2890 N 36 n sin 37 sa 807 m sd 751 m 38 a 358 ftsec2 up 39 a 496 ms2 up incline 310 a gsin 01 sin9 311 a a 644 ftsec2 b a 1610 ftsec2 312 a r 0257 ms2 b a 0513 ms2 313 12 s 314 T 1042 N 315 T1 39200 lb T100 392 lb 316 a 566 ms2 317 F 00206 lb 318 FA 4080 lb up 319 TA 750 N TB 554 N 320 μ 0429 321 a a 0 b a 1390 ms2 right 322 Not possible 323 μt 0555 324 a aA 1095 ms2 aB 0981 ms2 b aA aB 0667 ms2 325 B tanI a g sin θ x cos θ 326 n 660 327 k 5 lbin 328 T 1713 N 329 aA 1450 ms2 down incline aB 0725 ms2 up T 1054 N 330 x 201 m 331 u 2P P mgaL 332 aA 1364 ms2 right aB 932 ms2 down T 466 N 333 ap 237 ms2 down T 821 N 334 ax 32214 30x u 1447 ftsec 335 a a 0 b a 0714 ms2 left 336 Case a u 743 ms 337 a 1406 ms2 338 tan1 0 e 2 339 a h 555 m b h 1274 m 340 a u 0327 ms bt 00768 s y 001529 m 341 u 2100 ms 342 a 588 472 343 0 P 27 lb No motion 27 P 54 lb a aB 01789P 322 P 54 lb a 483 ftsec2 aB 0322P 966 344 u 1119 kms 345 T 1380 N a 0766 ms2 346 a T 852 N b T 1614 N 347 NA 1089 N NB 830 N 348 a N 1374 lb b 1610 ftsec2 349 R 1173 N a 721 ms2 350 a R 025 lb b R 0271 lb 351 0 453 352 P 4 lb side A 353 N 00241 lb 354 u 0540 355 N 863 revmin 356 w 1064 rads 357 a uB 542 ms b NA 241 N 358 6 337 degs 359 a 0818g F 2460 lb 360 a 220 ftsec2 361 uA 1407 ftsec uB 1638 ftsec 362 k cos20 363 s u0 mg 364 F 920 lb 365 NA 3380 N NB 1617 N 366 D 450 kN L 274 kN 367 T 176 N F 352 N 368 F 1659 N 369 v gr tan θ 370 w 501 rads 371 0 t 1414s T2 00707 00354t2 N 00707 00354t2 T1 00707 00354t2 N 00707 00354t2 1414 t 5 s 372 P 270 N P2 1962 N 373 p 3000 km j 600 ms2 374 Dynamic Fp 479 N F 1400 N Static Fp 589 N F 1019 N 375 u L2πgR N 2mg 376 u 1494 ftsec ωmin 0 vmax 345 ftsec 377 R 214 1913 cos θ N vB 206 ms 378 x 1188 mm N 253 N 379 F 439 N 380 T 252 lb N 0326 lb side B 381 N 14πμsμr2 1 382 Ω 477 rads 383 k 1 Rω2g2 W 99655 N 384 P 221 N N 1422 N 385 vs r0q0cos ωt eωt r r02 eωt ωt eωt v0 r0q0 cosh ωt 386 P 862 lb 387 a1 10 ms2 N1 N2 2 N T 283 N 388 a1 10 ms2 N1 N2 483 N T 523 N 389 a F 1562 lb b F 2260 lb c F 1562 lb 390 n 9620 ftsec θ 1139104 radsec 391 a 1153 ftsec2 s 272103 radsec2 392 T 253 N R 1028 N lower side 393 vmax π2 T mg3 sin θ 3 cos θ 2 394 v 552 ms 395 N 816 N R 387 N 396 κ r2μk lnv2 v02 r2rg Nmg 397 N 1 4r2x2 0 398 a U12 60 J b U12 235 J 399 vB 305 ms 3100 Uf 672 ftlb 3101 v 1718 ftsec 3102 k 974 lbin 3103 R 3340 lb 3104 u 2gh 3105 Q 1835 J a and b u 810 ms 3106 P 0400 hp 3107 a s 0663 m b s 0349 m 3108 a s 1853 m b s 1226 m 3109 a u 256 ms b x 989 mm 3110 a u 593 ftsec b u 655 ftsec 3111 P 0393 hp P 293 W 3112 v 1881 ms 3113 e 0764 3114 Q 903 kJ 3115 v 708 ftsec 3116 R 405 kN P 522 N 3117 R 405 kN ΔQ 1620 J 3118 e 0892 3119 At B P 1934 kW Halfway P 1367 kW F 6960 N 3120 a F 61200 lb b P 3270 hp c P 6530 hp d P 2670 hp 3121 v 1337 ftsec 3122 v 6460 ms 3123 a P 0 b P 287 W c P 0 3124 k 5mglh dd3 3125 a NB 4mg b NC Tmg c s 4R1 μk3 3126 v 530 ms 3127 v 1748 ftsec 3128 u 344 ms 3129 vc 359 ms 3130 δ 01445 m 3131 Fm 368 kW 3132 k 879 kNm 3133 v 780 ftsec 3134 u 1734 ms u 1889 ms 3135 δ 294 mm 3136 a P30 5 hp P60 16 hp b P 352 hp Pthres 317 hp v 709 mihr c u 0496 ms b xmax 1864 mm c xc 932 mm 3137 u 546 ftsec 3138 v 365 ftsec 3139 u 343 ms x 485 mm 3140 a vB 940 ms b δ 542 mm 3141 a NC 777 N b NC 1019 N c NE 353 N down 3142 y 0224 m 3143 vB 2gR kR2m3 22 vc 4gR kR2m3 22 N m5g kRm3 22 3144 W 881 lb 3146 v 1248 ftsec 3147 v0 6460 ms 3148 Uf 236 J Fav 338 N 3149 a u 384 ftsec b x 0510 in 3150 k 393 Nm ω 1370 ms θ 228 rads 3151 θ 422 rads 3152 vA 0616 ms vB 0924 ms 3153 vB 854 ftsec 3154 k 868 Nm t 1371 ms 3155 a u 306 ftsec b u 1641 ftsec 3156 v 0331 ms 3157 P 286 lb 3158 u 493 ms 3159 u 343 ftsec 3160 vB 26 300 kmh 3161 xmax 1059 mm umax 1493 ms x 272 mm 3162 r 204 mihr 3163 u μmf 3164 vB 351 kmh 3165 k 1551 Nm 3166 vBmax 0962 ms 3167 θ 438 3168 a m 0528 kg b u 1005 ms 3169 a 1143 ms 3170 a k 1119 Nm b u 0522 ms 3171 Jf 1045 rads 3172 v grπ2 4π 3173 17 Ns 3174 F 303 kN 3175 v2 1885i 74j 47k ms 3176 ΔE 13480 J n 999 3177 ΔR 568 N 3178 μt 0302 3179 vB 1652 ms down 3180 t 773 s 3181 a v 244 mihr b ΔE 2230 ftlb 3182 vc 1231 ms left 3183 R 423 lb 3184 v 210 ms 3185 x1 290 ms right x2 0483 ms left 3186 t 1218 min 3187 e 586 ftsec 3188 a 0 b v4 269 ms right 3189 T 2780 N 3190 v 1909 ms ΔE 1718103 J 3191 x1 205 ms left x1 0878 ms right 3192 v1 0417 ms v3 896 ms 3193 v7 000264 ms Fav 595 N 3194 v1 0 v3 242 ftsec v5 644 ftsec v7 0 3195 R 1496 kN 3196 v F0mb1 ext s F0mb t 1b ext 1 3197 t 4 min 33 sec 3198 v 342 ms 3199 v 1423 ms down incline t 825 s 3200 a u 1333 mihr b aA 978 ftsec2 aB 1956 ftsec2 c R 12150 lb 3201 v 0663 kmh 3202 v 01935 kmh 3203 v 1782 mihr θ 547 3204 v 1562 ftsec θ 502 3205 v 661 ms 3206 v1 0 v3 0468 ms v5 530 ms v7 0 3207 R 472 lb a 4660g d 0900 in 3208 v 520 knots 3209 s μAμB mBmA mC2 3210 Ry 559 lb Ry 218 lb 3211 R 430 N β 868 3212 a u 310 ms b u 723 ms c u 910 ms 3213 vB mAmB 2glmBmA 1 3214 v2 400 mms 3215 H0 693 kgm2s 3216 a G 8491 8491 kgms b H0 232k kgm2s c T 24 J 3217 H 389 Nms M 260 Nm 3218 H02 341 01333j 196 kgm2s 3219 Ho muci ak ĤO Fbi aj 3220 ω 5v3L 3221 a Ho mrz2gr ĤO mzgr b Ho 2mzgr ĤO 0 3222 δ 2m1m1 4m2 L1 3223 t 1508 s 3224 ωp 17723 mihr 3225 HB 1113k lbft 3226 dωdr 2απr 3227 ω 01721 rads CW 3228 ω ω04 n 34 3229 vB 47 850 ms up 58 980 ms 3230 A HO 0 B HOz 0120 kgm2s 3231 a Ho 0 b Ho 2mσ3 sin2 θ cos θg 3232 HO 2mguAk 3233 B HA 0 HD m1gβρ3 CCW C HA 0714mβρ CCW HD 1126mβρ CCW 3234 vL 88870 ftsec v 125700 ftsec 3235 ωd 277 rads CCW ω 521 3236 T0 0745 lb 3237 θ 529 3238 ω 300 rads U 534 J 3239 e 0829 n 312 3240 v 420 ms 3241 452 ms left x2 268 ms right 3242 F 1070 N left 3243 m1m2 e 3244 u u4 1 e2 3245 δ 1 e2πk 3246 vA 0633v vB 0733v 3247 e h2h14 3248 h 1094 in h2 743 in 3249 m 90 kg v 266 ms ΔE 2470 J 3251 a x d3 b x 0286d 3252 e 0333 3253 e 0434 3254 vn 1 e2n v1 3255 R 1613 m 3256 L2 eL1 3257 a h 1498 in b h 1625 in 3258 vB 217 kmh 3259 vA 673 ftsec at θA 635 vB 322 ftsec at θB 270 n 444 3260 θ 292104 3261 vA 683 ms at θA 180 vB 651 ms at θB 502 n 346 3262 v12 x 1672 ms vA x 1649 ms v22 x 699 ms v2 x 384 ms βd 3263 e hh h vx mh h h 3264 θ 823 223 3265 h 0385 m 3266 α 1137 786 3267 ω 1851 misec 3268 u 7569 ms 3269 See Prob 114 and its answer 3270 Δv 1987 misec 3271 a vrel 16227 mihr b vrel 18306 mihr 3272 Δh 880 km 3273 c 001196 τ 1 h 30 min 46 s 3274 vp 3745 mihr 3275 Δu 534 ms 3276 h 82 600 km 3277 a u 7544 ms b u 7912 ms c u 10398 ms d u 10668 ms 3278 Δv 3217 ms θ 7767 ms θ 90 3279 l 899 mi 3280 t 646 days 3281 Δτ 716 sec 3282 Δv 329 s 3283 vp 1683 ms vA 1609 ms Δv 1835 ms 3284 vB 10551 ftsec b 17833 mi 3285 v 1 h 36 min 25 s r τ τ 6 min 4 s 3286 tan β e sin θ1 e cos θ 3287 Δv 302 ftsec 3288 Δv 148 ftsec 3289 α 339 3290 β 1533 3291 τv 65869 h τnf 65468 h 3292 ΔνA R gR H 1 RR H 3293 ρ 00514 rads 3294 Δv 2940 ms 3295 t 1625 s 3296 h 922000 mi 3297 ΔvA 2370 ms ΔvB 1447 ms 3298 e 6572 km parallel to the xaxis 5301 e e 001284 v 7690 ms vp 7890 ms rmax 666106 m rmin 649106 m 3299 αrel kδ1m1 1m2 m 3300 μκ 0382 3301 G 91 kg ms G 3i kgms T 135 J T 15 J HO 45k kgm2s HO 15k kgm2s 3302 P 669 kN 3303 F 376 lb 3304 F 1940 kN 3305 xcrt 283 m vroll 246 ms 3307 vwall max a0mk 3308 a 1699 ms2 R 0 3309 a and b T 112 J 3310 T 3ma0 sin θ T2 90 N 3311 T0 mg a03 2 cos θ0 3312 Prel 01206 hp 3313 a and b h2 e2h1 3314 vA v2 2gl sin θ 2v0 cos θ 2gl sin θ12 3315 grel 9825 003382 cos2γ ms2 3317 Ut 754 J 3318 vB 287 ms vC 1533 ms 3319 P 1936 kW 3320 θ tan1gb 3321 k 318 kNm 3322 R 467 N 3323 vA 7451 ms e 00295 3324 δ mgR5 2μp k 3325 rmax 529105 km 3326 TB mvL r2 2g sin θ TC mvE r2 5g sin θ 3327 t 202 s 3328 u gr xmin 2R 3329 F 2650 lb 3330 v 655 ftsec x 0316 ft n 0667 3331 T 424 N 3332 a ac 1075 ms2 b at 1489 ms2 3333 vml 73 gl left 3334 δ mgk ddp 2ρ1 μH d 22ρ 3335 δ 255 in 3336 Fxy 428 lb 3337 s 228 m 3338 t 850 sec 3339 v1 1741 ms v2 0201 ms 3340 u 1829 mihr 3341 P30 3 hp P60 16 hp t 205 sec s 5900 ft 3342 r g 2ω2 cosh θ cos θ 3343 387 θ 658 3344 θ 809 05 sin θ cos θ 1 rads θmax 531 3345 μAlmax 386 ms at sB sRB 00767 m μBLmax 325 ms at sB sB0 00635 m 3346 t 340 s θ 663 3347 θmax 1072 rads at θ 656 θmin 408 3348 μmin 0622 at θ 1219 3349 θ 217 3350 t 1069 s θ 306 3351 t 0610 y 1396 ft 3352 umax 1283 ms at θ 1740 3353 t 0408 s 3354 vmax 569 ftsec at θ 508 Nmax 275 lb at θ 662 Chapter 4 41 r d7101 6j r d74 4j Fb FI 7m T 6mvr2 H0 4mυdk H0 Fdk 42 Hq 12 mυdk Hq Fq k 43 r d71 4j 6k r d74i 2j 6k Fb Fk k T 13mυ2 H0 mud12i 6j 2k H0 Fdj 44 Hq mud 7 721 24j 28k Hq 2Fd 7 21 3j 45 ay 519 ms2 46 T 583 lb 47 a 4 ms2 48 ac F 2m g sin θ 49 a 1519 ms2 410 Masscenter accelerations are identical 411 F 292 N 412 u 1137gr R 2229mg 413 H0 2mr2 α ujk 414 H0 33k kgm2s 415 g 1312 ftsec2 416 t 4m2ω2 M 417 r 807 rads 418 t 272 s 419 vA 1015 ms vB 1556 ms 420 v 0205 ms 421 v 0355 mihr n 950 422 r m1 m2 x1 m2l m0 m1 m2 423 x 0316 ft no 424 v 471 ms both spheres 425 a Pmin 9mg 8 b υ 3gr2 426 ΔQ 252 J It 1287 Ns 427 a a F 2m b b 2Fb mL2 428 v 0877 ms 429 v 727 kmh 430 v m0 m0 2m v0 u ubvob m0 m0 2m 431 vx 2gl δ 2 2gl 432 v 392 ftsec 433 ΔQ 0571ρgπ2 4108 a a g x L b r g x1 x L c v gL 4109 R pgx 4L 3x 2L x 4110 C 4340 N up D 3840 N down 4111 v gx R gL 32 x 4112 a hH g v h gH R 2ρgH 2h2 3H Chapter 5 51 vA 032i 008j ms2 aA 032i 076j ms2 52 vA hi bj aA bω2 hα i hω2 ba j 53 vA 1332i 219j ms2 aA 6421 i 9161 ms2 54 N 333 rev 55 θ θ0ω0t sin θ 0 θmax θ0ω0t2 at θ θ0 θmax 56 ω 0411 radsec αω 0344 radsec 57 v 5 ms a 50 ms2 58 b 1806 mm 59 N 300 rev 510 vA 1777i 270j ms aA 16341 i 457j ms2 511 θ θ0 1099 rad t 1667 s 512 Δθ 244 rad 513 θ 9 rad 514 θ 304 rads θ 346 rads2 515 α 395 rads2 516 t 01784 sec 517 a α 300 rads2 b aB 375 ms2 c ac 225 ms2 518 r 3 in 519 ω 2k radsec a 32 k radsec2 ac 2111 5j in sec2 520 v 0374i 01905j ms a 0751 0605j ms2 521 v 0223i 0789j ms a 3021 1683j ms2 522 v 00464i 01403j ms a 01966i 0246j ms2 523 θ 0596 rad 524 θ θ0 250 rev θ θ0 1875 rev 525 ω 246k radsec 526 aC 1496 ms2 527 N 513 revmin 528 NB 415 revmin 529 ωOA vd x2 d2 530 v 2ax 4b2 x2 531 t 667 sec 532 v 628 ms 533 v rω sin θ a rα sin θ rω2 cos θ 3x 2L 1 3x3 4L2 534 535 ω 12 radsec uO 34 ftsec 536 u u0 21 sin θ a u02 r toward O 537 aB 789 ms2 down 538 ωOA hv h2 s2 539 v 7x2 cot θ2 540 vO 12 ms ω 1333 rads CCW 541 v v x r x r2 542 v rhaω x2 h2 543 ax ω2 rsin θ 544 v x xr2 r2 1 545 v 12 vA tan θ 546 v 2s b2 L2 2bL cos θ L tan θ 547 ω 43 radsec CCW α 1 radsec2 CCW 548 ω 1795 radsec CW 549 ω 1056 rads CW α 0500 rads2 CCW 550 ωCB 630 rads 552 α t02 2π r2 553 β 628 cos θ 0278 1939 cos θ radsec 554 ω 0825 radsec CW 555 vC vB 2 8 sec2 θ 2 556 αk2 1923 rads 557 ωAB rAωl cos θ 1 r2l2 sin2 θ αAB rAω2 l r2 l2 1 sin θ 1 r2 l2 sin2 θ32 558 α 01408 radsec2 CCW 559 vB 1386i 12j ms 560 a N 917 revmin CCW b N 458 revmin CCW c N 458 revmin CW 561 vA 1672i 107 253j kmh vB 105 585j kmh vD 108 929j kmh vC 1672i 107 257j kmh 562 vCD 0579 ms 563 ω 665 radsec CW 564 v 071 04j ms vp 03i ms 565 589 mms 566 ωAB 096 rads CCW 567 ω 0375 rads CCW 568 vO 06 ms vF 0849 ms 569 vAB 2371 310j insec 570 ω 849 ms right α 261 rads CW 571 ωOA 333k rads 572 vAB 12i j ms vp 12i 08j ms 573 vB 438 ms ω 323 rads CCW 574 vBC 277 radsec CCW 575 vp 0900 ms 576 ω ωO CW ωA 258ωO down 577 ωAB 0966ωCD CCW vB 2roq 60 578 ω 0295 radsec CCW 579 ωBC 3 radsec CW 580 vA 904 insec vC 699 insec 581 ω 1394 rads CCW vA 0408 ms down 582 vB 1255 rads CW vC 1155 ms 583 ω 859 rads CCW 584 ωAB 1938 radsec CW 585 vD 9 ms 586 vC 624 ftsec 587 αk2 1923 rads CCW 588 ΔB 1725 rads CCW ωBC 4 rads CCW 589 ωCA 0429k rads 590 vB 397 ms 591 05 m above G vA vB 233 ms 592 05 m below G vA 1949 ms vB 266 ms 593 ωB 859 rads CCW 594 015 m below P vp 03 ms vA 0806 ms 595 vA 277 mms 596 a vA 20j insec vB 40j insec b vA 15j insec vB 75j insec 597 vA 0408 ms down 598 a ωA vC r CW ωOA μ r CCW c vA v vB 2v vC v all right vD vp 0 599 vA 904 insec vC 699 insec 5100 vA 0707 ms vP 1581 ms 5101 ωBD 12 rads CCW ωAD 1333 rads CCW 5102 vB 0884 ms ωB 320 rads CCW 5103 ωBC 277 radsec CCW 5104 ωAB ω CW vB 258ω down 5105 ωAB 0966ω CCW vB 1414τω 60 5106 vC 01386 ms ωC 0289 rads CW 5107 ω 15 rads CW vp 1897 ms 5108 v 231 ms ω 1333 rads CW 5109 v 1071 mihr vF 698 ftsec 5110 ωCA 0429k rads 5111 vC vB 2 sec2 θ 2 8 5112 ωAB 1414 radsec CCW ωBD 377 radsec CW 5113 ωAB 1938 radsec CW 5114 ω 450 nms t 747 rads CCW 5115 ωAD 125 rads CCW ωBD 75 rads CCW 5116 vA 0278 ms 5117 vA 595 ms 5118 ω 110 radsec CW 5119 ω 1073 rads CW 5120 a ωB 360 revmin b ωD0 600 revmin 5121 aL 958 ms2 aB 909 ms2 5122 a α 00833 radsec2 CCW b aC 0625 ftsec2 up c d 15 ft 5123 a aA 02 ms2 b aA 439 ms2 c aC 62 ms2 5124 a 5 ms2 5125 θ sin1F R v0 11 R r0 R2 r2 5126 v0 061 ms a0 18i ms2 5127 α 0286 rads2 CCW aA 0655 ms2 down 5128 aAB 002791 ms2 5129 aA 266 ms2 5130 θ 367 mihr 5131 aAB ω2 5132 ωAB 4k rads2 aA 161 ms2 5133 a B 246 ms2 left 5134 aC 0267i 3j ftsec2 aD 2i 0733j ftsec2 aL j ftsec2 5135 αAB 864ms2 CCW aG 682ms2 up 5136 α 2Lo7r CW 5137 aA 241 270j ftsec2 aD 2651 736j ftsec2 5138 ωOA 396 rads2 CCW 5139 α 800 rads2 CW apr 890 ms2 105 5140 a αB 239 ms2 ω 362 rads2 CW 5141 aΑ 833i 10j ms2 ap 833j ms2 aAB 1389i 333j ms2 5142 αAB 242 aBC 0 5143 aBC 208 radsec2 CCW ω2l2 5144 αOB lrl 5145 aOB 576 radsec2 CW 5146 ap 362 ms2 5147 aAB 1602 rads2 CW aBC 1331 rads2 CCW 5148 α 00986 rads2 CW 5149 aAB 0711j ftsec2 5150 aOA 0 ap 480i 360j ms2 5151 aAB 366 rads2 CCW aB 1984 ms2 345 5152 αAB 1688 rads2 CCW 5153 aA 489 ms2 right aAB 0467 rads2 CCW 5154 αBD 469 rads2 CW 5155 aD 0568 ms2 down 5156 aE 0285 ms2 right 5157 vA 011i 025j ms β 682 5158 acar 04j ms2 aA 035i 03j ms2 γ 1394 5159 vA 333i 45j ftsec aA 345i 1267j ftsec2 5160 vA 438i 158j ftsec aA 487i 382j ftsec2 5161 aA 1006 ftsec2 5162 vA 34i ms aA 2i 0667j ms2 5163 a αcφ 0 b αcar 00203 ms2 5164 ω 0j no 5165 αcar 2aωi 5166 a aAB 176i 070j ftsec2 b aA 1042i 570j ftsec2 5167 αAB 1042i 570e ftsec2 5168 vp 469k ms 5169 aPC 0634 rads CCW vmax 0483 ms 120 5170 s 01350 in 5171 vrad 271i 0259j ms arad 0864i 00642j ms2 5172 vrad 1136i 0537j ms arad 0854i 000918j ms2 5173 αrad ω2ω2 4ω2 5176 ω 4 rads CW a 640 rads2 CCW 5177 vrad 223i 657j ftsec arad 1606i 270j ftsec2 5178 vu1 393 ms and arad 1522 ms2 1911 ωPC 1429 rads CCW aPC 1700 rads2 CW 5179 vu1 771 ms and arad 1566 ms2 1911 ωPC 1046 rads CW aPC 1190 rads2 CW 5180 a vrad 5250i 5190j kmh b vrad 7380i kmh 5181 α 1565 ftsec2 5182 α 5 rads CCW arad 8660i mms2 5183 ω 1653 rads2 CCW 5184 vrad 26 220i kmh arad 802j ms2 5185 t 1Kk tan1 ω0Kk 5186 vp 427 ftsec 5187 α 075 rads2 CCW 5188 ω ω1 ωω tan θ α ω ωω tan θ1 ωω tan θ tan θ 5189 θ 60 5190 ωAB 1203 rads CCW 5191 aC 831 ftsec2 up 5192 vA 12491 1891k ftsec vB 176i 1089j ftsec 5193 ωAB 0354 radsec CW v0 788 insec 5194 aB 525 insec2 left 5195 ωBC 2 rads CW 5196 a aA 808 ftsec2 b aA 1744 ftsec2 0966dd 5197 ωOA 5198 vB 288 mms 5199 ωB cos θ D r cos θ u sin 2θ 5200 αDE 245 rads2 CCW 5201 ωIB 324 rads CW 5202 αB 412 rads2 CW 5203 Δvw1 5031 871j kmh 5204 α 625 rads2 CW 5205 αmax 2 ms2 at θ 1095 ω1max 2 ms at θ 251 5206 α 461 5207 ωABmax 654 rads at θ 202 5208 ωABmax 747 rads at θ 215 ωABmax 886 rads at θ 234 5209 ωBCmax 1122 rads at θ 1821 ωBCmax 1015 rads at θ 203 ωBCmax 1183 rads at θ 216 2 cos θ β 5210 α2 2 cos β cos θ β 5211 βθ 2 cos θ 15 4 cos θ 5212 vA rωsin θ 1 cos θlr2 sin2 θ vAmax 696 ftsec at θ 723 5213 θ 723 Chapter 6 61 α g3 62 α 3g 63 B 15 lb 64 P μk M m g cos θ 65 α 0268g 66 d μk h2 67 a and b P mg cb 68 P 3M mg 69 α 1306 ms2 right 610 α 0706 ms2 right 611 FA 1110 N Ox 45 N right Oy 667 N down 612 α 414 ms2 613 l 273 N Ax 1834 N right Ay 1557 N up 614 a NA 1280 lb 40 NB 1920 lb 60 615 a NB 1908 lb 596 NB 1292 lb 404 615 a NA 1920 lb 60 NB 1280 lb 40 a NB 2550 lb 796 NB 652 lb 204 616 A 1192 N 617 A 1389 N down 618 α 161 ftsec2 619 TA 1299 lb TB 390 lb α 805 radsec2 620 P 11871b θ 492 B 410 lb Bet 417 lb 621 l 341 s 623 M 1960 Nm CCW 623 P 783 N M 287 Nm 624 μ 0598 625 D 234 N α 587 rads2 CW 626 N 257 kN 627 a θ 513 b θ 248 a 59 g 628 a θ tan1 u2gr b Slips first if μ b2h and μ tan β u2 gr μ tan θ1 μ tan θ Tips first if μ b2h and tan β b2h u2 gr 2h1 b2h tan θ 629 a D 1714 N b D 2178 N 630 θ 0964 nose up 631 B 1883 N 632 A 202 kN 633 R 490 N 634 FA FB 245 N 635 α 1193 rads2 CCW FA 769 N 636 IO 1453 lbftsec2 637 α 912 radsec2 CW 638 a α g2r CW O mg2 b α 2g3r CW O mg3 639 R 357 lb 640 A 563 N 641 α 8g3πb CW Ox 32mg9π2 left Oy 1 329π2 mg up 642 β π2 α 8g3bπ CW β π α 8g3bπ CW 643 A A α 32g5b B B α 32g7b 644 α 15g47 π CW 645 α 0389 gb CW 646 M ωρ24π 12πr4 4lt 32 r2 rl ρ2 647 Ox 2mg sin θ Ot mω22 cos θ 648 a α 785 rads2 CCW b α 628 rads2 CCW 649 R 18 lb 650 a α 846 rads2 CCW b α 1116 rads2 CCW 651 Oτ 866 lb at all times 652 Mτ 01045 lbft Mrot 0836 lbft 653 M mr2α R 2π mr 2α2 ω4 654 x l2 3 73l CW 655 x l6 α 32 gb CW 656 α 67 gl 127 π 5 3 CW 657 t 786s 658 O 1mg cos2 θ 100 sin2 θ 659 α g2r CCW A 0593 mg 660 b 407 mm R 1678 N 661 A 221 N B 1103 N 662 FA 1083 N FB 1416 N 663 αB 255 rads2 CCW 664 T 987 N A 1007 kN 665 α 384 radsec2 CW t 349 sec 666 a F mla6 b A 106 mla c ω αβ 667 a MA 1098 Nm CCW b MB 551 Nm CCW 668 R 1013 N 669 a μk 01880 b l 531 670 a μk 0229 b θ 546 671 aB P2m 3i j 672 aA P10m 37i 9j 673 α 488 rads2 CW ω0 5j ms2 674 α 861 rads2 CCW α0 5j ms2 675 α 0310 rads2 aq 687 ms2 aq 274 ms2 676 α0 702 ms2 α 908 rads2 CCW 677 α 0775 678 α 1380 ftsec2 F 1714 lb 679 α 1331 ftsec2 F 0693 lb 680 α0 241 ms2 up incline α 938 rads2 CW 681 ω 12α0k0 682 Ax αr sin θ μz 0 Bx g2r sin θ μ 12 tan θ 683 F 3mg N 1316 mg 684 T 208 N 685 T 213 mg 686 N m g ρ2ω2R r 687 α 228 rads2 CW μk 0275 688 α 212 rads2 CW a 0425 ms2 right 689 α 02985 rads2 CCW a 1027 ms2 right F 1762 N 690 α 12bg3 CW TA 3mgb2 h2 up 691 BC 403 N tension 692 B 364 N 693 α0 373 ms2 down incline 694 M3 355 Nm CCW 695 αA 1145g down incline 696 α 5α7 CW ω 107 g1 cos θ α sin θ 697 αA 593 ms2 left 698 α 84α65L CCW 699 ω 297 radsec CCW 6100 αA 14109 g down incline 6101 s 1866 ft 6102 α Fk02 α cos θ g sin θ Ox mgsin θ α cos θ 1 F2k02 On m g cos θ α sin θ 1 2π2k02 2α F2k02 α 531 6103 v 1173 ms 6104 N mgb 6gb sin θL 13 sin2 θ CW 6105 A 347 lb 6106 a m 350M b α 000581 gb CCW μkmin 0589 6107 α 3g2b CW 6108 α 181 8 radsec2 CCW RA 1128 lb right RB 03591 lb 255 aA 650 ftsec2 down aB 569 ftsec2 down incline 6109 ω 24g7L CW ωA 314 gL 48g7L 6110 ω 7L48g 6111 v 297 ftsec 6112 ω 0839 gb CW 6113 v 301 ms 6114 θ 332 6115 O 9127 mg up 6116 A vA 2gx sin θ B vB gx sin θ 6117 h 545 mm 6118 ω 1319 rads 6119 N 3240 revmin 6120 k 926 Nm ω 242 rads CW 6121 l0 900 mm 6122 ω 331 rads 6123 M 284 lbin CW 6124 v 6gb sin θ2 6125 ω 459 rads 6126 vA 245 ms 6127 ΔE 0435 ftlb 6128 N 346 lb 716 Problem Answers 6129 a m 1196 kg b ω 436 rads CW 6130 ω 223 rads CW 6131 ω sqrt12kb3e4bc2b3e4bcos3θ2 6132 δ 211 motor shaft turns CW 6133 ω 236 rads 6134 k 240 lbin 6135 ω 951 rads 6136 a k 933 Nm b ω 1484 rads CW 6137 a ω 2 sqrt39π7 g85π7 r b ω sqrt39π7 g215π7 r 6138 P mg sin θ M0rω 6139 v 229 ms 6140 ω 311 radsec 6141 N 3720 revmin 6142 ω sqrt3g2b22 1 6143 u 802 ftsec 6144 υA sqrt3gl 6145 υy 1325 ms 6146 υ 595 ms P m2 m0 g cos θbmm0 CCW 6147 b ω 6148 α Mmb2cos2 θ 13 CW 6149 α 9g cos θ2b 6150 θ 708 6151 α 337 radsec2 CW 6152 α 2P5m tanθ2 g 6153 α M2mbsqrt1h2b2 g 6154 α M mgb cos θ a sin θMA CCW 6155 θ 643 6156 α P2 cos2 θ 1mb8 cos2 θ 1 CW 6157 α P2m g 6158 α 38 Pm 3g2 6159 N 1330 revmin 6160 α 1353 rads2 CCW 6161 α 1054 radsec2 CCW 6162 M 0 MB 1144 kNm CW 6163 α 273 rads2 CW 6164 a 0341 ms2 down incline 6165 a H0 0587 kgm2s CW b H0 0373 kgm2s CW 6166 a ω 1190 rads CW b ω 72 rads CCW 6167 ω 1093 rads 6168 υ 0379 ms up ω 560 rads CW 6169 H 266104 kgm2s 6170 ω 234 radsec CW 6171 t lωe 6172 ω 1202 radsec CW 6173 a ω2 1166 radsec CW b ω2 1220 radsec CW 6174 υx MMm υm υy mMm υm ω 12π mL m4M7m CCW 6175 t m12m1 m2 rωμk g 6176 t m12m1 m2 rωμs g 6177 ωx2 3mυ1MmL CW QΔτ M2Mm mυ1 right 6178 ωx2 7259 υ1b CW υ2x2 1859 υ1 right ωyx2 2759 υ1 down 6179 N2 204 revs 6180 ω2max 1718 υ1L CW x0291L 6181 μk 0204 υ 3 ms 6182 τ 0750 001719 N 6183 RA 272 lb RB 1877 lb both up 6184 υ 488 ms 6185 N2 259 revmin 6186 Ω2 2Il Id0 Iω2Il Id 6187 a ω2 657 rads b and c ω3 1757 rads 6188 t 2v0g7μk cos θ 2 sin θ t 5v0μk7μk 2 tan θ ω 5v0μk7μk 2 tan θ Problem Answers 717 6189 t 2rωg2 sin θ 7μk cos θ υ 2rωsin θ μk cos θ2 sin θ 7μk cos θ ω 2rωsin θ μk cos θ2 sin θ 7μk cos θ 6190 N 478 revsec 6191 a ωx1 MtIl I Im ωwf IlIm MtIlI Il 6192 α 0974 radsec 6193 I 345 kgm2 6194 N 638 revmin 6195 ω 1096 revmin ΔE 1298 J 6196 v sqrtgL4 sin2 θ 3gL cos θ 6197 ω 0308 rads CCW 6198 υ u3 1 2 cos θ η 00202 6199 υ rsqrtl2 r2 sqrt2ghk2 r2 6200 Ω 1135 rads 6201 hmin 12 b mgP c μb hmax 12 b mgP c μb 6202 α 0604 rads2 CCW 6203 N 504 revmin ΔE 981 J 6204 a 222 ms2 right 6205 H0 υr m2 m0 2 r212r2 CW 6206 t sqrtL3g 0θ dθsqrtcos θ cos β Chapter 7 71 Finite rotations cannot be added as proper vectors 72 Infinitesimal rotations add as proper vectors 73 υx 6 ftsec R 281 in aP 11080 insec2 74 aB 1285 ms2 75 υA 08i 15j 2k ms υB 262 ms 77 N2 440 revmin 78 υ ωl cos θ i d cos θ l sin θ j l sin θ k a ω2h sin θ cos θ d cos2 θ i lj h sin2 θ d cos θ sin θ k 6207 ω 3b sqrt2gha 6208 μk 23 tan θ 6209 ω 415 rads CCW 6210 a Fo 688 lb b Fo 766 lb 6211 M 2mrω1 6212 B 2130 lb 6213 t 1206 s 6214 a α 494 rads CW b α 625 rads CW 6215 β 1926 6216 b L1 rL 6217 r rω0sqrtl0 mω2 6218 Mx 460 lbin My 1611 lbin Mz 1841 lbin 6219 x Iβ 6220 ω 250 rads CW 6221 ωmax 235 rads CW at θ 1888 6222 υ1 1558 l cos 45 a ftsec 6223 υ2max 757 ftsec at θ 482 6224 θ 1217 6225 ωmax 230 ωmax 0389 rads 6226 Ox 3mg4 sin θ 3 cos θ 2 Oy mg4 3 cos θ 12 6227 ωmax 399 ω 450 rads CW 6228 ωmax 0680 rads at θ 224 ωmax 459 6229 ωmin 0910 rads at θ 740 ω 1586 rads at θ 90 6230 t 285 sec υA 757 ftsec 79 ω pj ωyk α pωy i 710 α 12i radsec2 ap 358j 80k insec2 711 α 12π2 j radsec2 υ 5π4i 6j 3k insec a 5π225j 18k insec2 712 α 50ππ2 j k radsec2 713 a ω 265 rads b ω 1732 rads 714 ω 04i 269k rads α 08j rads2 715 α 15i 08j 26k rads2 716 25 rads α 3j rads2 717 ω 0693j 240k rads α 1386i rads2