Slide - Cinética Plana de Corpos Rígido - 2024-1

Parte 6 - Cinética Plana de Corpos Rígidos Equações do movimento Σ F = mā Soma das forças que atuam sobre o corpo rígido (Diagrama de Corpo Rído) = Força Resultante (Diagrama de Forças Resultantes) m: massa do corpo rígido ā: aceleração do centro de massa G do corpo rígido Σ M_G = I̅α Soma dos momentos das forças com relação ao eixo que passa pelo centro de massa do corpo rígido (Diagrama de Corpo Rígido) = Momento resultante em torno do eixo que passa pelo centro de massa do CR (Diagrama de Forças (Momentos) Resultantes) SAMPLE PROBLEM 3/18 The system shown is released from rest with the lightweight slender bar OA in the vertical position shown. The torsional spring at O is undeflected in the initial position and exerts a restoring moment of magnitude kθ0 on the bar, where θ 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 kθ0 = 13 N·m/rad: (a) Determine the speed vA of particle A when θ reaches 90°. (b) Plot vA as a function of θ over the range 0 ≤ θ ≤ 90°. Identify the maximum value of vA and the value of θ 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 vA and ωB when θ = 90°. Noting V9 - V1 = that the speed of point C is always v9/u, and further noting that the speed of 0 L cylinder B is one-half the speed of point C at θ = 90°, we conclude that at θ = 90°, v B = 0.797 m/s V1 - = 0+0+0 = Establishing datums at the initial altitudes of bodies A and B, and with state 1 at S0 (!!!) 10 L θ = 0 and state 2 at θ = 90°, we write T1 + V1 + U1/2 = T2 + V2 (-) } 15 {1) = 0. = -. J kθ02 Solution (c). .SetRA [7: 0+0 = 0+0+...2gyc = Finally, because: , - Cme. (-) {15.) 44 0+0 = 1 (0.6(092) t = Upon substitution of the given quantities, we vary 0 to produce the plot of vA versus 8. The maximum value of vA is seen to be The absolute-value signs reflect the fact that vA is known to be positive. Lorem ipsum degree. Ass. Helpful Hints 11 3/143 Point P1 on the 2-kg cylinder has an initial velocity v, = 0.8 m/s as it passes position A. Neglect the mass of the pulleys and cable and determine the - = 0. distance y of point P below A when the 3-kg cylin- der has acquired an upward velocity of 0.6 m/s. Problem 3/155 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 v 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 negli- gible. The sliders have equal masses, and the mo- tion is restricted to y ≥ 0. Problem 3/166 3/167 The mechanism is released from rest with θ = 180°, where the uncompressed spring of stiffness k = 900 N/m is just touching the underside of the 4-kg collar. Determine the angle θ corresponding to the maximum compression of the spring. Mo- tion is in the vertical plane, and the mass of the links may be neglected. Problem 3/167 SAMPLE PROBLEM 6/6 The drum A is given a constant angular acceleration α of 3 rad/s^2 and causes the 70-kg spool B to roll on the horizontal surface by means of the connecting cable, which wraps around the inner hub of the spool. The radius of gyration k of the spool about its mass center G is 250 mm, the coefficient of static friction between the spool and the horizontal surface is 0.35. Determine the tension T in the cable and the friction force F exerted by the horizontal surface on the spool. Solution. The free-body diagram and the kinetic diagram of the spool are drawn as shown. The correct direction of the friction force F may be assigned in this problem by observing from both diagrams that with counterclockwise angular acceleration, a moment sum about point G (and also about point D) must be counterclockwise. A point on the connecting cable has an acceleration ar = rα = (0.25)(3) = 0.75 m/s^2, which is also the horizontal component of the acceleration of point D on the spool. It will be assumed initially that the spool rolls without slipping, in which case it has a counterclockwise angular acceleration α = (ar)/(0.45) = 0.75/0.30 = 2.5 rad/s^2. The acceleration of the mass center G is, therefore, ar = (rα)(5/2) = 1.125 m/s^2. With the kinematics determined, we now apply the three equations of motion: [ΣFx = max] F − T = 70(−1.125) (a) [ΣFy = may] N − 70(9.81) = 0 N = 687 N (b) [ΣMG = Iα] (0.750) − T(0.150) = 700(250)(2.5) (c) Solving (a) and (b) simultaneously gives F = 75.8 N t = 154.6 N Ans. To establish the validity of our assumption of no slipping, we see that the surfaces are capable of supporting a maximum friction force F(max) = μsN = 0.636807 = 171.7 N. Since only 75.8 N of friction force is required, we conclude that our assumption of rolling without slipping is valid. If the coefficient of static friction had been 0.1, for example, then the friction force would have been limited to (0.1687) = 68.7 N, which is less than 75.8 N, and the spool would slip. In this event, the alternate relation ar = rα would no longer hold. With (ar) known, the angular acceleration would be α = (ar)/(r/yD). Using this relation along with F = μsN = 68.7 N, we would then resolve the three equations of motion for the unknowns T, F, and α. Alternatively, with point C as a moment center in the case of pure rolling, we may see Eq. 6/2 and obtain T directly. Thus, [ΣMC = α + mra] 0.37 (700)(0.25)(2.5) + 701.125)(0.45) T = 154.6 N Ans. Helpful Hints The relation between ar and a is the kinematic constraint which accompanies the assumption that the spool rolls without slipping. Be careful not to make the mistake of using mr^2 for I of the spool, which is not a uniform circular disk. Our principles of relative acceleration are a necessity here. Hence, the relation (αGH)m = GDo should be recognized. SAMPLE PROBLEM 6/3 The concrete block weighing 644 lb is elevated by the hoisting mechanism shown, where the cables are securely wrapped around the respective drums. The drums, which are fastened together and turn as a single unit about their mass center at O, have a combined weight of 322 lb and a radius of gyration about O of 18 in. If a constant tension P of 400 lb is maintained by the power unit at A, determine the vertical acceleration of the block and the resultant force on the bearing at O. Solution I. The free-body and kinetic diagrams of the drums and concrete block are drawn showing all forces which act, including the components Ox and Oy of the bearing reaction. The resultant of the force system on the drums for centripetal rotation is the couple I α = I Oα, where [I = km^2] I = IO = 18 ( ) 32.2 = 22.5 lb-ft-sec^2 Taking moments about the mass center O for the pulley in the sense of the angular acceleration α gives [ΣMG = Iα] 400 ( )240̶̈ − T( )12 = 22.5α (a) The acceleration of the block is described by [ΣFy = may] T − 644 644 = ma (b) From α, rα we have a = (12)(12)α. With this substitution, Eqs. (a) and (b) are combined to give T = 71 lb α = 3.67 rad/sec^2 α = 3.67 ft/sec^2 Ans. The bearing reaction is computed from its components. Since a = 0, we use the equilibrium equations [ΣFx = 0] Ox − 400 cos 45° = 0 Ox = 283 lb (b) (−ΣFy = 0] Oy − 322 − 44410 sin 45° = 0 Oy = 1322 lb O = O (283) + (1322) lb = 1352 lb Ans. Solution II. We may use a more condensed approach by drawing the free-body diagram of the entire system, thus eliminating reference to T, which becomes internal to the new system. From the kinetic diagram for this system, we see that the moment sum about O must equal the resultant couple Iα for the drums, plus the moment of the resultant ma for the block. Thus, from the principle of Eq. 6/5 we have [ΣMo = Iα + m αt] 0.400T + 0.22 − 644 12 400 (12)α 144 212 = 22.5α + 644 (12) With a (=12)(12)α, the solution gives, as before, α = 3.67 ft/sec^2 Thus, We may equate the force sums on the entire system to the sums of the resultants. Thus, [ΣFx = Σma]/ Ox − 322 − 644 400 sin 45° = 0 Oy = 1322 lb [ΣFy = Σma] O ay − 400 cos 45° = 0 O Ox = 283 lb O | o Helpful Hints Be alert to the fact that the tension T is not 644 lb. If it were, the block would not accelerate. Do not overlook the need to express k0 in feet when using it in ft/sec^2. SAMPLE PROBLEM 6/4 The pendulum has a mass of 7.5 kg with center of mass at G and has a radius of gyration about the pivot O of 250 mm. If the pendulum is released from rest at 0 = 0, determine the total force supported by the bearing at the instant when 0 = 60°. Friction in the bearing is negligible. Solution. The free-body diagram of the pendulum in a general position is shown along with the corresponding kinetic diagram, where the components of the resultant force have been drawn through G. The normal component 0, is found from a force equation in the l-direction, which involves the normal acceleration ra?. Since the angular velocity w of the pendulum is found from the integral of the angular acceleration and since O, depends on the tangential acceleration ra, it follows that a should be obtained first. To this end with lo = kg’m, the moment equation about O gives 12MO - lca = 0] 7.5(9.81)(0.25) cos 0 = (0.25)(7.5)a = 28.2 cos 0 rad/s? and for 237062 60° Io doa = a a J0a do = fo a a do = Jo 2 28.2 cos 0 d0 w? = w| = 4.88 (rad/s)? The remaining two equations of motion applied to the 60° position yield Jo - 7.5(0.8) sin 60° = 7.5(0.25)(4.88) O, = 15.52 N -7 _[2a , - O, + 7.5(9.81) cos 60° = 7.5(0.25)(28.2) cos 60° OQ= 10.37 N {-(155.2) + (10.37) = 15.56 N Ans. The proper sense for 0, may be observed at the outset by applying the moment equation ZMo = la to agree with a. The force OQ about is clockwise the center of percussion Q,, shown in the lower figure, which avoids the necessity of computing a. First, we must obtain the distance q, which is q= kg/f’ q= (0.25) 9Oy 0.348 m 0.250 ZMQ =0 Oj(0.34) 7.5(9.81)cos 60470,348204.50) = 0 Q = 10.37 N Ans. O 0 0 e Review the theory again and satisfy yourself that ZMO = lq, + Ira + me?a, = mra?e. 2 Note especially here that the force summations are taken in the positive direction of the acceleration components of the mass center G. Problem 6/44 6/44 If the system is released from rest while in the horizontal position shown, determine the angular acceleration of the lightweight right-angle shaft. The sphere of radius r has mass m. Neglect friction at the bearing 0. r 6/67 The robotic device consists of the stationary pedestal OA, arm AB pivoted at A, and arm BC pivoted at B. The rotation axes are normal to the plane of the figure. Estimate (a) the moment M, applied to arm AB required to rotate it about joint A at 4 rad/s2 counterclockwise from the position shown with joint B locked and (b) the moment M applied to arm BC required to rotate it about joint B at the same rate with joint A locked. The mass of arm AB is 25 kg and that of BC is 4 kg, with the stationary portion of joint A excluded entirely and the mass of joint B divided equally between the two arms. Assume that the centers of mass G and G2 are in the geometric centers of the arms and model the arms as slender rods. 700 1% G, 380 B 90° с G, 45° A R 6/67 6/94 The robotic device of Prob. 6/67 is repeated here. Member AB is rotating about joint A with a counterclockwise angular velocity of 2 rad/s, and this rate is increasing at 4 rad/s2. Determine the moment My exerted by arm AB on arm BC if joint B is held in a locked condition. The mass of arm BC is 4 kg, and the arm may be treated as a uniform slender rod. 700 1 G1 380 B 90° с 45 6/94 6/105 The connecting rod AB of a certain internal-combustion engine weighs 1.2 lb with mass center at G and has a radius of gyration about G of 1.12 in. The piston and piston pin A together weigh 1.80 lb. The engine is running at a constant speed of 3000 rev/min, so that the angular velocity of the crank is 3000(2x/60 = 100n rad/sec. Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle 0 = 90°. (Suggestion: Use the alternative moment relation, Eq. 6/3, with B as the moment center.) Problem 6/105 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS (m = mass of body shown) BODY Circular Cylindrical Shell Half Cylindrical Shell Circular Cylinder Semicylinder Rectangular Parallelepiped MASS CENTER — x̄ = 2r/π — x̄ = 4r/3π — MASS MOMENTS OF INERTIA I_xx = 1/2mr^2 + 1/12ml^2 I_xx = I_yy = 1/2mr^2 + 1/12ml^2 I_xx = 1/2mr^2 + 1/12ml^2 I_xx = I_yy = 1/2mr^2 + 1/12ml^2 I_xx = 1/12m(b^2 + l^2) I_xx, z and I_yy, x = 3/10mr^2 I_xx, y = 1/3mb^2 + 1/3ml^2 I_xx = I_yy = 1/2ml^2 I_yz = 3/10mr^2 + 1/3ml^2 I_xx, y1, z and I_yy, x = 1/12m(b^2 + l^2) I_xx = 1/2mr^2 + 1/12ml^2 I_xx, y = I_yy, z = 1/3mb^2 + 1/3ml^2 I_xx = 1/12m(b^2 + l^2) I_xx = I_yy = 1/2mr^2 + 1/12ml^2 I_xx, y = I_yy, z = 1/3mb^2 + 1/3ml^2 I = mr^2 I_zx = (1/2)(1 - 4/9π^2)mr^2 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m = mass of body shown) BODY Spherical Shell Hemispherical Shell Sphere Hemisphere Uniform Slender Rod MASS CENTER — x̄ = r/2 — x̄ = 3r/8 — MASS MOMENTS OF INERTIA I_zz = 2/3mr^2 I_xx = I_yy = I_zz = 2/3mr^2 I_yz = I_zx = 5/12mr^2 I_zz = 2/5mr^2 I_xx = I_yy = I_zz = 3/5mr^2 I_yz = I_zx = 83/320mr^2 I_yy = 1/12ml^2 I_xyz = 1/3ml^2 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m = mass of body shown) BODY Quarter-Circular Rod Elliptical Cylinder Conical Shell Half Conical Shell Right-Circular Cone MASS CENTER x̄ = ȳ — x̄ = 2h/3 x̄ = 4r/3π z̄ = 2h/3 x̄ = 3h/4 MASS MOMENTS OF INERTIA I_xx = I_yy = 1/2mr^2 I_zz = mr^2 I_xx = 1/12ma^2 + 1/12ml^2 I_yy = 1/12mb^2 + 1/12ml^2 I_zz = 1/12m(a^2 + b^2) I_yy, xz = 1/12mb^2 + 1/3ml^2 I_yy = 1/2mr^2 + 1/2mh^2 I_xyz = 1/3ml^2 + 1/2mh^2 I_zz = 1/3mr^2 I_yy = 1/2mr^2 I_yx, wx = I_yxh, y = 1/5mr^2 + 1/2mh^2 I_zz = mr^2 I_zx = (1/2)(1 - 16/9π^2)mr^2 I_yy = 3/20mr^2 + 1/3mh^2 I_yy, yz = 35/120mr^2 + 1/10mh^2 I_zz = 3/10mr^2 I_yy = 3/20mr^2 + 3/80mh^2 TABLE D/4 PROPERTIES OF HOMOGENEOUS SOLIDS Continued (m = mass of body shown) BODY Half Cone Semellipsoid Elliptic Paraboloid Rectangular Tetrahedron Half Torus MASS CENTER x̄ = r/7 z̄ = 3h/4 z̄ = 3c/8 z̄ = 2c/3 x̄ = a/4 ȳ = b/4 z̄ = c/4 x̄ = a^2 + 4R^2/2πR MASS MOMENTS OF INERTIA I_xx = I_yy = 3/20mr^2 + 3/8mh^2 I_zz = 3/10mr^2 Ī_zz = (3/10 - 1/π^2)mr^2 I_xx = 1/5m(b^2 + c^2) I_yy = 1/5m(a^2 + c^2) I_zz = 1/5m(a^2 + b^2) Ī_xx = 1/5m(a^2 + 19/64c^2) Ī_yy = 1/5m(a^2 + 19/64c^2) I_xx = 1/5mb^2 + 1/2mc^2 I_yy = 1/5ma^2 + 1/2mc^2 I_zz = 1/6m(a^2 + b^2) Ī_xx = 1/6m(a^2 + 3/2c^2) Ī_yy = 1/6m(a^2 + 3/2c^2) I_xx = 1/10m(b^2 + c^2) I_yy = 1/10m(a^2 + c^2) I_zz = 1/10m(a^2 + b^2) Ī_xx = 3/80m(a^2 + c^2) Ī_yy = 3/80m(a^2 + b^2) Ī_zz = 3/80m(a^2 + b^2) I_xx = I_yy = 1/2mR^2 + 5/9ma^2 I_zz = mR^2 + 3/4ma^2 14 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 d𝐔 exercido pela força F sobre a partícula que está posicionada a r da origem O de um referencial fixo. d𝐔 = 𝐅 ∙ d𝐫 dr: tangente à trajetória ‘.’ denotaProduto escalar dU = Ft ds Ft: componente tangencial de F ds: módulo de d𝐫 onde, F = F_xi + F_yj + F_zk r = xi + yj + zk dr = dxi + dyj + dzk U = ∫_1^1 F . dr = ∫_1^2 (F_xdx + F_ydy + F_zdz) Da Figura 3/3, U = ∫_s1^s2 F_t ds Cálculo do trabalho - Alguns exemplos muito usuais: Trabalho de uma força sobre o corpo, trabalho de força elástica e trabalho da força peso Exemplo 1 - Trabalho da força P exercido sobre o corpo. P e α são constantes. O trabalho da força P sobre o corpo é, U_1-2 = ∫_1^2 F . dr = ∫_1^2 [(Pcos(α))i + (Psen(α))j] . dxi = ∫_x1^x2 Pcos(α) dx = Pcos(α)(x_2 - x_1) = PLcos(α) Exemplo 2 - Trabalho exercido pela força elástica (da mola) sobre a partícula. Mola indeformada F = kx F = - kx i k: coeficiente de elasticidade da mola U_{1-2} = \int_{1}^{2} F. dr = \int_{1}^{2} (- kxi). (dxi) = - \int_{x_{1}}^{x_{2}} kx dx = \frac{1}{2}k(x_{1}^{2} - x_{2}^{2}) onde x_{1} e x_{2} são os comprimentos da mola a partir de sua posição indeformada - vide Figura acima. Exemplo 3 - Trabalho exercido pela força peso sobre uma partícula de massa m. dr = dxi + dyj U_{1-2} = \int_{1}^{2} F. dr = \int_{1}^{2} (- mg j). (dxi + dy j) = - mg \int_{y_{1}}^{y_{2}} dy = - mg(y_{2} - y_{1}) onde y_{1} e y_{2} são posições do bloco nos estados 1 e 2, respectivamente, a partir de uma referência comum - vide Figura acima. Trabalho e a Energia Cinética Para o Movimento Curvilíneo U_{1-2} = \int_{1}^{2} F. dr = \int_{s_{1}}^{s_{2}} F_{t} ds F = ma U_{1-2} = \int_{1}^{2} F. dr = \int_{1}^{2} ma. dr a. dr = a_{t} ds Regra da cadeia: a_{t} = \frac{dv}{ds} \frac{ds}{dt} = \frac{dv}{ds} v \quad ou \quad a_{t} ds = v dv Portanto, U_{1-2} = \int_{1}^{2} F. dr = \int_{v_{1}}^{v_{2}} mv dv = \frac{1}{2} m(v_{2}^{2} - v_{1}^{2}) ou U_{1-2} = T_{2} - T_{1} = \Delta T 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 igual a variação da energia cinética dessa partícula para essas posições. 19 Potência P=F.v 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 3/11 Calculate the velocity v of the 50-kg crate when it reaches the bottom of the chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coefficient of kinetic friction is 0.30. Solution. The free-body diagram of the crate is drawn and includes the normal force R and the kinetic friction force F calculated in the usual manner. The work done by the weight is positive, whereas that done by the friction force is negative. The total work done on the crate during the motion is [U = F_{R}] \quad U_{1-2} = 50(9.81)(10 sin 15°) - 142.1(10) = -151.9 J The work-energy equation gives [T_{1} + U_{1-2} = T_{2}] \quad \frac{1}{2} m(v_{1}^{2}) + U_{1-2} = \frac{1}{2} m(v_{2}^{2}) \frac{2}{2} (50)(4)^{2} - 151.9 = \frac{1}{2} (50)(v_{2}^{2}) v_{2} = 3.15 m/s Ans. Since the net work done is negative, we obtain a decrease in the kinetic energy. SAMPLE PROBLEM 3/12 The flatbed truck, which carries an 80-kg crate, starts from rest and attains a speed of 72 km/h in a distance of 75 m on a level road with constant acceleration. Calculate the work done by the friction force acting on the crate during this interval if the static and kinetic coefficient of friction between the crate and the truck bed are (a) 0.30 and 0.28, respectively, or (b) 0.25 and 0.20, respectively. Solution. If the crate does not slip on the bed, its acceleration will be that of the truck, which is [v^{2} = 2as] the work-energy equation gives [v_{2}^{2} = 2as] a = \frac{(72/3.6)^{2}}{2(75)} = 2.67 m/s^{2} Case (a). This acceleration requires a friction force on the block of [F = 80(2.67) = 213 N which is less than the maximum possible value of \mu_s N_0 = 0.30(80)(9.81) = 235 N Therefore, the crate does not slip and the work done by the actual static friction force of 213 N is [U = F_8] U_{1-2} = 213(75) = 16 000 J \quad or \quad 16 kJ Ans. Case (b). For \mu_s = 0.25, the maximum possible friction force is 0.25(80)(9.81) = 1962 N, which is slightly less than the value of 213 N required for no slipping. Therefore, we conclude that the crate slips, and the friction force is governed by the kinetic coefficient and is F = 0.2080(9.81) = 1570 N. The acceleration becomes [F = ma] a = -F/m = 157.0/80 = 1.962 m/s^{2} The distances traveled by the crate and the truck are in proportion to their accelerations. Thus, the crate has a displacement of (1.962/2.67)(75) = 55.2 m, and the work done by kinetic friction is [U = F_s] U_{1-2} = 157.05(55.2) = 8660 J \quad or \quad 8.66 kJ Ans. 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 x1 = 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 system 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 x1 = 0.233 m to x2 = 0.233 + 1.2 = 1.433 m, the work done by the spring force acting on the block is Ua2=12k(x12−x22) Ua2=12180(0.2332−(0.233+1.2)2)=−800JUa2=12k(x12−x22) Ua2=12180(0.2332−(0.233+1.2)2)=−800J 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 (1.2)2 + (0.9)2 = 0.96 = 1.0 m. Thus, the work done is 300(0.96) = 180 J. We now apply the work–energy equation to the system and get T1 + Ua2 = T2≡ 0 − 800 + 180 = 12(50)v2v=2.00m/sT1 + Ua2 = T2≡ 0 − 800 + 180 = 12(50)v2v=2.00m/s 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 tension 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 solution 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 required 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 x1 and x2, 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 kinetic 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, ΣF = 0⇒ T − 693μk − 800 sin 30° = 0 T = 693μk + 400ΣF = 0⇒ T − 693μk − 800 sin 30° = 0 T = 693μk + 400 The power output of the winch gives the tension in the cable P = Tv⇒ T = Pb = 6560/4 = 825 lb Substituting T gives 825 = 693μk + 400 μk = 0.613 When the power is increased, the tension momentarily becomes P = Tv⇒ T = Pb = 8(550)/4 = 1100 lb and the corresponding acceleration is given by ΣFz = ma⇒ 1100 = 693(0.613) − 800 sin 30° + 80032.2 a= 11.07ft/sec2ΣFz = ma⇒ 1100 = 693(0.613) − 800 sin 30° + 80032.2 a= 11.07ft/sec2 Helpful Hints Note the conversion from horsepower to ft·lb/sec. As the speed increases, the acceleration will drop until the speed stabilizes 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 deformation. Specify the required stiffness k for each of the two springs behind the bumper. The springs are undeformed at the start of impact. Problem 3/101 3/134 The spring attached to the 10-kg mass is nonlinear, having the force-deflection relationship shown in the figure, and is unstretched when x = 0. If the mass is moved to the position x = 100 mm and released from rest, determine its velocity v when x = 0. Determine the corresponding velocity v' if the spring were linear according to F = 4x, where x is in meters and the force F is in kilonewtons. Force F, kN Linear, F = 4x Nonlinear, F = 4x - 120x^3 Deflection x, m 10 kg μs = 0.25 μk = 0.20 Problem 3/134 25 Potencial Gravitacional 𝑉𝑔 = 𝑚𝑔ℎ Vg é o potencial gravitacional ou energia potencial gravitacional da partícula de massa m. 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 = ∫0^x kx dx = 1/2 kx^2 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, ΔVe = 1/2 k(x2^2 - x1^2) que representa o trabalho realizado de uma força externa sobre a mola. Portanto, o trabalho realizado da mola sobre a partícula é igual a -ΔVe, onde, x1 e x2 são os comprimentos da mola a partir de sua posição indeformada. 27 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, 28 −∆𝑉𝑔: trabalho realizada pela força peso para elevar a partícula de massa m de uma elevação h, −∆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 stiffness of 12 lb/ft and 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, UR12 = 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( ) = -12 ft-lb The initial and final elastic (spring) potential energies are V1 = 1/2 k(s1^2) = 1/2(12)(24/12)^2 = 48 ft-lb V2 = 1/2 k(s2^2) = 1/2(12)( )^2 = 8.24 ft-lb Substitution into the alternative work-energy equation yields [T1 + V1 + UR12 = T2 + V2] 0 + 48 + 0 = 1/2 ( )(v2^2) - 12 + 8.24 v2 = 23.6 ft/sec Ans. 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 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. 3/2a. The only nonpotential 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. UAC = 2500(0.9) = 150 J We define a datum at position A so that the initial and final gravitational potential 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)( ) + 1.2^2 = 97.2 J Substitution into the alternative work-energy equation 3/2a gives [T1 + VA + UAC = TC + VC] 0 + 0 + 10.8 + 150 = 1/2(10)(vC^2) + 58.9 + 97.2 vC = 0.974 m/s Ans. Helpful Hints • 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 depends only on the direction of the motion and does not require knowledge of the shape of the path. • Do not hesitate to use subscripts tailored to the problem at hand. Here we use A and C rather than 1 and 2. • The reactions of the guide on the slider are normal to the direction of motion and do no work. 33 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. 34 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. Figura 3/13 Impulso: área sobre a curva m(v_1)_x + ∫_{t_1}^{t_2} ΣF_x dt = m(v_2)_x m(v_1)_y + ∫_{t_1}^{t_2} ΣF_y dt = m(v_2)_y m(v_1)_z + ∫_{t_1}^{t_2} ΣF_z dt = m(v_2)_z Figura 3/12 36 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 3/19 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 v_1 = 50 ft/sec, and just after impact its velocity is v_2 = 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: 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 ∫_{t_1}^{t_2} R_x dt, the average impact force R_x is constant, so that it can be brought outside the integral sign, resulting in R_x ∫_{t_1}^{t_2} 1 dt = R_x Δt. The linear impulse in the y-direction has been similarly treated. SAMPLE PROBLEM 3/20 A 2-lb particle moves in the vertical y-z plane (y 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 = 3/2 t^2 i^ + 3j - t^3 k^ - 4tk^, 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} Don't forget that ΣF includes all ex- ternal forces acting on the particle, including the weight. 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 b and moving at 3 mi/hr overtakes car A and is coupled to it. Determine (a) the common velocity 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 u of the block when t = 0.4 sec. The coefficient of kinetic friction is μk = 0.30. v_0 = 2 ft/sec 20 lb μk = 0.30 0.2 0.4 t, sec t, sec t, sec 0.2 0.4 t, sec 16 t, sec 16 8 8 P, lb P lb lb 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. μs = 0.60, μk = 0.40 4 P, N t, s t, s t, s 10 kg P P 100 Problem 3/205 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 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 at the skip at t = 8 s. Treat the skip as a particle. Solution. The stated variation of P with the 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 s. The impulse-momentum equation applied consistently in the positive-x direction gives m1v1 |_t=4 + \int_0^t F_x dt = m2v2 |_t=s 150(-4) + \int_0^4 (2)(600 + 2600(t - 150)(9.81) sin 30 dt = 150(0) -2.46 = f^4 + 2.46 = 6.46 s -2.46 = t^4 + .76 m/s Ans. Part (b). Applying the momentum equation to the entire 8-s interval gives m1v1 |_t=s + \int_4^s 0 dt = m2v2 |_t=s 150(-4) + \int_4^8(2)(600 + 1609 -1509.81) sin 30(8) = 150(0) v2 + 47.6 m/s The same result is obtained by analyzing the interval from 0 to 8 s. 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 thetriangular area for the P-variation for the first 4 s, doubled for the force of 2P. 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| 0.060(600) + 4(12)cos30 + sin30 = (4 + 0.050)v2 v1 + 10.261 + 13.893 m/s The final velocity and its direction are given by [v2 = \sqrt{ vx^2 + vy^2} vg = (10.261^2) + (13.893) = 16.83 m/s [tan θ = v_y/v_x] [tan θ = v_y/v_x] = 1.199 = 52.4° Ans. The final velocity and its direction are given by v2 = \sqrt{ (10.261 ufgt)^2 + (13.839)^2} = 16.83 m/s [v \sqrt{ (10.261)^2 + .4603 + 600 m/s } = 673] v = 16.83 m/s θ = -- θ2 = 52.4° Ans. Helpful Hint Working with the vector form of the principle of conservation of linear momentum is clearly equivalent to working with the component form. v 41 3.7 Impulso Angular e Quantidade de Movimento Angular HO: quantidade de movimento angular da partícula de massa m com relação ao ponto O. HO é o momento da quantidade de movimento linear mv. Ele é perpendicular a G = m v e a r. 42 Taxa de Variação Temporal da Quqntidade de Movimento Angular 43 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 P. Determine the angular momentum HO about point O and the time derivative ḢO. Solution. We begin with the definition of angular momentum and write ḢO = r x mv = (3i + 6j + 4k) x 2(5j) = -40i + 30k N∙m/s = ... From Eq. 3/31, ḢO = MO = r x F = (3i + 6j + 4k) x 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 mUAvA rA = mUBvB vB = rA/ vA vB = (6000)(10^4)/740 rB 75(10^6) vB = 59 200 m/s Ans. 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. Problem 3/226 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 v = 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. Problem 3/235 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 \( \theta_{z} \) of the assembly just after impact. The pivot at 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. \[(H_O)_1 = (H_O)_2\] \[mvl = (m + 2m)(l)\theta_2 l + 4m(2l)2\] \[\theta_{z} = \frac{v_1}{12l} \mathrm{CW} \quad \text{Ans.}\] 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 \(\theta_{z}\). SAMPLE PROBLEM 3/27 A small mass particle is given an initial velocity v0 tangent to the horizontal rim of a smooth hemispherical bowl at a radius r0 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, \[(I_O \omega)_1 = (I_O \omega)_2 \quad mv_{0}r_0 = mvr \cos \theta\] Also, energy is conserved so that E1 = E2. Thus \[T_1 + V_1 = T_2 + V_2\] \[\frac{1}{2} mv_{0}^{2} + mgh = \frac{1}{2} mv_{0}^{2} + 0\] \[v = \sqrt{v_{0}^{2} + 2gh}\] Eliminating v and substituting \(r^{2} = r_{0}^{2} - h^2\) give \[v_{0}r_{0} = \sqrt{v_{0}^{2} + 2gh}(r_{0} - \frac{h^2}{r_{0}}) \cos \theta\] \[\theta = \cos^{-1}\]\[ 1\left/\sqrt{1 + \frac{2gh}{v_{0}^{2}} \left(\frac{1 - \frac{h^2}{r_{0}^2}}{1}\right)}\right.\right]\quad \text{Ans.}\] Helpful Hint 1 The angle θ is measured in the plane tangent to the hemispherical surface at B.