F = 30 lb. 60 in. 60 in.

Problem 18.1 horizontal force F = 30 lb is applied to the 30-lb refrigerator as shown. Friction is negligible. (a) (b) What is the magnitude of the refrigerator s acceleration? What normal forces are eerted on the refrigerator b the floor at and B? F 60 in 8 in 14 in 14 in ssume that the refrigerator rolls without tipping. We have the following equations of motion. ( ) 30 lb F : (30 lb) = 3. ft/s a F : + B 30 lb = 0 F = 30 lb 60 in. 30 lb MG : (30 lb)(3 in.) (14 in.) + B(14 in.) = 0 Solving we find (a) a = 4. ft/s 8 in. (b) = 80.7 lb, B= 149.3 lb B Since >0 and B>0 then our assumption is correct. Problem 18. Solve Problem 18.1 if the coefficient of kinetic friction at and B is µ k = 0.1. ssume sliding without tipping ( ) 30 lb F : (30 lb) (0.1)( + B) = 3. ft/s a F : + B 30 lb = 0 MG : (30 lb)(3 in.) (14 in.) + B(14 in.) F = 30 lb 60 in. 30 lb (0.1)( + B)(8 in.) = 0 8 in. Solving, we find µ µb (a) a = 0.98 ft/s N B (b) = 57.7 lb, B= 17 lb 469

Problem 18.3 s the 800-lb airplane begins its takeoff run at t = 0, its propeller eerts a horizontal force T = 1000 lb. Neglect horizontal forces eerted on the wheels b the runwa. T (a) What distance has the airplane moved at t = s? (b) what normal forces are eerted on the tires at and B? 4 ft 3 ft 5 ft W B ft The unknowns are N,N B,a. The equations of motion are: F : T = W g a, F : N + N B W = 0 M G : N B (ft) N (5ft) + T(1ft) = 0 Putting in the numbers for T,W, and g and solving we find N = 943 lb, N B = 1860 lb, a = 11.5 ft/s. (a) (b) The distance is given b d = 1 at = 1 (11.5 ft/s )( s) = 3 ft. d = 3 ft. The forces were found to be N = 943 lb, N B = 1860 lb. Problem 18.4 The Boeing 747 begins its takeoff run at time t = 0. The normal forces eerted on its tires at and B are N = 175 kn and N B = 800 kn. If ou assume that these forces are constant and neglect horizontal forces other than the thrust T, how fast is the T 3 m 5 m airplanes moving at t = 4 s? (See ctive Eample 18.1.) B m 6 m are: The unknowns are T,W,a. The equations of motion F : T = W g a, F : N + N B W = 0, M G : N B (m) N (4 m) T(m) = 0. Putting in the numbers for N and N B and solving, we find a =.31 m/s, T = 700 kn, W = 980 kn. The velocit is then given b v = at = (.31 m/s )(4 s) = 9.3 m/s. v = 9.3 m/s. 470

Problem 18.5 The crane moves to the right with constant acceleration, and the 800-kg load moves without swinging. (a) (b) What is the acceleration of the crane and load? What are the tensions in the cables attached at and B? 5 5 1 m B 1.5 m 1.5 m From Newton s second law: F = 800a N. The sum of the forces on the load: 5 5 F = F sin 5 + F B sin 5 800a = 0. F F B F = F cos 5 + F B cos 5 800g = 0. The sum of the moments about the center of mass: MCM = 1.5F cos 5 + 1.5F B cos 5 1.0 m 1.5 m mg 1.5 m F sin 5 F B sin 5 = 0. Solve these three simultaneous equations: a = 0.858 m/s, F = 3709 N, F B = 4169 N 471

Problem 18.6 The total weight of the go-cart and driver is 40 lb. The location of their combined center of mass is shown. The rear drive wheels together eert a 4- lb horizontal force on the track. Neglect the horizontal forces eerted on the front wheels. (a) (b) What is the magnitude of the go-cart s acceleration? What normal forces are eerted on the tires at and B? 15 in 6 in 4 in 16 in 60 in B ( ) 40 lb F : (4 lb) = 3. ft/s a F : N + N B (40 lb) = 0 MG : N (16 in.) + N B (44 in.) + (4 lb)(15 in.) = 0 40 lb Solving we find (a) a = 3. ft/s 4 lb (b) N = 18 lb, N B = 58 lb N N B 47

Problem 18.7 The total weight of the biccle and rider is 160 lb. The location of their combined center of mass is shown. The dimensions shown are b = 1 in.,c = 16 in., and h = 38 in. What is the largest acceleration the biccle can have without the front wheel leaving the ground? Neglect the horizontal force eerted on the front wheel b the road. Strateg: You want to determine the value of the acceleration that causes the normal force eerted on the front wheel b the road to equal zero. h B b c Given: b = 1 in., c= 16 in., h= 38 in. Find: a so that N = 0 ) 160 lb F : F B = ( 3. ft/s a 160 lb F : N + N B (160 lb) = 0 MG : N b + N B c F B h = 0 N = 0 h Solving we find N B = 160 lb, F B = 67.4 lb, a = 13.6 ft/s b c F B N N B 473

Problem 18.8 The moment of inertia of the disk about O is I = 0 kg-m.tt = 0, the stationar disk is subjected to a constant 50 N-m torque. 50 N-m (a) What is the magnitude of the resulting angular acceleration of the disk? (b) How fast is the disk rotating (in rpm) at t = 4s? O (a) M = Iα α = M I α =.5 rad/s. = 50 N-m 0 kg-m =.5 rad/s. (b) The angular velocit is given b ω = αt = (.5 rad/s )(4 s) = 10 rad/s ( )( ) 1 rev 60 s = 95.5 rpm. π rad 1 min ω = 95.5 rpm. Problem 18.9 The 10-lb bar is on a smooth horizontal table. The figure shows the bar viewed from above. Its moment of inertia about the center of mass is I = 0.8 slug-ft. The bar is stationar when the force F = 5lb is applied in the direction parallel to the ais. t that instant, determine ft ft B (a) the acceleration of the center of mass, and (b) the acceleration of point. F (a) F = ma a = F m = 5lb ( ) = 16.1 ft/s / 10 lb 3. ft/s a G = (16.1 ft/s )j. (b) M G : F l = Iα α = Fl I = (5 lb)(4 ft) (0.8 slug-ft ) = 1.5 rad/s. a = a G + α r /G = (16.1 ft/s )j + ( 1.5 rad/s )k ( ft)i a = (41.1 ft/s )j. 474

Problem 18.10 The 10-lb bar is on a smooth horizontal table. The figure shows the bar viewed from above. Its moment of inertia about the center of mass is I = 0.8 slug-ft. The bar is stationar when the force F = 5 lb is applied in the direction parallel to the ais. t that instant, determine the acceleration of point B. ft ft B F = ma a = F = m F 5 lb = 16.1 ft/s / 10 lb 3. ft/s MG : F l Fl (5 lb)(4 ft) = 1.5 rad/s. = Iα α = = I (0.8 slug-ft ) ab = ag + α rb/g = (16.1 ft/s )j + ( 1.5 rad/s )k ( ft)i ab = ( 8.9 ft/s )j. Problem 18.11 The moment of inertia of the astronaut and maneuvering unit about the ais through their center of mass perpendicular to the page is I = 40 kg-m. thruster can eert a force T = 10 N. For safet, the control sstem of his maneuvering unit will not allow his angular velocit to eceed 15 per second. If he is initiall not rotating, and at t = 0, he activates the thruster until he is rotating at 15 per second, through how man degrees has he rotated at t = 10 s? T 300 mm First find the angular acceleration. MG : T d = I α α= Td (10 N)(0.3 m) = = 0.075 rad/s. I (40 kg-m ) To reach maimum angular velocit it takes π rad (15 /s) ω 180 = 3.49 s ω = αt t = = α (0.075 rad/s ) During this time, the astronaut has rotated through θ1 = 1 1 αt = (0.075 rad/s )(3.49 s) = 0.457 rad. fter this time, the astronauts turns at the fied rate. He rotated an additional angle given b π rad (10 s 3.49 s) = 1.704 rad. θ = ωt = (15 /s) 180 The total rotation is then θ = θ1 + θ = (0.457 + 1.704) rad 180 π rad = 14. θ = 14. c 008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 475

Problem 18.1 The moment of inertia of the helicopter s rotor is 40 slug-ft. The rotor starts from rest. t t = 0, the pilot begins advancing the throttle so that the torque eerted on the rotor b the engine (in ft-lb) is given as a function of time in seconds b T = 00t. (a) How long does it take the rotor to turn ten revolutions? (b) What is the rotor s angular velocit (in rpm) when it has turned ten revolutions? Find the angular acceleration T = Iα α = 00t T = = 0.476t I 40 Now answer the kinematics questions α = 0.476t, (a) (b) ω = 0.38t, When it has turned 10 revolutions, π rad (10 rev) = 0.0794t 3 rev The angular velocit is ω = 0.38(9.5) = 0.4 rad/s θ = 0.0794t 3. 1 rev π rad t = 9.5 s. 60 s 1 min = 195 rpm ω = 195 rpm. Problem 18.13 The moments of inertia of the pulles are I = 0.005 kg-m, IB = 0.045 kg-m, and IC = 0.036 kg-m. 5 N-m counterclockwise couple is applied to pulle. Determine the resulting counterclockwise angular accelerations of the three pulles. The tension in each belt changes as it goes around each pulle. The unknowns are TB, TBC, α, αb, αc. We will write three dnamic equations and two constraint equations M : (5 N-m) TB (0.1 m) = (0.005 kg-m )α 100 mm MB : TB (0. m) TBC (0.1 m) = (0.045 kg-m )αb 100 mm MC : TBC (0. m) = (0.036 kg-m )αc B 00 mm C 00 mm (0.1 m)α = (0. m)αb (0.1 m)αb = (0. m)αc. Solving, we find TB = 4. N, TBC = 14.1 N, α = 313 rad/s, αb = 156 rad/s, αc = 78.1 rad/s. 476 c 008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the

Problem 18.14 The moment of inertia of the windtunnel fan is 5 kg-m. The fan starts from rest. The torque eerted on it b the engine is given as a function of the angular velocit of the fan b T = 140 0.0ω N-m. (a) (b) When the fan has turned 60 revolutions, what is its angular velocit in rpm (revolutions per minute)? What maimum angular velocit in rpm does the fan attain? Strateg: B writing the equation of angular motion, determine the angular acceleration of the fan in terms of its angular velocit. Then use the chain rule: α = dω dt = dω dθ dθ dt = dω dθ ω. M : (140 N-m) (0.0 N-m/s )ω = (5 kg-m )α ( ) ( ) 140 0.0 α = 5 rad/s 5 rad/s4 ω = (0.6 rad/s ) (0.0000889 rad/s 4 )ω (a) α = ω dω dθ = (0.6 rad/s ) (0.0000889 rad/s 4 )ω ω 0 ωdω (0.6 rad/s ) (0.0000889 rad/s 4 )ω = Solving we find ( )( ) 1 rev 60 s ω = 59.1 rad/s = 565 rpm π rad 1 min 60(π)rad dθ 0 (b) The maimum angular velocit occurs when the angular acceleration is zero α = (0.6 rad/s ) (0.0000889 rad/s 4 )ω = 0 ( )( ) 1 rev 60 s ω = 83.7 rad/s = 799 rpm π rad 1 min 477

Problem 18.15 The moment of inertia of the pulle about its ais is I = 0.005 kg-m. If the 1-kg mass is released from rest, how far does it fall in 0.5 s? 100 mm Strateg: Draw individual free-bod diagrams of the pulle and the mass. The two free-bod diagrams are shown. The five unknowns are T, O, O, α, a. We can write four dnamic equations and one constraint equation, however, we onl need to write two dnamic equations and the one constraint equation. M O : T(0.1 m) = (0.005 kg-m )α, F : T (1 kg)(9.81 m/s ) = (1 kg)a, a = (0.1 m)α. Solving we find T = 3.7 N, a = 6.54 m/s, α = 65.4 rad/s. Now from kinematics we know d = 1 at = 1 (6.54 m/s )(0.5 s) d = 0.818 m. 478

Problem 18.16 The radius of the pulle is 15 mm and the moment of inertia about its ais is I = 0.05 kg-m. If the sstem is released from rest, how far does the 0-kg mass fall in 0.5 s? What is the tension in the rope between the 0-kg mass and the pulle? 4 kg 0 kg The free-bod diagrams are shown. We have si unknowns T 1,T,O,O,a,α. We have five dnamic equations and one constraint equation available. We will use three dnamic equations and the one constraint equation M O : (T 1 T )(0.15 m) = (0.05 kg-m )α, F 1 : T 1 (4 kg)(9.81 m/s ) = (4 kg)a, F : T (0 kg)(9.81 m/s ) = (0 kg)a, a = (0.15 m)α. Solving we find T 1 = 6.3 N, T = 80.8 N,a= 5.77 m/s, α = 46. rad/s. From kinematics we find d = 1 at = 1 (5.77 m/s )(0.5 s) = 0.71 m. d = 0.71 m, T = 80.8 N. 479

Problem 18.17 The moment of inertia of the pulle is 0.4 slug-ft. The coefficient of kinetic friction between the 5-lb weight and the horizontal surface is µ k = 0.. Determine the magnitude of the acceleration of the 5-lb weight in each case. 5 lb 6 in 5 lb 6 in lb lb (a) (b) The free-bod diagrams are shown. (a) T = lb. ( ) 6 (T 1 T ) 1 ft = (0.4 slug-ft )α, ( ) 5lb T 1 (0.)(5 lb) = 3. ft/s a, ( ) 6 a = 1 ft α. Solving we find T 1 = 1.09 lb, α = 1.14 rad/s, a = 0.570 ft/s (b) T = lb. ( ) ( ) 6 6 (T 1 T ) 1 ft = (0.4 slug-ft )α, a = 1 ft α, ( ) ( ) 5lb lb T 1 (0.)(5 lb) = 3. ft/s a, T ( lb) = 3. ft/s a. Solving we find T 1 = 1.09 lb, T = 1.97 lb, α = 1.10 rad/s, a = 0.550 ft/s Note that (b) has more inertia than (a) and therefore has to accelerate more slowl. 480

Problem 18.18 The 5-kg slender bar is released from rest in the horizontal position shown. Determine the bar s counterclockwise angular acceleration (a) at the instant it is released, and (b) at the instant when it has rotated 45. 1. m (a) The free-bod diagram is shown. M O : mg L = 1 3 ml α α = 3g L = 3(9.81 m/s ) = 1.3 rad/s. (1. m) α = 1.3 rad/s. (b) The free-bod diagram is shown. M O : mg L cos 45 = 1 3 ml α α = 3g L cos 45 = 3(9.81 m/s ) cos 45 (1. m) α = 8.67 rad/s. Problem 18.19 The 5-kg slender bar is released from rest in the horizontal position shown. t the instant when it has rotated 45, its angular velocit is 4.16 rad/s. t that instant, determine the magnitude of the force eerted on the bar b the pin support. (See Eample 18.4.) 1. m First find the angular acceleration. M O : mg L cos 45 = 1 3 ml α α = 3 g L cos 45 = 3(9.81 m/s ) cos 45 = 8.67 rad/s (1. m) Using kinematics we find the acceleration of the center of mass. a G = a O + α r G/O ω r G/O a G = 0 + (8.67)k (0.6)( cos 45 i sin 45 j) (4.16) (0.6)( cos 45 i sin 45 j) = (11.0i + 3.66j) m/s. From Newton s second law we have F : O = ma = (5 kg)(11.0 m/s ) = 55.1 N F : O mg = ma O = m(g + a ) = (5 kg)(9.18 m/s + 3.66 m/s ) = 67.4 N The magnitude of the force in the pin is now O = O + O = (55.1 N) + (67.4 N) = 87.0 N. O = 87.0 N. 481

Problem 18.0 The 5-kg slender bar is released from rest in the horizontal position shown. Determine the magnitude of its angular velocit when it has fallen to the vertical position. 1. m Strateg: : Draw the free-bod diagram of the bar when it has fallen through an arbitrar angle θ and appl the equation of angular motion to determine the bar s angular acceleration as a function of θ. Then use the chain rule to write the angular acceleration as α = dω dt = dω dθ dθ dt = dω dθ ω. First find the angular acceleration. M O : mg L cos θ = 1 3 ml α α = 3g L cos θ Using the hint we have α = ω dω dθ = 3g L cos θ ω 0 90 3g ω dω = cos θ dθ 0 L 1 ω = 3g ] 90 L sin θ = 3g 0 L ω = 3g L = 3(9.81 m/s ) = 4.95 rad/s. (1. m) ω = 4.95 rad/s. Problem 18.1 The object consists of the -kg slender bar BC welded to the 3-kg slender bar BDE. The ais is vertical. (a) (b) What is the object s moment of inertia about point D? Determine the object s counterclockwise angular acceleration at the instant shown. 0. m 0. m B C D 0.4 m 0. m E The free-bod diagram is shown (a) I D = 1 1 (kg)(0.4 m) + ( kg)(0.4 m) + 1 1 (3kg)(0.6 m) + (3 kg)(0.1 m) I D = 0.467 kg-m. (b) M D :[( kg)(0.4 m) + (3 kg)(0.1 m)](9.81 m/s ) = (0.467 kg-m )α α = 3.1 rad/s. 48

Problem 18. The object consists of the -kg slender bar BC welded to the 3-kg slender bar BDE. The ais is vertical. t the instant shown, the object has a counterclockwise angular velocit of 5 rad/s. Determine the components of the force eerted on it b the pin support. 0. m 0. m B D 0.4 m 0. m E C The free-bod diagram is shown. The moment of inertia about the fied point D is I D = 1 1 (kg)(0.4 m) + ( kg)(0.4 m) + 1 1 (3kg)(0.6 m) + (3 kg)(0.1 m) = 0.467 kg-m. The angular acceleration is given b M D :[( kg)(0.4 m) + (3 kg)(0.1 m)](9.81 m/s ) = (0.467 kg-m )α 10.8 N-m α = 0.467 kg-m = 3.1 rad/s. From Newton s Second Law we have F : D = ( kg)(0.4 m)(5 rad/s) + (3 kg)(0.1 m)(5 rad/s) F : D (5 kg)(9.81 m/s ) = ( kg)(0.4 m)(3.1 rad/s ) (3 kg)(0.1 m)(3.1 rad/s ) Solving, we find D = 7.5 N, D = 3.6 N. Problem 18.3 The length of the slender bar is l = 4 m and its mass is m = 30 kg. It is released from rest in the position shown. (a) (b) If = 1 m, what is the bar s angular acceleration at the instant it is released? What value of results in the largest angular acceleration when the bar is released? What is the angular acceleration? The moment of inertia about the fied point is I = 1 1 ml + m. The angular acceleration can be found M fied point : mg = Iα = m 1 (l + 1 )α α = 1g l + 1 (b) m l To find the critical value for we differentiate and set equal to zero to get dα d = d ( ) 1g 1g d l + 1 = l + 1 88g (l + 1 ) = = 1g(l 1 ) (l + 1 ) = 0 l = (4m) = 1.15 m. = 1.15 m. 1 1 (a) Using the given numbers we have α = 1(9.81 m/s )(1 m) (4 m) + 1(1 m) = 4.0 rad/s. α = 4.0 rad/s. The corresponding angular acceleration is α = 1(9.81 m/s )(1.15 m) (4 m) + 1(1.15 m) = 4.5 rad/s α = 4.5 rad/s. 483

Problem 18.4 Model the arm BC as a single rigid bod. Its mass is 30 kg, and the moment of inertia about its center of mass is I = 360 kg-m. Point is stationar. If the hdraulic piston eerts a 14-kN force on the arm at B what is the arm s angular acceleration? 1.80 m 1.40 m B 0.30 m C 0.80 m 0.70 m The moment of inertia about the fied point is I = I G + md = (360 kg-m ) + (30 kg)([1.10 m] + [1.80 m] ) = 1780 kg-m. The angle between the force at B and the horizontal is ( ) 1.5 m θ = tan 1 = 47.0. 1.4 m The rotational equation of motion is now M : (14 kn) sin θ(1.4 m) (14 kn) cos θ(0.8 m) (30 kg)(9.81 m/s )(1.80 m) = (1780 kg-m )α. Solving, we find α = 0.581 rad/s. α = 0.581 rad/s counterclockwise. Problem 18.5 The truck s bed weighs 8000 lb and its moment of inertia about O is 33,000 slug-ft. t the instant shown, the coordinates of the center of mass of the bed are (10, 1) ft and the coordinates of point B are (15, 11) ft. If the bed has a counterclockwise angular acceleration of 0. rad/s, what is the magnitude of the force eerted on the bed at B b the hdraulic clinder B? O B 30 The rotational equation of motion is M O : F sin 30 (15 ft) + F cos 30 (11 ft) (8000 lb)(10 ft) = (33,000 slug-ft )(0. rad/s ) Solving for F we find F = 5090 lb. 484

Problem 18.6 rm BC has a mass of 1 kg and the moment of inertia about its center of mass is 3 kg-m. Point B is stationar and arm BC has a constant counterclockwise angular velocit of rad/s. t the instant shown, what are the couple and the components of force eerted on arm BC at B? 300 mm C 40 B 700 mm Since the angular acceleration of arm BC is zero, the sum of the moments about the fied point B must be zero. Let M B be the couple eerted b the support at B. Then i j k M B + r CM/B mg = M B + 0.3 cos 40 0.3 sin 40 0 = 0. 0 117.7 0 M B = 7.05k (N-m) is the couple eerted at B. From Newton s second law: B = ma,b mg = ma where a, a are the accelerations of the center of mass. From kinematics: a = α r CM/O ω r CM/O = ( )(i0.3 cos 40 + j0.3 sin 40 ) = 0.919i 0.771j (m/s ), where the angular acceleration is zero from the problem statement. Substitute into Newton s second law to obtain the reactions at B: B = 11.0 N, B = 108.5 N. 485

Problem 18.7 rm BC has a mass of 1 kg and the moment of inertia about its center of mass is 3 kg-m. t the instant shown, arm B has a constant clockwise angular velocit of rad/s and arm BC has counterclockwise angular velocit of rad/s and a clockwise angular acceleration of 4 rad/s. What are the couple and the components of force eerted on arm BC at B? 300 mm 40 B C 700 mm Because the point B is accelerating, the equations of angular motion must be written about the center of mass of arm BC. The vector distances from to B and B to G, respectivel, are r B/ = r B r = 0.7i, M B B r G/B = 0.3 cos(40 )i + 0.3 sin(40 )j = 0.98i + 0.198j (m). The acceleration of point B is B mg a B = α r B/ ω B r B/ = ω B (0.7i) (m/s ). The acceleration of the center of mass is a G = a B + α BC r G/B ωbc r G/B i j k a G =.8i + 0 0 4 0.9193i 0.7713j 0.98 0.198 0 =.948i 1.691j (m/s ). From Newton s second law, B = ma G = (1)(.948) = 35.37 N, B mg = ma G, B = (1)( 1.691) + (1)(9.81) = 97.43 N From the equation of angular motion, M G = Iα BC. The moment about the center of mass is i j k M G = M B + r B/G B = 0.98 0.198 0 35.37 97.43 0 = M B k 9.1k (N-m). Note I = 3 kg-m and α BC = 4k (rad/s ), from which M B = 9.1 + 3( 4) = 17.1 N-m. 486

Problem 18.8 The space shuttle s attitude control engines eert two forces F f = 8 kn and F r = kn. The force vectors and the center of mass G lie in the plane of the inertial reference frame. The mass of the shuttle is 54,000 kg, and its moment of inertia about the m ais through the center of mass that is parallel to the z F f ais is 4.5 10 6 kg-m. Determine the acceleration of G the center of mass and the angular acceleration. (You can ignore the force on the shuttle due to its weight). 18 m 1m m 5 6 F r Newton s second law is F = (Ff cos 5 F r cos 6 )i (F f sin 5 + F r sin 6 )j = ma. Setting F f = 8000 N, F r = 000 N and m = 54,000 kg and solving for a, we obtain a = 0.1108i 0.0168j (m/s ). The equation of angular motion is M = (18)(Ff sin 5 ) ()(F f cos 5 ) (1)(F r sin 6 ) + ()(F r cos 6 ) = Iα where I = 4.5 10 6 kg-m. Solving for α the counterclockwise angular acceleration is α = 0.00047 rad/s. Problem 18.9 In Problem 18.8, suppose that F f = 4 kn and ou want the shuttle s angular acceleration to be zero. Determine the necessar force F r and the resulting acceleration of the center of mass. zero: The total moment about the center of mass must equal M = (18)(Ff sin 5 ) ()(F f cos 5 ) (1)(F r sin 6 ) + ()(F r cos 6 ) = 0 Setting F f = 4000 N and solving F r = 306 N. From Newton s second law F = (Ff cos 5 F r cos 6 )i (F f sin 5 + F r sin 6 )j = 54,000a, we obtain a = 0.0313i 0.0109j (m/s ). 487

Problem 18.30 Points B and C lie in the plane. The ais is vertical. The center of mass of the 18- kg arm BC is at the midpoint of the line from B to C, and the moment of inertia of the arm about the ais through the center of mass that is parallel to the z ais is 1.5 kg-m. t the instant shown, the angular velocit and angular acceleration vectors of arm B are ω B = 0.6k (rad/s) and α B = 0.3k (rad/s ). The angular velocit and angular acceleration vectors of arm BC are ω BC = 0.4k (rad/s) and α BC = k (rad/s). Determine the force and couple eerted on arm BC at B. z 760 mm 15 B 50 C 900 mm The acceleration of point B is a B = a + α B r /B ω B r /B or i j k a B = 0 0 0.3 0.76 cos 15 0.76 sin 15 0 B mg (0.6) (0.76 cos 15 i 0.76 sin 15 j) 50 B = 0.33i 0.149j (m/s ) M B The acceleration of the center of mass G of arm BC is a G = a B + α BC r G/B ωbc r G/B a B = 0.33i 0.149j i j k + 0 0 0.45 cos 50 0.45 sin 50 0 (0.4) (0.45 cos 50 i + 0.45 sin 50 j), or a G = 1.059i + 0.374j (m/s ). The free bod diagram of arm BC is: Newton s second law is F = B i + (B mg)j = ma G : B i + [ B (18)(9.81) ] j = 18( 1.059i + 0.374j). Solving, we obtain B = 19.1 N,B = 183.3 N. The equation of angular motion is MG = I BC α BC : or (0.45 sin 50 )B (0.45 cos 50 )B + M B = (1.5)() Solving for M B, we obtain M B = 6.6 N-m. 488

Problem 18.31 Points B and C lie in the plane. The ais is vertical. The center of mass of the 18- kg arm BC is at the midpoint of the line from B to C, and the moment of inertia of the arm about the ais through the center of mass that is parallel to the z ais is 1.5 kg-m. t the instant shown, the angular velocit and angular acceleration vectors of arm B are ω B = 0.6k (rad/s) and α B = 0.3k (rad/s ). The angular velocit vector of arm BC is ω BC = 0.4k (rad/s). If ou want to program the robot so that the angular acceleration of arm BC is zero at this instant, what couple must be eerted on arm BC at B? From the solution of Problem 18.30, the acceleration of point B is a B = 0.33i 0.149j (m/s ).Ifα BC = 0, the acceleration of the center of mass G of arm BC is a G = a B ω BC r G/B = 0.33i 0.149j (0.4) (0.45 cos 50 i + 0.45 sin 50 j) = 0.370i 0.05j (m/s ). From the free bod diagram of arm BC in the solution of Problem 18.30. Newton s second law is F = B i + (B mg)j = ma G : B i + [B (18)(9.81)]j = 18( 0.370i 0.05j). Solving, we obtain B = 6.65 N, B = 17.90 N. The equation of angular motion is MG = I BC α BC = 0: (0.45 sin 50 )B (0.45 cos 50 )B + M B = 0. Solving for M B, we obtain M B = 5.3 N-m. Problem 18.3 The radius of the -kg disk is R = 80 mm. Its moment of inertia is I = 0.0064 kg-m.it rolls on the inclined surface. If the disk is released from rest, what is the magnitude of the velocit of its center two seconds later? (See ctive Eample 18.). R 30 There are four unknowns (N, f, a, α), three dnamic equations, and one constraint equation. We have M G : fr = Iα, F : mg sin 30 f = ma a = rα Solving, we find a = mgr sin 30 I + mr = (kg)(9.81 m/s )(0.08 m) sin 30 0.0064 kg-m + ( kg)(0.08 m) = 3.7 m/s. From the kinematics we have v = at = (3.7 m/s )( s) = 6.54 m/s. v = 6.54 m/s. 489

Problem 18.33 The radius of the -kg disk is R = 80 mm. Its moment of inertia is I = 0.0064 kg-m. What minimum coefficient of static friction is necessar for the disk to roll, instead of slip, on the inclined surface? (See ctive Eample 18..) R 30 There are five unknowns (N,f,a,α,µ s ), three dnamic equations, one constraint equation, and one friction equation. We have M G : fr = Iα, F : mg sin 30 f = ma, F : N mg cos 30 = 0, a = Rα, f = µ s N. Putting in the numbers and solving, we find N = 17.0 N,f= 3.7 N, a= 3.7 m/s,α= 40.9 rad/s, µ s = 0.19. Problem 18.34 thin ring and a homogeneous circular disk, each of mass m and radius R, are released from rest on an inclined surface. Determine the ratio v ring /v disk of the velocities of the their centers when the ave rolled a distance D. R D R D There are four unknowns (N,f,a,α), three dnamic equations, and one constraint equation. We have M G : fr = Iα, F : mg sin θ f = ma, a = Rα, Solving, we find a = mgr sin θ I + mr For the ring I ring = mr a ring = g sin θ For the disk I disk = 1 mr a disk = g 3 sin θ The velocities are then v ring = a ring D = gd sin θ,v disk = a disk D = The ratio is v ring v disk = gd sin θ = 3/4 4 gd sin θ 3 v ring v disk = 3/4 4 gd sin θ 3 490

Problem 18.35 The stepped disk weighs 40 lb and its moment of inertia is I = 0. slug-ft. If the disk is released from rest, how long does it take its center to fall 3 ft? (ssume that the string remains vertical.) 4 in 8 in The moment about the center of mass is M = RT. From the equation of angular motion: RT = Iα, from which T = Iα R. From the free bod diagram and Newton s second law: F = T W = ma, where a is the acceleration of the center of mass. From kinematics: a = Rα. Substitute and solve: T W a = ( ). I R + m W The time required to fall a distance D is t = D D(I + R m) = a R. W For D = 3 ft, R = 4 1 = 0.3333 ft, W = 40 lb, m = W g I = 0. slug-ft, t = 0.676 s = 1.4 slug, 491

Problem 18.36 The radius of the pulle is R = 100 mm and its moment of inertia is I = 0.1 kgm. The mass m = 5 kg. The spring constant is k = 135 N/m. The sstem is released from rest with the spring unstretched. t the instant when the mass has fallen 0. m, determine (a) the angular acceleration of the pulle, and (b) the tension in the rope between the mass and the pulle. R k m The force in the spring is k. There are five unknowns (O,O,T,a,α), four dnamic equations, and one constraint equation. M O : (k)r TR = Iα, F : T mg = ma, a = Rα Solving we find (a) α = R(mg k) I + mr = (0.1 m)([5 kg][9.81 m/s ] [135 N/m][0. m]) 0.1 kg-m + (5 kg)(0.1 m) α = 14.7 rad/s. (b) T = m(g Rα) = (5 kg)(9.81 m/s [0.1 m][14.7 rad/s ]) T = 41.7 N. 49

Problem 18.37 The radius of the pulle is R = 100 mm and its moment of inertia is I = 0.1 kg-m. The mass m = 5 kg. The spring constant is k = 135 N/m. The sstem is released from rest with the spring unstretched. What maimum distance does the mass fall before rebounding? R Strateg: ssume that the mass has fallen an arbitrar distance. Write the equations of motion of the mass and the pulle and use them to determine the acceleration a of the mass as a function of. Then appl the chain rule: dv dt = dv d d dt = dv d v. k m The force in the spring is k. There are five unknowns (O,O,T,a,α), four dnamic equations, and one constraint equation. M O : (k)r TR = Iα, F : T mg = ma, a = Rα Solving we find a = R (mg k) I + mr = v dv d 0 0 Thus R (mg k) vdv = 0 I + mr d = R I + mr mg 1 k = 0 = 0or = mg k The maimum distance is (mg k) d = 0 0 = mg k = (5 kg)(9.81 m/s ) 135 N/m = 0.77 m = 0.77 m. 493

Problem 18.38 The mass of the disk is 45 kg and its radius is R = 0.3 m. The spring constant is k = 60 N/m. The disk is rolled to the left until the spring is compressed 0.5 m and released from rest. k R (a) (b) If ou assume that the disk rolls, what is its angular acceleration at the instant it is released? What is the minimum coefficient of static friction for which the disk will not slip when it is released? + 0 = 0.5 mg k = 600 N/m m = 45 kg F s O R = 0.3 m I 0 = 1 mr =.05 N-m,F s = k F : F s f = ma 0 N f F : N mg = 0 M0 : fr = I 0 α Rolling implies a 0 = Rα We have, at = 0.5 m k f = ma 0 N mg = 0 Rf = I 0 α a 0 = Rα Four eqns, four unknowns (a 0,α,N,f) (a) Solving f = 100 N, N = 441.5 N α = 14.81 rad/s (clockwise) a 0 = 4.44 m/s (b) for impending slip, f = µ s N µ s = f/n = 100/441.5 µ s = 0.7 494

Problem 18.39 The disk weighs 1 lb and its radius is 6 in. It is stationar on the surface when the force F = 10 lb is applied. (a) (b) If the disk rolls on the surface, what is the acceleration of its center? What minimum coefficient of static friction is necessar for the disk to roll instead of slipping when the force is applied? F There are five unknowns (N,f,a,α,µ s ), three dnamic equations, one constraint equation, and one friction equation. F : F f = ma, F : N mg = 0, ( ) 1 M G : fr = mr α, a = rα, f = µ s N. Solving, we find (a) a = F 3m = (10 lb) ( ) = 8.94 ft/s. a = 17.9 ft/s. 1 lb 3 3. ft/s (b) µ s = F (10 lb) = 3mg 3(1 lb) = 0.78 µ s = 0.78. 495

Problem 18.40 4-lb sphere with radius R = 4in is placed on a horizontal surface with initial angular velocit ω 0 = 40 rad/s. The coefficient of kinetic friction between the sphere and the surface is µ k = 0.06. What maimum velocit will the center of the sphere attain, and how long does it take to reach that velocit? ω 0 Strateg: The friction force eerted on the spinning sphere b the surface will cause the sphere to accelerate to the right. The friction force will also cause the sphere s angular velocit to decrease. The center of the sphere will accelerate until the sphere is rolling on the surface instead of slipping relative to it. Use the relation between the velocit of the center and the angular velocit of the sphere when it is rolling to determine when the sphere begins rolling. Given W = 4 lb, g= 3. ft/s,m= W/g, R = 4/1 ft, µ k = 0.06 We have F : µ k N = ma W F : N mg = 0 MG : µ k NR = 5 mr a W µ R W Solving we find α = 5µ kg R = 14.49 rad/s, a = µ k g = 1.93 ft/s From kinematics we learn that α = 14.49 rad/s,ω= (14.49 rad/s )t (40 rad/s) a = 1.93 ft/s, v = (1.93 ft/s )t when we reach a stead motion we have v = Rω (1.93 ft/s )t = (0.33 ft)[(14.49 rad/s )t (40 rad/s)] Solving for the time we find t = 1.97 s v = 3.81 ft/s µ k N N W 496

Problem 18.41 soccer plaer kicks the ball to a teammate 8 m awa. The ball leaves the plaer s foot moving parallel to the ground at 6 m/s with no angular velocit. The coefficient of kinetic friction between the ball and the grass is µ k = 0.3. How long does it take the ball to reach his teammate? The radius of the ball is 11 mm and its mass is 0.4 kg. Estimate the ball s moment of inertia b using the equation for a thin spherical shell: I = 3 mr. Given µ = 0.3, r = 0.11 m, g = 9.81 m/s, v 0 = 6 m/s The motion occurs in two phases. (a) Slipping. F : µn = ma mg F : N mg = 0 µν MG : µnr = 3 mr α N Solving we find a = µg v = v 0 µgt, s = v 0 t 1 µgt α = 3µg R ω = 3µg R t When it stops slipping we have v = Rω v 0 µgt = 3 µgt t = v 0 5µg = 0.765 s v = 3.6 m/s, s= 3.67 m (b) Rolling Stead motion a = 0, v = 3.6 m/s, s = (3.6 m/s)(t 0.765 s) + 3.67 m When it reaches the teammate we have 8m= (3.6 m/s)(t 0.765 s) + 3.67 m t = 1.97 s 497

Problem 18.4 The 100-kg clindrical disk is at rest when the force F is applied to a cord wrapped around it. The static and kinetic coefficients of friction between the disk and the surface equal 0.. Determine the angular acceleration of the disk if (a) F = 500 N and (b) F = 1000 N. Strateg: First solve the problem b assuming that the disk does not slip, but rolls on the surface. Determine the friction force, and find out whether it eceeds the product of the coefficient of friction and the normal force. If it does, ou must rework the problem assuming that the disk slips. 300 mm F Choose a coordinate sstem with the origin at the center of the disk in the at rest position, with the ais parallel to the plane surface. The moment about the center of mass is M = RF Rf, from which RF Rf = Iα. From which W F f = RF Iα R = F Iα R. f N From Newton s second law: F f = ma, where a is the acceleration of the center of mass. ssume that the disk rolls. t the point of contact a P = 0; from which 0 = a G + α r P/G ω r P/G. a G = a i = α Rj ω Rj i j k = 0 0 α ω Rj = Rαi ω Rj, 0 R 0 from which a = 0 and a = Rα. Substitute for f and solve: a = F (m + IR ). (b) For F = 1000 N the acceleration is a = 4F 3m = 4000 300 = 13.33 m/s. The friction force is f = F ma = 1000 1333.3 = 333.3 N. The drum slips. The moment equation for slip is RF + Rµ k gm = Iα, from which α = RF + Rµ kgm I = F mr + µ kg R = 53.6 rad/s. For a disk, the moment of inertia about the polar ais is I = 1 mr, from which a = 4F 3m = 000 300 = 6.67 m/s. (a) For F = 500 N, the friction force is f = F ma = F 3 = 500 3 = 167 N. Note: µ k W = 0. mg= 196. N, the disk does not slip. The angular velocit is α = a R = 6.67 0.3 =. rad/s. 498

Problem 18.43 The ring gear is fied. The mass and moment of inertia of the sun gear are ms = 30 kg and IS = 40 kg-m. The mass and moment of inertia of each planet gear are mp = 38 kg and IP = 0.60 kg-m. If a couple M = 00 N-m is applied to the sun gear, what is the latter s angular acceleration? Ring gear 0.18 m. 0.86 m M 0.50 m Planet gears (3) Sun gear F MS = 00 N-m Sun Gear: Planet Gears: er M0 : MS 3RF = IS αs Mc : Gr F r = IP αp G r C IP R F O et Ft : F + G = mp act 3 small disks Is Ms F F From kinematics act = rαp αp rp = RαS We have 5 eqns in 5 unknowns. Solving, αs = 3.95 rad/s (counterclockwise) Problem 18.44 In Problem 18.43, what is the magnitude of the tangential force eerted on the sun gear b each planet gear at their point of contact when the 00 N-m couple is applied to the sun gear? See the solution to Problem 18.43. Solving the 5 eqns in 5 unknowns ields αs = 3.95 rad/s, G = 9.63 N, agt = 0.988 m/s, αp = 5.49 rad/s, F = 7.9 N. F is the required value. c 008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 499

Problem 18.45 The 18-kg ladder is released from rest in the position shown. Model it as a slender bar and neglect friction. t the instant of release, determine (a) the angular acceleration of the ladder and (b) the normal force eerted on the ladder b the floor. (See ctive Eample 18.3.) 30 4 m The vector location of the center of mass is r G = (L/) sin 30 i + (L/) cos 30 j = 1i + 1.73j (m). Denote the normal forces at the top and bottom of the ladder b P and N. The vector locations of and B are r = L sin 30 i = i (m), r B = L cos 30 j = 3.46j (m). The vectors r /G = r r G = 1i 1.73j (m), r B/G = r B r G = 1i + 1.73j (m). The moment about the center of mass is P B M = r B/G P + r /G N, i j k i j k M = 1 1.73 0 + 1 1.73 0 P 0 0 0 N 0 = ( 1.73P + N)k (N-m). From the equation of angular motion: (1) 1.73 P + N = Iα. From Newton s second law: () P = ma, (3) N mg = ma, where a, a are the accelerations of the center of mass. mg N From kinematics: a G = a + α r G/ ω r G/. The angular velocit is zero since the sstem was released from rest, i j k a G = a i + 0 0 α = a i 1.73αi αj 1 1.73 0 = (a 1.73α)i αj (m/s ), from which a = α. Similarl, i j k a G = a B + α r G/B, a G = a B + 0 0 α 1 1.73 0 = a B j + 1.73αi + αj, from which a = 1.73α. Substitute into (1), () and (3) to obtain three equations in three unknowns: 1.73P + N = Iα, P = m(1.73)α, N mg = mα. Solve: (a) α = 1.84 rad/s, P = 57.3 N, (b) N = 143.47 N 500

Problem 18.46 The 18-kg ladder is released from rest in the position shown. Model it as a slender bar and neglect friction. Determine its angular acceleration at the instant of release. 30 4 m 0 Given m = 18 kg, L= 4m,g= 9.81 m/s,ω = 0 First find the kinematic constraints. We have N a = a G + α r /G = a i + a j + αk ([ L ] [ ] ) L sin 30 i + cos 30 j 30 = (a α L ) cos 30 i + (a α L ) sin 30 j a B = a G + α r B/G ([ ] L = a i + a j + αk sin 30 i + [ L ] ) cos 30 j = (a + α L ) cos 30 i + (a + α L ) sin 30 j The constraints are a i = a α L cos 30 = 0 mg N B 0 a B (sin 0 i + cos 0 j) = (a + α L ) cos 30 sin 0 + (a + α L ) sin 30 cos 0 The dnamic equations: F : N + N B sin 0 = ma F : N B cos 0 mg = ma MG : N ( L cos 30 ) + N B cos 0 ( L sin 30 ) + N B sin 0 ( L cos 30 ) = 1 1 ml α Solving five equations in five unknowns we have α =.35 rad/s CCW lso a = 4.07 ft/s,a = 5.31 ft/s,n = 43.7 N, N B = 86. N 501

Problem 18.47 The 4-kg slender bar is released from rest in the position shown. Determine its angular acceleration at that instant if (a) the surface is rough and the bar does not slip, and (b) the surface is smooth. 1 m 60 (a) The surface is rough. The lower end of the bar is fied, and the bar rotates around that point. M B : mg L cos θ = 1 3 ml α α = 3g L cos θ = 3(9.81 m/s ) cos 60 (1 m) α = 7.36 rad/s. (b) The surface is smooth. There are four unknowns (N, a,a,α), three dnamic equations, and one constraint equation (the component of the acceleration of the point in contact with the ground is zero). F :0= ma, F : N mg = ma, M G : N L cos θ = 1 1 ml α a + α L cos θ = 0 Solving, we find α = 6g cos θ L(1 + 3 cos θ) = 6(9.81 m/s ) cos 60 (1 m)(1 + 3 cos 60 ) = 16.8 rad/s. α = 16.8 rad/s. 50

Problem 18.48 The masses of the bar and disk are 14 kg and 9 kg, respectivel. The sstem is released from rest with the bar horizontal. Determine the bar s angular acceleration at that instant if (a) the bar and disk are welded together at, (b) the bar and disk are connected b a smooth pin at. O 1. m 0.3 m Strateg: In part (b), draw individual free-bod diagrams of the bar and disk. (a) L = 1. m R = 0.3 m m B = 14 kg O is a fied point For the bar m D = 9kg O O C G m o g O mb g I G = 1 1 m BL = 1 1 (14)(1.) = 1.68 N-m I OB = I G + m B ( L ) I OB = 6.7 N-m For the disk: I = 1 m DR = 1 (9)(0.3) = 0.405 N-m I OD = I + m 0 L = 13.37 N-m The total moment of inertia of the welded disk and bar about O is I T = I OB + I OD = 0.09 N-m F : O = O = ma G F : O m B g m D g = (m B + m D )a G ( ) L M0 : m B g Lm D g = I T α We can solve the last equation for α without finding the location and acceleration of the center of mass, G. Solving, α = 9.38 rad/s (clockwise) (b) In this case, onl the moment of inertia changes. Since the disk is on a smooth pin, it does not rotate. It acts onl as a point mass at a distance L from point O. In this case, I OD = m DL and I T = I OB + I OD = 19.68 N-m We now have ( ) L M0 : m B g Lm D g = I T α Solving α = 9.57 rad/s (clockwise) 503

Problem 18.49 The 5-lb horizontal bar is connected to the 10-lb disk b a smooth pin at. The sstem is released from rest in the position shown. What are the angular accelerations of the bar and disk at that instant? O 3 ft 1 ft Given g = 3. ft/s,w bar = 5lb, W disk = 10 lb, O m bar = W bar g, m disk = W disk g O L = 3ft,R= 1ft W bar W disk The FBDs The dnamic equations MO : m bar g L L = 1 3 m bar L α bar MGdisk : R = 1 m disk R α disk F : m disk g = m disk a disk Kinematic constraint α bar L = a disk α disk R Solving we find α disk = 3.58 rad/s,α bar = 1.5 rad/s,a disk = 34.0 m/s, = 0.556 N Thus α disk = 3.58 rad/s CCW, α bar = 1.5 rad/s CW 504

Problem 18.50 The 0.1-kg slender bar and 0.-kg clindrical disk are released from rest with the bar horizontal. The disk rolls on the curved surface. What is the bar s angular acceleration at the instant it is released? 40 mm 10 mm The moment about the center of mass of the disk is M = fr, from the equation of angular motion, Rf = I d α d. From Newton s second law: f B W d = m d a d. Since the disk rolls, the kinematic condition is a d = Rα d. Combine the epressions and rearrange: f = Iα d /R, Iα d /R B W d = m d a d, from which B + W d = (Rm d + I d /R)α d. The moment about the center of mass of the bar is B B B B W b W d f N M b = ( ) ( ) L L + B, from which ( ) ( ) L L + B = I b α b. From Newton s second law W b + B = m b a b, where a b is the acceleration of the center of mass of the bar. The kinematic condition for the bar is a CM = α b (( ) ) L i = ( ) L α b j, from which a b = ( ) L α b. Similarl, a D = a CM + α b ((L/)i), from which a d = Lα b. From which: α d = Lα b /R. Substitute to obtain three equations in three unknowns: ( B + W d = Rm d + I )( d L ) α b, R R ( ) ( ) L L + B = I b α b, W b + B = m b ( L ) α b. Substitute known numerical values: L = 0.1 m, R = 0.04 m, m b = 0.1 kg, W b = m b g = 0.981 N, m d = 0. kg,w d = m d g = 1.96 N, I b = (1/1)m b (L ) = 1. 10 4 kg-m, I d = (1/)m d R = 1.6 10 4 kg-m. Solve: α b = 61.3 rad/s, = 0.368 N,B = 0.45 N. 505

Problem 18.51 The mass of the suspended object is 8 kg. The mass of the pulle is 5 kg, and its moment of inertia is 0.036 kg-m. If the force T = 70 N, what is the magnitude of the acceleration of? T 10 mm Given m = 8kg, m B = 5kg, I B = 0.036 kg-m T T R = 0.1 m, g= 9.81 m/s,t= 70 N The FBDs The dnamic equations FB : T + T m B g B = m B a B m B g B F : B m g = m a B MB : T R + TR = I B α B Kinematic constraints a B = a,a B = Rα B Solving we find a = 0.805 m/s m g We also have a B = 0.805 m/s, α B = 6.70 rad/s, T = 68.0 N, B = 84.9 N 506

Problem 18.5 The suspended object weighs 0 lb. The pulles are identical, each weighing 10 lb and having moment of inertia 0.0 slug-ft. If the force T = 15 lb, what is the magnitude of the acceleration of? T 4 in 4 in Given g = 3. ft/s, W = 0 lb, W disk = 10 lb, I= 0.0 slug-ft T W disk T m = W g, m disk = W disk g,r= 4 ft, T = 15 lb 1 The FBDs The dnamic equations F1 : T + T T 1 m disk g = m disk a 1 T 4 W disk T 1 F : T 4 + T 1 T 3 m disk g = m disk a F3 : T 3 m g = m a M1 : TR T R = Iα 1 T 3 M : T 1 R T 4 R = Iα The kinematic constraints W a 1 = Rα 1, a = Rα,a 1 = Rα,a = a Solving we find a = 3.16 ft/s We also have a 1 = 6.3 ft/s,a = 3.16 ft.s,α 1 = 19.0 rad/s,α = 9.48 rad/s T 1 = 16.8 lb, T = 13.7 lb, T 3 =.0 lb, T 4 = 16. lb 507

Problem 18.53 The -kg slender bar and 5-kg block are released from rest in the position shown. If friction is negligible, what is the block s acceleration at that instant? (See Eample 18.5.) 1 m L = 1m, m = kg M = 5kg 55 ssume directions for B, B, I G = 1 1 m BL F : B = ma G (1) F : B mg = m B a G () MG : ( ) ( ) L L cos θ B + sin θ B = I G α (3) m L cosθ F : B = Ma 0 (4) F : N B Mg = 0 (5) θ G From kinematics, ω = 0 (initiall) a 0 = a G + αk r 0/G mg B O (L / ) sin θ where r 0/G = L cos θi L sin θj B From the diagram a 0 = a 0 i { a0 = a G + (αl/) sin θ (6) 0 = a G + (αl/) cos θ (7) We know θ = 55, I G = 0.167 kg-m, L = 1m, m = kg, M = 5 kg. We have 7 eqns in 7 unknowns (a G,a G,a 0,α,B,B,N), Solving, we get B M B O M g B = 5.77 N, (opposite the assumed direction) B = 13.97 N, N a G =.88 m/s, a G =.83 m/s α = 9.86 rad/s, N = 63.0 N a 0 = 1.15 m/s. (to the right) 508

Problem 18.54 The -kg slender bar and 5-kg block are released from rest in the position shown. What minimum coefficient of static friction between the block and the horizontal surface would be necessar for the block not to move when the sstem is released? (See Eample 18.5.) This solution is ver similar to that of Problem 18.53. We add a friction force f = µ s N and set a 0 = 0. L = 1m m = kg M = 5kg I G = 1 1 ml = 0.167 kg-m θ G mg L F : B = ma G (1) B F : B mg = ma G () B MG : ( ) ( ) L L cos θ B + sin θ B = I G α (3) Mg (These are the same as in Problem 18.53) Note: In Prob 18.53, B = 5.77 N (it was in the opposite direction to that assumed). This resulted in a 0 to the right. Thus, friction must be to the left F : B µ s N = ma 0 = 0 (4) M B B F : N B Mg = 0 (5) From kinematics, µ s N a 0 = a G + α r 0/G = 0 N O = a G + (αl/) sin θ (6) O = a G + (αl/) cos θ (7) Solving 7 eqns in 7 unknowns, we get B = 6.91 N, B = 14.78 N, a G = 3.46 m/s, a G =.4 m/s N = 63.8 N, α = 8.44 rad/s µ s = 0.108 509

Problem 18.55 s a result of the constant couple M applied to the 1-kg disk, the angular acceleration of the 0.4-kg slender bar is zero. Determine M and the counterclockwise angular acceleration of the rolling disk. 40 1 m 0.5 m M There are seven unknowns (M,N,f,O,O,a,α), si dnamic equations, and one constraint equation. We use the following subset of those equations. M G rod : O (0.5 m) cos 40 O (0.5 m) sin 40 = 0, F rod : O = (0.4 kg)a, F rod : O (0.4 kg)(9.81 m/s ) = 0, M G disk : M f(0.5 m) = 1 (1kg)(0.5 m) α, F disk : O f = (1 kg)a, a = (0.5 m)α. Solving, we find O = 3.9 N, O = 3.9 N, f = 11.5 N,a= 8.3 m/s, α = 3.9 rad/s,m= 3.91 N-m. M = 3.91 N-m, α= 3.9 rad/s. 510

Problem 18.56 The slender bar weighs 40 lb and the crate weighs 80 lb. t the instant shown, the velocit of the crate is zero and it has an acceleration of 14 ft/s toward the left. The horizontal surface is smooth. Determine the couple M and the tension in the rope. 6 ft M 3 ft 6 ft There are si unknowns (M,T,N,O,O,α), five dnamic equations, and one constraint equation. We use the following subset of the dnamic equations. M O : M (40 lb)(1.5 ft) T cos 45 (6ft) = 1 3 T sin 45 (3ft) ( ) 40 lb 3. ft/s (45 ft )α, ( ) F : T cos 45 80 lb = 3. ft/s (14 ft/s ) The constraint equation is derived from the triangle shown. We have L = 45 ft, d = 6 ft, θ = 63.4. = L cos θ + d L sin θ ( ) ẋ = L sin θ L cos θ sin θ d L sin θ θ Since the velocit ẋ = 0, then we know that the angular velocit ω = θ = 0. Taking one more derivative and setting ω = 0, we find ( ẍ = L sin θ L cos θ sin θ d L sin θ ( ) = L sin θ L cos θ sin θ d L sin θ α Solving these equations, we find that ) θ (14 ft/s ) α = 1.56 rad/s, M = 40 lb-ft, T= 49. lb. 511

Problem 18.57 The slender bar weighs 40 lb and the crate weighs 80 lb. t the instant shown, the velocit of the crate is zero and it has an acceleration of 14 ft/s toward the left. The coefficient of kinetic friction between the horizontal surface and the crate is µ k = 0.. Determine the couple M and the tension in the rope. 6 ft M 3 ft 6 ft There are seven unknowns (M,T,N,O,O,α,f), five dnamic equations, one constraint equation, and one friction equation. We use the following subset of the dnamic equations. M O : M (40 lb)(1.5 ft) T cos 45 (6ft) T sin 45 (3ft) = 1 3 ( ) 40 lb 3. ft/s (45 ft )α, ( ) F : T cos 45 80 lb + (0.)N = 3. ft/s (14 ft/s ) F : T sin 45 + N (80 lb) = 0. The constraint equation is derived from the triangle shown. We have L = 45 ft, d = 6 ft,θ = 63.4. = L cos θ + d L sin θ ( ) ẋ = L sin θ L cos θ sin θ d L sin θ θ Since the velocit ẋ = 0, then we know that the angular velocit ω = θ = 0. Taking one more derivative and setting ω = 0, we find ( ) ẍ = L sin θ L cos θ sin θ d L sin θ θ (14 ft/s ) ( ) = L sin θ L cos θ sin θ d L sin θ α Solving these equations, we find that α = 1.56 rad/s,n = 168 N, M = 470 lb-ft, T= 59.9 lb. 51

Problem 18.58 Bar B is rotating with a constant clockwise angular velocit of 10 rad/s. The 8-kg slender bar BC slides on the horizontal surface. t the instant shown, determine the total force (including its weight) acting on bar BC and the total moment about its center of mass. We first perform a kinematic analsis to find the angular acceleration of bar BC and the acceleration of the center of mass of bar BC. First the velocit analsis: 0.4 m B 10 rad/s 0.4 m 0.8 m C v B = v + ω B r B/ = 0 + ( 10k) (0.4i + 0.4j) = ( 4i + 4j) v C = v B + ω BC r C/B = ( 4i + 4j) + ω BC k (0.8i 0.4j) = ( 4 + 0.4 ω BC )i + (4 + 0.8 ω BC )j Since C stas in contact with the floor, we set the j component to zero ω BC = 5 rad/s. Now the acceleration analsis. a B = a + α B r B/ ω B r B/ = 0 + 0 (10) (0.4i + 0.4j) = ( 40i 40j) a C = a B + α BC r C/B ω BC r C/B = ( 40i 40j) + α BC k (0.8i 0.4j) ( 5) (0.8i 0.4j) = ( 60 + 0.4α BC )i + ( 30 + 0.8α BC )j Since C stas in contact with the floor, we set the j component to zero α BC = 37.5 rad/s. Now we find the acceleration of the center of mass G of bar BC. a G = a B + α BC r G/B ω BC r G/B = ( 40i 40j) + (37.5)k (0.4i 0.j) ( 5) (0.4i 0.j) = ( 4.5i 0j) m/s. The total force and moment cause the accelerations that we just calculated. Therefore F = ma G = (8 kg)( 4.5i 0j) m/s = ( 340i 160j) N, M = Iα = 1 1 (8kg)([0.8 m] + [0.4 m] )(37.5 rad/s ) = 0 N-m. F = ( 340i 160j) N, M= 0 N-m counterclockwise. 513

Problem 18.59 The masses of the slender bars B and BC are 10 kg and 1 kg, respectivel. The angular velocities of the bars are zero at the instant shown and the horizontal force F = 150 N. The horizontal surface is smooth. Determine the angular accelerations of the bars. 0.4 m B C F 0.4 m 0. m Given B m B = 10 kg, m BC = 1 kg, g= 9.81 m/s L B = 0.4 m, L BC = 0.4 + 0. m,f= 150 N m B g B B The FBDs The dnamic equations M : m B g L B + B L B = 1 3 m BL B α B m BC g F FBC : B F = m BC a BC N FBC : B m BC g + N = m BC a BC MBCG : (B F)(0. m) + (B + N)(0.1 m) = 1 1 m BCL BC α BC The kinematic constraints a BC = α B L B + α BC (0.1 m) a BC = α BC (0. m) α B L B + α BC (0. m) = 0 Solving we find α B = 0.6 rad/s,α BC = 41. rad/s α B = 0.6 rad/s CCW, α BC = 41. rad/s CW We also find a BC = 8.3 m/s, a BC = 4.1 m/s N = 44 N, B = 51. N, B = 76.5 N, 514

Problem 18.60 Let the total moment of inertia of the car s two rear wheels and ale be I R, and let the total moment of inertia of the two front wheels be I F. The radius of the tires is R, and the total mass of the car, including the wheels, is m. If the car s engine eerts a torque (couple) T on the rear wheels and the wheels do not slip, show that the car s acceleration is a = RT R m + I R + I F. Strateg: diagrams. Isolate the wheels and draw three free-bod The free bod diagrams are as shown: We shall write three equations of motion for each wheel and two equations of motion for the bod of the car: We shall sum moments about the ales on each wheel. Rear Wheel: F m B g G F = F + f R = m R a, F G F = N R m R g F = 0, ( MRale = Rf R T = I R α = I R a ) R F m R g G m F g Front Wheel: F = G + f F = m F a, F G T f R N R f F N F F = N F m F g G = 0, ( MFale = Rf F = I F α = I F a ) R Car Bod: F = F G = m B a, F = F + G m B g = 0. Summing the equations for all three bodies, we get N R + N F = (m B + m R + m F )g = mg. Summing the equations for all three bodies in the direction, we get f R + f F = (m B + m R + m F )a = ma. (1) From the moment equations for the wheels, we get f F = I F a/r and f R = I R a/r + T/R. Substituting these into Eq. (1), we get a = RT/(mR + I R + I F ) as required. 515

Problem 18.61 The combined mass of the motorccle and rider is 160 kg. Each 9-kg wheel has a 330- mm radius and a moment of inertia I = 0.8 kg-m. The engine drives the rear wheel b eerting a couple on it. If the rear wheel eerts a 400-N horizontal force on the road and ou do not neglect the horizontal force eerted on the road b the front wheel, determine (a) the motorccle s acceleration and (b) the normal forces eerted on the road b the rear and front wheels. (The location of the center of mass of the motorccle not including its wheels, is shown.) 73 mm 649 mm B 1500 mm In the free-bod diagrams shown, m w = 9 kg and m = 160 18 = 14 kg. Let a be the motorccle s acceleration to the right and let α be the wheels clockwise angular acceleration. Note that a = 0.33α. (1) Front Wheel: F = B + f F = m ω a, () F = B + N F m ω g = 0, (3) B M = ff (0.33) = I α. (4) Rear Wheel: M mg B F = + f R = m ω a, (5) B B F = + N R m ω g = 0, (6) M m wg m wg M = M fr (0.33) = Iα. (7) f R f F N R N F Motorccle: F = B = ma, (8) F = B mg = 0, (9) M = M + ( + B )(0.73 0.33) + B (1.5 0.649) (0.649) = 0. (10) Solving Eqs (1) (10) with f R = 400 N, we obtain (a) a=.39 rad/s and (b) N R = 455 N, N F = 1115 N. 516

Problem 18.6 In Problem 18.61, if the front wheel lifts slightl off the road when the rider accelerates, determine (a) the motorccle s acceleration and (b) the torque eerted b the engine on the rear wheel. See the solution of Problem 18.61. We set N F = 0 and replace Eq. (4) b f F = 0. Then solving Eqs. (1) (10), we obtain (a) a = 9.34 m/s, (b) M = 516 N-m. Problem 18.63 The moment of inertia of the vertical handle about O is 0.1 slug-ft. The object B weighs 15 lb and rests on a smooth surface. The weight of the bar B is negligible (which means that ou can treat the bar as a two-force member). If the person eerts a 0.-lb horizontal force on the handle 15 in above O, what is the resulting angular acceleration of the handle? 6 in B O 1 in Let α be the clockwise angular acceleration of the handle. The acceleration of B is: F a B = a + α B r B/ : i j k a B i = (6/1)αi + 0 0 α B 1 0.5 0 15 in. 6 in. O β C we see that α B = 0 and a B = (6/1)α (1). The free bod diagrams of the handle and object B are as shown. Note that β = arctan(6/1) = 6.6. Newton s second law for the object B is β C 15 lb C cos β = (15/3.)a B, () The equation of angular motion for the handle is N (15/1)F (6/1)C cos β = (0.1)α (3). Solving Equations (1) (3) with F = 0. lb, we obtain α = 1.06 rad/s a = (6/1) B a B i 517

Problem 18.64 The bars are each 1 m in length and have a mass of kg. The rotate in the horizontal plane. Bar B rotates with a constant angular velocit of 4 rad/s in the counterclockwise direction. t the instant shown, bar BC is rotating in the counterclockwise direction at 6 rad/s. What is the angular acceleration of bar BC? 4 rad/s B 6 rad/s a BC C The FBD Given m = kg,l= 1m,θ= 45 B The kinematics a B = a + α B r B/ ω B r B/ = 0 + 0 (4 rad/s) (1m)i = (16 m/s )i B G θ a G = a B + α BC r G/B ω BC r G/B = (16 m/s )i + α BC k (0.5 m)(cos θi sin θj) (6 rad/s) (0.5 m)(cos θi sin θj) = ( 16 m/s + [0.5 m sin θ]α BC [18 m/s ] cos θ)i + ([0.5 m cos θ]α BC + [18 m/s ] sin θ)j Our kinematic constraints are a = 16 m/s + [0.5 m sin θ]α BC [18 m/s ] cos θ a = [0.5 m cos θ]α BC + [18 m/s ] sin θ The dnamic equations F : B = ma F : B = ma MG : B (0.5 m) sin θ B (0.5 m) cos θ = 1 1 m(1.0 m) α BC Solving we find α BC = 17.0 rad/s CCW 518

Problem 18.65 Bars OQ and PQ each weigh 6 lb. The weight of the collar P and friction between the collar and the horizontal bar are negligible. If the sstem is released from rest with θ = 45, what are the angular accelerations of the two bars? Q ft ft θ O P Let αoq and αp Q be the clockwise angular acceleration of bar OQ and the counterclockwise angular acceleration of bar PQ. The acceleration of Q is aq = a0 + α0q rq/0 i = 0 cos 45 j 0 sin 45 k αoq 0 Q α OQ α PQ G 45 P O Q = αoq sin 45 i αoq cos 45 j. Q The acceleration of P is O 6 lb Q Q 6 lb ap = aq + αp Q rp /Q N i ap i = αoq sin 45 i αoq cos 45 j + 0 cos 45 j 0 sin 45 Equating i and j components, From the diagrams: The equation of angular motion of bar OQ is M0 = I0 αoq : Q ( sin 45 ) Q ( cos 45 ) + 6 cos 45 = 13 (6/3.)() αoq ap = αoq sin 45 αp Q sin 45 0 = αoq cos 45 + αp Q cos 45 (1) The equations of motion of bar PQ are (). The acceleration of the center of mass of bar PQ is ag = aq + αp Q rg/q = αoq sin 45 i i αoq cos 45 j + 0 cos 45 k αp Q. 0 j 0 sin 45 k αp Q. 0 F = Q = (6/3.)aG (6) F = N Q 6 = (6/3.)aG (7) M = (N + Q + Q )(cos 45 ) = 1 (6/3.)() αp Q 1 (5). (8). Solving Equations (1) (8), we obtain αoq = αp Q = 6.83 rad/s Hence, ag = αoq sin 45 + αp Q sin 45 (3); ag = αoq cos 45 + αp Q cos 45 (4). c 008 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copright laws as the 519

Problem 18.66 In Problem 18.65, what are the angular accelerations of the two bars if the collar P weighs lb? In the solution of Problem 18.65, the free bod diagram of bar PQ has a horizontal component P to the left where P is the force eerted on the bar b the collar. Equations (6) and (8) become F = Q P = (6/3.)a G M = (N P + Q + Q )(cos 45 ) = 1 1 (6/3.)() α PQ and the equation of motion for the collar is P = (/3.)a P solving equations (1 9), we obtain α OQ = α PQ = 4.88 rad/s. Problem 18.67 The 4-kg slender bar is pinned to - kg sliders at and B. If friction is negligible and the sstem is released from rest in the position shown, what is the angular acceleration of the bar at that instant? Epress the acceleration of B in terms of the acceleration of, a B = a + α B r B/ : a B cos 45 i a B sin 45 i j k j = a j + 0 0 α B 0.5 1. 0, 1. m or a B cos 45 = 1.α B, (1); and a B sin 45 = a + 0.5α B, (). We epress the acceleration of G in terms of the acceleration of, a G = a + α B r G/ : 45 0.5 m B i j k a G = a G i + a G j = a j + 0 0 α B 0.5 0.6 0, or a G = 0.6α B, (3); and a G = a + 0.5α B, (4); The free bod diagrams are as shown. The equations of motion are Slider : a G α B B a B N = 0 (5), and ()(9.81) + = a, (6); Slider B : P [B + B + ()(9.81)] cos 45 = 0, (7); and [()(9.81) B + B ] cos 45 = a B, (8); Bar: + B k = 4a G (9); B (4)(9.81) B B N and + B (4)(9.81) = 4a G (10); ()(9.81) (L/)[(B ) cos β + (B ) sin β] = 1 1 (4)L α B (11), B P where L = (0.5) + (1.) m and β arctan(0.5/1.) =.6. Solving Equations (1) (11), we obtain α B = 5.18 rad/s. B ()(9.81) 50

Problem 18.68 The mass of the slender bar is m and the mass of the homogeneous disk is 4m. The sstem is released form rest in the position shown. If the disk rolls and the friction between the bar and the horizontal surface is negligible, show that the disk s angular acceleration is α = 6g/95R counterclockwise. R R For the bar: The length of the bar is L = 5R. ppl Newton s second law to the free bod diagram of the bar: B = ma G, B + N mg = ma G, where a G,a G are the accelerations of the center of mass of the bar. The moment about the bar center of mass is RB RN R B = I B α B. N mg B B B B 4 mg N D f For the disk: ppl Newton s second law and the equation of angular motion to the free bod diagram of the disk. f B = 4ma D, N D 4mg B = 0, RB + Rf = I D α D From kinematics: Since the sstem is released from rest, ω B = ω D = 0. The acceleration of the center of the disk is a D = Rα D i. The acceleration of point B in terms of the acceleration of the center of the disk is i j k a B = a D + α D r B/D = a D + 0 0 α D = Rα D i Rα D j. R 0 0 The acceleration of the center of mass of the bar in terms of the acceleration of B is i j k a G = a B + α B r G/B ωb r G/B = a B + 0 0 α B R R 0 = a B + Rα B i Rα B j, ( a G = R α D α B ) i R(α D + α B )j. The acceleration of the center of mass of the bar in terms of the acceleration of is i j k a G = a + a B r G/ = a + 0 0 α B R R 0 = a Rα B i + Rα B j. From the constraint on the motion, a = a i. Equate the epressions for a G, separate components and solve: α B = α D. Substitute to obtain a G = 5R 4 α D, a G = R α D. Collect the results: (3) RB RN R B = I B α D, (4) f B = 4Rmα D, (5) N D 4mg B = 0, (6) RB + Rf = I D α D. From (1), (), and (3) B = mg ( 9mR 16 + I ) B α D. 4R From (1), (4) and (6), ( ID B = R + 1Rm ) α D. 4 Equate the epressions for B and reduce to obtain α D = ( mg ) ( 93Rm 16 1 + I D R + I ). B 4R For a homogenous clinder of mass 4m, I D = R m. For a slender bar of mass m about the center of mass, I B = 1 1 ml = 5 1 mr. Substitute and reduce: a D = 6g 95R. (1) B = 5Rm 4 α D, () B + N mg = Rm α D, 51

Problem 18.69 Bar B rotates in the horizontal plane with a constant angular velocit of 10 rad/s in the counterclockwise direction. The masses of the slender bars BC and CD are 3 kg and 4.5 kg, respectivel. Determine the and components of the forces eerted on bar BC b the pins at B and C at the instant shown. 10 rad/s B C 0. m D 0. m 0. m First let s do the kinematics 0. m Velocit v B = v + ω B r B/ B G C = 0 + (10 rad/s)k (0. m)j 0. m = ( m/s)i 10 rad/s v C = v B + ω BC r C/B = ( m/s)i + ω BC k (0. m)i = ( m/s)i + (0. m)ω BC j 0.4 m D v D = v C + ω CD r D/C C = ( m/s)i + (0. m)ω BC j + ω CD k (0. m)(i j) B = ( [ m/s] + [0. m]ω CD )i + (0. m)(ω BC + ω CD )j B C C Since D is pinned we find ω CD = 10 rad/s, ω BC = 10 rad/s cceleration D a B = a + α B r B/ ω B r B/ D = 0 + 0 (10 rad/s) (0. m)j = (0 m/s )j a C = a B + α BC r C/B ω BC r C/B = (0 m/s )j + α BC k (0. m)i ( 10 rad/s) (0. m)i = (0 m/s )i + ([0. m]α BC 0 m/s )j The FBDs The dnamics FBC : B + C = (3 kg)( 10 m/s ) FBC : B + C = (3 kg)( 40 m/s ) a D = a C + α CD r D/C ω CD r D/C MG1 : (C B )(0.1 m) = 1 1 (3kg)(0. m) ( 00 rad/s ) = (0 m/s )i + ([0. m]α BC 0 m/s )j + α CD k MD : C (0. m) + C (0. m) = 1 3 (4.5 kg)( [0. m]) (00 rad/s ) (0. m)(i j) (10 rad/s) (0. m)(i j) = ( 40 m/s + [0. m]α CD )i + ([0. m][α BC + α BC ])j Since D is pinned we find α BC = 00 rad/s, α CD = 00 rad/s Now find the accelerations of the center of mass G. a G = a B + α BC r G1/B ω BC r G1/B Solving we find B = 0 N, B = 50 N C = 190 N, C = 70 N = (0 m/s )j + ( 00 rad/s )k (0.1 m)i ( 10 rad/s) (0.1 m)i = ( 10i 40j) m/s 5

Problem 18.70 The -kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar slides on the smooth bar. t the instant shown, r = 1. m,ω = 0.4 rad/s, and the collar is sliding outward at 0.5 m/s relative to the bar. If ou neglect the moment of inertia of the collar (that is, treat the collar as a particle), what is the bar s angular acceleration? ω Strateg: Draw individual free-bod diagrams of the bar and collar and write Newton s second law for the collar in terms of polar coordinates. r m Diagrams of the bar and collar showing the force the eert on each other in the horizontal plane are: the bar s equation of angular motion is e θ N e r M0 = I 0 α: Nr = 1 3 ()() α (1) r N In polar coordinates, Newton s second law for the collar is O [( d ) ( r F = ma: Neθ = m dt rω e r + rα + dr ) ] dt ω e θ. Equating e θ components, ( N = m rα + dr ) dt ω = (6)[rα + (0.5)(0.4)] (). Solving Equations (1) and () with r = 1. m gives α = 0.55 rad/s Problem 18.71 In Problem 18.70, the moment of inertia of the collar about its center of mass is 0. kg-m. Determine the angular acceleration of the bar, and compare our answer with the answer to Problem 18.70. Let C be the couple the collar and bar eert on each other: The bar s equation of angular motion is M0 = I 0 α: Nr C = 1 3 ()() α (1). The collar s equation of angular motion is M = Iα: C = 0.α (). From the solution of Problem 18.70, the e θ component of Newton s second law for the collar is N = (6)[rα + (0.5)(0.4)] (3) Solving Equations (1) (3) with r = 1. m gives α = 0.50 rad/s. 53

Problem 18.7 The ais L 0 is perpendicular to both segments of the L-shaped slender bar. The mass of the bar is 6 kg and the material is homogeneous. Use integration to determine the moment of inertia of the bar about L 0. 1 m L O m Let be the bar s cross-sectional area. The bar s mass is m = 6kg= ρ(3 m), soρ = kg/m. dm d For the horizontal part (Fig. a), I h = dm = m 0 ρd = 8 16 ρ = 3 3 kg-m. L O d dm LO r For the vertical part (Fig. b), (a) (b) I v = r dm = m 1 ( + )ρ d 0 = 13 3 ρ = 6 3 kg-m. Therefore I 0 = I h + I v = 14 kg-m. Problem 18.73 Two homogenous slender bars, each of mass m and length l, are welded together to form the T- shaped object. Use integration to determine the moment of inertia of the object about the ais through point O that is perpendicular to the bars. O l l Divide the object into two pieces, each corresponding to a slender bar of mass m; the first parallel to the -ais, the second to the -ais. B definition l I = r dm + r dm. 0 m For the first bar, the differential mass is dm = ρdr. ssume that the second bar is ver slender, so that the mass is concentrated at a distance l from O. Thus dm = ρd, where lies between the limits l l. The distance to a differential d is r = l +. Thus the definition becomes l 1 I = ρ r dr + ρ 0 l (l + )d [ r 3 ] l I = ρ 3 + ρ 0 ] [l 1 + 3 3 1 ( 1 = ml 3 + 1 + 1 ) = 17 1 1 ml 54

Problem 18.74 The slender bar lies in the plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the z ais. m 50 1 m 1m I z = ρ d 0 The densit is ρ = 6kg 3m = kg/m m + ρ[(1 m+ s cos 50 ) + (s sin 50 ) ] ds 0 m I z = 15.1 kg-m 50 1 m Problem 18.75 The slender bar lies in the plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the ais. 1m I = ρ d 0 The densit is ρ = 6kg 3m = kg/m m + ρ[(1 m+ s cos 50 ) ] ds 0 m I = 1.0 kg-m 50 1 m 55

Problem 18.76 The homogeneous thin plate has mass m = 1 kg and dimensions b = 1 m and h = m. Determine the mass moments of inertia of the plate about the,, and z aes. Strateg: The mass moments of inertia of a thin plate of arbitrar shape are given b Eqs. (18.37) (18.39) in terms of the moments of inertia of the cross-sectional area of the plate. You can obtain the moments of inertia of the triangular area from ppendi B. b h m = 1 kg rea = 1 bh ρ = mass/rea h dm = ρd From ppendi B, I = 1 36 bh3 rea = 1 (1)() = 1m ρ = 1 kg/m I = 1 36 hb3 I = ρ d = ρ d b I = ρi = 1 36 h b3 = 1 3 ()(1)3 I = 0.667 kg-m I z = I + I I z =.667 + 0.667 kg-m I z = 3.333 kg-m I = ρi,i = ρi ( ) 1 I = 1 (1)() 3 =.667 kg-m 36 56

Problem 18.77 The brass washer is of uniform thickness and mass m. (a) (b) Determine its moments of inertia about the and z aes. Let R i = 0, and compare our results with the values given in ppendi C for a thin circular plate. R i R o (a) The area moments of inertia for a circular area are R i R o I = I = πr4 4. For the plate with a circular cutout, I = π 4 (R4 o R4 i ) The area mass densit is m, thus for the plate with a circular cut, m = m π(ro R i ), from which the moments of inertia I (-ais) = m(r4 o R4 i ) 4(R o R i ) = m 4 (R o + R i ) I (z-ais) = I (-ais) = m (R o + R i ). (b) Let R i = 0, to obtain I -ais = m 4 R o, I (z-ais) = m R o, which agrees with table entries. 57

Problem 18.78 The homogenous thin plate is of uniform thickness and weighs 0 lb. Determine its moment of inertia about the ais. 1 = 4 4 ft I = r dm. m The definition of the moment of inertia is The distance from the -ais is, where varies over the range 4 4. Let τ = m = W be the area mass densit. The mass g of an element d is dm = W d. Substitute into the definition: g I -ais = W g The area is 4 4 ( ) 4 d 4 = W [ 4 3 ] +4 g 3 5 = W 0 4 g [68.667]. 4 ) ] 4 = (4 d = [4 3 = 1.333 ft 4 4 1 4 The moment of inertia about the -ais is I (-ais) = W g (3.) = 0 3.17 (3.) = 1.99 slug-ft. Problem 18.79 Determine the moment of inertia of the plate in Problem 18.78 about the ais. The differential mass is dm = W d d. The distance g of a mass element from the -ais is, thus I = W g = W 3g +4 4 +4 4 4 4 d d 0 ) 3 (4 d 4 = W [64 4 3 + 30 ] 4 3g 5 7 448 4 = W 3g [34.057]. From the solution to Problem 18.78, = 1.333 ft. Thus the moment of inertia about the -ais is I -ais = W 3g (34.057) (1.333) = W g (3.657) =.7 slug-ft. 58

Problem 18.80 The mass of the object is 10 kg. Its moment of inertia about L 1 is 10 kg-m. What is its moment of inertia about L? (The three aes are in the same plane.) 0.6 m 0.6 m L L 1 L The strateg is to use the data to find the moment of inertia about L, from which the moment of inertia about L can be determined. I L = (0.6) (10) + 10 = 6.4 kg-m, from which I L = (1.) (10) + 6.4 = 0.8 kg-m Problem 18.81 n engineer gathering data for the design of a maneuvering unit determines that the astronaut s center of mass is at = 1.01 m, = 0.16 m and that her moment of inertia about the z ais is 105.6 kg-m. The astronaut s mass is 81.6 kg. What is her moment of inertia about the z ais through her center of mass? The distance from the z ais to the z ais is d = + = 1.057 m. The moment of inertia about the z ais is I z -ais = d m + I z-ais = (1.0457)(81.6) + 105.6 = 0.7 kg-m Problem 18.8 Two homogenous slender bars, each of mass m and length l, are welded together to form the T-shaped object. Use the parallel-ais theorem to determine the moment of inertia of the object about the ais through point O that is perpendicular to the bars. O l l Divide the object into two pieces, each corresponding to a bar of mass m. Bdefinition I = l 0 r dm. For the first bar, the differential mass is dm = ρdr, from which the moment of inertia about one end is l [ r I 1 = ρ r 3 ] l dr = ρ 0 3 0 = ml 3. is the moment of inertia about the center of the bar. From the parallel ais theorem, the moment of inertia about O is I 0 = ml 3 + l m + ml 1 = 17 1 ml For the second bar l [ r I = ρ l r 3 ] l dr = ρ 3 l = ml 1 59

Problem 18.83 Use the parallel-ais theorem to determine the moment of inertia of the T-shaped object in Problem 18.98 about the ais through the center of mass of the object that is perpendicular to the two bars. ( ) l m + lm = = 3 m 4 l. The location of the center of mass of the object is Use the results of Problem 18.98 for the moment of inertia of a bar about its center. For the first bar, I 1 = ( ) l m + ml 4 1 = 7 48 ml. For the second bar, I = ( ) l m + ml 4 1 = 7 48 ml. The composite: I c = I 1 + I = 7 4 ml Check: Use the results of Problem 18.98: I c = ( ) 3l (m) + 17 4 1 ml ( 9 = 8 + 17 ) ml = 7 1 4 ml. check. Problem 18.84 The mass of the homogeneous slender bar is 30 kg. Determine its moment of inertia about the z ais. 0.8 m 0.6 m m The densit is ρ = 30 kg 3m = 10 kg/m I z = 1 3 (10 kg)(1.0 m) + 1 (0 kg)( m) 1 + (0 kg)[(1.6 m) + (0.8 m) ] I z = 74 kg-m 530

Problem 18.85 The mass of the homogeneous slender bar is 30 kg. Determine the moment of inertia of the bar about the z ais through its center of mass. = First locate the center of mass (10 kg)(0.3 m) + (0 kg)(1.6 m) 30 kg = 1.167 m (10 kg)(0.4 m) + (0 kg)(0.8 m) = = 0.667 m 30 kg Using the answer to 18.100 I z = (74 kg-m ) (30 kg)(1.167 + 0.667 )m I z = 19.8 kg-m Problem 18.86 The homogeneous slender bar weighs 5 lb. Determine its moment of inertia about the z ais. 4 in The Bar s mass is m = 5/3. slugs. Its length is L = L 1 + L + L 3 = 8 + 8 + 8 + π(4) = 31.9 in. The masses of the parts are therefore, M 1 = L ( )( ) 1 8 5 L m = = 0.0390 slugs, 31.9 3. M = L L m = ( (64) 31.9 )( ) 5 = 0.0551 slugs, 3. M 3 = L ( )( ) 3 4π 5 L m = = 0.061 slugs. 31.9 3. The center of mass of part 3 is located to the right of its center C a distance R/π = (4)/π =.55 in. The moment of inertia of part 3 about C is r dm = m 3 r = (0.061)(4) = 0.979 slug-in. m 3 The moment of inertia of part 3 about the center of mass of part 3 is therefore I 3 = 0.979 m 3 (.55) = 0.58 slug-in. The moment of inertia of the bar about the z ais is 8 in I (z ais) = 1 3 m 1L 1 + 1 3 m L + I 3 + m 3 [(8 +.55) + (4) ] = 11.6 slug-in = 0.0803 slug-ft. 531

Problem 18.87 Determine the moment of inertia of the bar in Problem 18.86 about the z ais through its center of mass. In the solution of Problem 18.86, it is shown that the moment of inertia of the bar about the z ais is I (z ais) = 11.6 slug-in. The and coordinates of the center of mass coincide with the centroid of the bar: = 1L 1 + L + 3 L 3 L 1 + L + L 3 = (4)(8) + (4) [ 8 + 8 + 8 + (4) ] π(4) π 8 + 8 + 8 + π(4) = 6.58 in, = 1L 1 + L + 3 L 3 L 1 + L + L 3 = 0 + (4) 8 + 8 + π(4)(4) 8 + 8 + 8 + π(4) = 3.00 in. The moment of inertia about the z ais is ( ) 5 I (z ais) = I (z ais) ( + ) = 3.44 slug-in. 3. Problem 18.88 The rocket is used for atmospheric research. Its weight and its moment of inertia about the z ais through its center of mass (including its fuel) are 10,1000 lb and 10,00 slug-ft, respectivel. The rocket s fuel weighs 6000 lb, its center of mass is located at = 3 ft, = 0, and z = 0, and the moment of inertia of the fuel about the ais through the fuel s center of mass parallel to z ais is 00 slug-ft. When the fuel is ehausted, what is the rocket s moment of inertia about the ais through its new center of mass parallel to z ais? Denote the moment of inertia of the empt rocket as I E about a center of mass E, and the moment of inertia of the fuel as I F about a mass center F. Using the parallel ais theorem, the moment of inertia of the filled rocket is I R = I E + E m E + I F + F m F, about a mass center at the origin ( R = 0). Solve: I E = I R E m E I F F m F. The objective is to determine values for the terms on the right from the data given. Since the filled rocket has a mass center at the origin, the mass center of the empt rocket is found from From which E = ( ) 186.335 ( 3) = 4.5 ft 14.4 is the new location of the center of mass. Substitute: I E = I R E m E I F F m F = 1000 515.5 00 1677.01 = 3810 slug-ft 0 = m E E + m F F, from which ( ) mf E = F. m E Using a value of g = 3 ft/s, m F = W F g = 6000 = 186.34 slug, 3. m E = (W R W F ) g = 10000 6000 3. = 14.3 slug. 53

Problem 18.89 The mass of the homogeneous thin plate is 36 kg. Determine the moment of inertia of the plate about the ais. 0.4 m 0.4 m 0.3 m 0.3 m Divide the plate into two areas: the rectangle 0.4 m b 0.6 m on the left, and the rectangle 0.4 m b 0.3 m on the right. The mass densit is ρ = m. The area is = (0.4)(0.6) + (0.4)(0.3) = 0.36 m, from which ρ = 36 0.36 = 100 kg/m. The moment of inertia about the -ais is I -ais = ρ ( ) 1 ( ) 1 3 (0.4)(0.6 3 ) + ρ 3 (0.4)(0.3) 3 = 3.4 kg-m Problem 18.90 Determine the moment of inertia of the 36-kg plate in Problem 18.89 about the z ais. The basic relation to use is I z-ais = I -ais + I -ais. The value of I -ais is given in the solution of Problem 18.89. The moment of inertia about the -ais using the same divisions as in Problem 8.89 and the parallel ais theorem is I -ais = ρ from which ( ) 1 ( ) 1 3 (0.6)(0.4) 3 + ρ 1 (0.3)(0.4) 3 + (0.6) ρ(0.3)(0.4) = 5.76 kg-m, I z-ais = I -ais + I -ais = 3.4 + 5.76 = 9 kg-m 533

Problem 18.91 The mass of the homogeneous thin plate is 0 kg. Determine its moment of inertia about the ais. 1000 mm 400 mm 400 mm 00 mm 00 mm Break the plate into the three regions shown. = (0. m)(0.8 m) + (0. m)(0.4 m) 1000 + 1 (0.4 m)(0.6 m) = 0.36 m 0 kg ρ = = 55.6 kg/m 0.36 m Using the integral tables we have I = 1 3 (0. m)(0.8 m)3 + 1 1 (0. m)(0.4 m)3 + (0. m)(0.4 m)(0.6 m) 800 00 400 + 1 36 (0.6 m)(0.4 m)3 + 1 (0.6 m)(0.4 m)(0.667 m) = 0.1184 m 4 I ais = (55.6 kg/m )(0.1184 m 4 ) = 6.58 kg-m 00 534

Problem 18.9 The mass of the homogeneous thin plate is 0 kg. Determine its moment of inertia about the ais. See the solution to 18.91 I = 1 3 (0.8 m)(0. m)3 + 1 1 (0.4 m)(0. m)3 + (0. m)(0.4 m)(0.3 m) + 1 36 (0.4 m)(0.6 m)3 + 1 (0.6 m)(0.4 m)(0.6 m) = 0.055 m 4 I ais = (55.6 kg/m )(0.055 m 4 ) = 3.07 kg-m Problem 18.93 The thermal radiator (used to eliminate ecess heat from a satellite) can be modeled as a homogeneous thin rectangular plate. The mass of the radiator is 5 slugs. Determine its moments of inertia about the,, and z aes. 3 ft 6 ft 3 ft ft The area is = 9(3) = 7 ft. The mass densit is ρ = m = 5 7 = 0.185 slugs/ft. The moment of inertia about the centroid of the rectangle is ( ) 1 I c = ρ 9(3 3 ) = 3.75 slug-ft, 1 ( ) 1 I c = ρ 3(9 3 ) = 33.75 slug-ft. 1 Use the parallel ais theorem: I -ais = ρ( + 1.5) + I c = 65 slug-ft, I -ais = ρ(4.5 3) + I c = 45 slug-ft. I z-ais = I -ais + I -ais = 110 slug-ft 535

Problem 18.94 The mass of the homogeneous thin plate is kg. Determine the moment of inertia of the plate about the ais through point O that is perpendicular to the plate. 80 mm 10 mm 30 mm O 130 mm 30 mm B determining the moments of inertia of the area about the and aes, we will determine the moments of inertia of the plate about the and aes, then sum them to obtain the moment of inertia about the z ais, which is I 0. The areas are 1 = 1 (130)(80) mm, = π(10) mm. Using ppendi B, I = 1 [ ] 1 1 (130)(80)3 4 π(10)4 + (30) O 1 100 mm 30 mm = 5.6 10 6 mm 4, I = 1 [ ] 1 4 (80)(130)3 4 π(10)4 + (100) = 40.79 10 6 mm 4. Therefore I ( ais) = I ( ais) = Then m 1 I = 150 kg-mm, m 1 I = 16700 kg-mm. I (z ais) = I ( ais) + I ( ais) = 18850 kg-mm. I (z ais) = 0.0188 kg-m. 536

Problem 18.95 The homogeneous cone is of mass m. Determine its moment of inertia about the z ais, and compare our result with the value given in ppendi C. (See Eample 18.10.) Strateg: Use the same approach we used in Eample 18.10 to obtain the moments of inertia of a homogeneous clinder. R h z The differential mass ( m ) dm = πr dz = 3m V R h r dz. The moment of inertia of this disk about the z-ais is 1 mr. The ( ) R radius varies with z, r = z, from which h I z-ais = 3mR h 5 h 0 z 4 dz = 3mR h 5 [ z 5 ] h 5 0 = 3mR 10 Problem 18.96 Determine the moments of inertia of the homogeneous cone in Problem 18.95 about the and aes, and compare our results with the values given in ppendi C. (See Eample 18.10.) The mass densit is ρ = m V = 3m πr. The differential h element of mass is dm = ρπr dz.. The moment of inertia of this elemental disk about an ais ( ) through its center of mass, parallel to the 1 - and -aes, is di = r dm. Use the parallel ais theorem, 4 ( ) 1 I = r dm + z dm. m 4 m Noting that r = R z, then h ( πr 4 ) r dm = ρ h 4 z 4 dz, ( πr ) and z dm = ρ h z 4 dz. Substitute: ( πr 4 ) h ( πr I = ρ 4h 4 z 4 ) h dz + ρ 0 h z 4 dz, 0 ( 3mR I = 4h 5 + 3m )[ z 5 ] h h 3 5 0 ( 3 = m 0 R + 3 ) 5 h = I. 537

Problem 18.97 The homogeneous object has the shape of a truncated cone and consists of bronze with mass densit ρ = 800 kg/m 3. Determine the moment of inertia of the object about the z ais. 60 mm 180 mm z 180 mm Consider an element of the cone consisting of a disk of thickness dz: We can epress the radius as a linear function of zr = az + b. Using the conditions that r = 0atz = 0 and r = 0.06 m at z = 0.36 m to evaluate a and b we find that r = 0.167 z. From ppendi C, the moment of inertia of the element about the z ais is r (I z ) element = 1 mr = 1 [ρ(πr )dz]r = 1 ρπ(0.167z)4 dz. z dz z We integrate this result to obtain the mass moment of inertia about the z ais for the cone: 0.36 [ 1 z 5 ] 0.36 I (z ais) = 0.18 ρπ(0.167)4 5 0.18 = 1 [ z 5 ] 0.36 (800)π(0.167)4 5 0.18 = 0.0116 kg-m. Problem 18.98 Determine the moment of inertia of the object in Problem 18.97 about the ais. Consider the disk element described in the solution to Problem 18.97. The radius of the laminate is r = 0.167z. Using ppendi C and the parallel ais theorem, the moment of inertia of the element about the ais is (I ) element = 1 4 mr + mz = 1 4 [ρ(πr )dz]r + [ρ(πr )dz]z = 1 4 ρπ(0.167z)4 dz + ρπ(0.167z) z dz. Integrating the result, I ( ais) = 1 0.36 0.36 4 ρπ(0.167)4 z 4 dz + ρπ(0.167) z 4 dz 0.18 0.18 = 0.844 kg-m. 538

Problem 18.99 The homogeneous rectangular parallelepiped is of mass m. Determine its moments of inertia about the,, and z aes and compare our results with the values given in ppendi C. a Consider a rectangular slice normal to the -ais of dimensions b b c and mass dm. The area densit of this slice is ρ = dm. The moment of inertia about the ais of the centroid of a thin bc plate is the product of( the area ) densit and the area moment ( ) of inertia 1 1 of the plate: di = ρ bc 3, from which di = c dm. B 1 1 smmetr, the moment of inertia about the z ais is ( ) 1 di z = b dm. 1 z c B smmetr, the argument can be repeated for each coordinate, to obtain I = m 1 (a + c ) I z = m 1 (b + a ) b Since the labeling of the - - and z-aes is arbitrar, di = di z + di, where the -ais is normal to the area of the plate. Thus ( ) 1 (b di = + c ) dm, 1 from which ( ) 1 I = (b + c ) 1 m dm = m 1 (b + c ). Problem 18.100 The sphere-capped cone consists of material with densit 7800 kg/m 3. The radius R = 80 mm. Determine its moment of inertia about the ais. z R 4R Given ρ = 7800 kg/m 3, R = 0.08 m Using the tables we have I = 3 ( ρ 1 ) 10 3 πr [4R] R + 5 (ρ 3 πr3 ) R I = 0.0535 kg-m 539

Problem 18.101 Determine the moment of inertia of the sphere-capped cone described in Problem 18.100 about the ais. The center of mass of a half-sphere is located a distance 3R from the geometric center of the circle. 8 ( I = ρ 1 )( 3 3 πr [4R] 5 [4R] + 3 ) 0 R + (ρ 3 ) 5 πr3 R ( ρ )( ) 3R 3 πr3 + (ρ 3 )( 8 πr3 4R + 3R ) 8 I =.08 kg-m Problem 18.10 The circular clinder is made of aluminum (l) with densit 700 kg/m 3 and iron (Fe) with densit 7860 kg/m 3. Determine its moment of inertia about the ais. l I = 1 [(700 kg/m )π(0.1 m) (0.6 m)](0.1 m) z 600 mm z 600 mm Fe 00 mm, + 1 [(7860 kg/m )π(0.1 m) (0.6 m)](0.1 m) I = 0.995 kg-m Problem 18.103 Determine the moment of inertia of the composite clinder in Problem 18.10 about the ais. First locate the center of mass = [(700 kg/m 3 )π(0.1 m) (0.6 m)](0.3 m) + [(7860 kg/m 3 )π(0.1 m) (0.6 m)](0.9 m) (700 kg/m 3 )π(0.1 m) (0.6 m) + (7860 kg/m 3 )π(0.1 m) (0.6 m) = 0.747 m [ 1 I = [(700 kg/m 3 )π(0.1 m) (0.6 m)] 1 (0.6 m) + 1 ] (0.1 m) 4 + [(700 kg/m 3 )π(0.1 m) (0.6 m)]( 0.3 m) [ 1 + [(7680 kg/m 3 )π(0.1 m) (0.6 m)] 1 (0.6 m) + 1 ] (0.1 m) 4 + [(7680 kg/m 3 )π(0.1 m) (0.6 m)](0.9 m ) I = 0.1 kg-m 540

Problem 18.104 The homogeneous machine part is made of aluminum allo with mass densit ρ = 800 kg/m 3. Determine the moment of inertia of the part about the z ais. 0 mm z 10 mm 40 mm 40 mm We divide the machine part into the 3 parts shown: (The dimension into the page is 0.04 m) The masses of the parts are m 1 = (800)(0.1)(0.08)(0.04) = 1.075 kg, 0.08 m 0.1 m 1 m = (800) 1 π(0.04) (0.04) = 0.81 kg, m 3 = (800)π(0.0) (0.04) = 0.141 kg. 0.1 m + 0.1 C m 3 0.04 m 0.0 m Using ppendi C and the parallel ais theorem the moment of inertia of part 1 about the z ais is I (z ais)1 = 1 1 m 1[(0.08) + (0.1) ] + m 1 (0.06) = 0.00573 kg-m. The moment of inertia of part about the ais through the center C that is parallel to the z ais is 1 m R = 1 m (0.04) The distance along the ais from C to the center of mass of part is 4(0.04)/(3π) = 0.0170 m. Therefore, the moment of inertia of part about the z ais through its center of mass that is parallel to the ais is 1 m (0.04) m (0.0170) = 0.000144 kg-m. Using this result, the moment of inertia of part about the z ais is I (z ais) = 0.000144 + m (0.1 + 0.017) = 0.00544 kg-m. The moment of inertia of the material that would occup the hole 3 about the z ais is I (z ais)3 = 1 m 3(0.0) + m 3 (0.1) = 0.0005 kg-m. Therefore, I (z ais) = I (z ais)1 + I (z ais) I (z ais)3 = 0.00911 kg-m. 541

Problem 18.105 Determine the moment of inertia of the machine part in Problem 18.104 about the ais. We divide the machine part into the 3 parts shown in the solution to Problem 18.104. Using ppendi C and the parallel ais theorem, the moments of inertia of the parts about the ais are: I ( ais)1 = 1 1 m 1[(0.08) + (0.04) ] = 0.0007168 kg-m I ( ais) = m [ 1 1 (0.04) + 1 4 (0.04) ] = 0.0001501 kg-m I ( ais)3 = m 3 [ 1 1 (0.04) + 1 4 (0.0) ] Therefore, = 0.000038 kg-m. I (ais) = I ( ais)1 + I ( ais) I ( ais)3 = 0.000834 kg-m. Problem 18.106 The object shown consists of steel of densit ρ = 7800 kg/m 3 of width w = 40 mm. Determine the moment of inertia about the ais L 0. 0 mm O Divide the object into four parts: 100 mm Part (1): The semi-clinder of radius R = 0.0 m, height h 1 = 0.01 m. Part (): The rectangular solid L = 0.1 mbh = 0.01 m b w = 0.04 m. Part (3): The semi-clinder of radius R = 0.0 m, h 1 = 0.01 m Part (4): The clinder of radius R = 0.0 m, height h = 0.03 m. Part (1) m 1 = ρπr h 1 = 0.049 kg, 10 mm 30 mm Part (4) L 0 m 4 = ρπr h = 0.94 kg, I 4 = ( 1 ) m4 (R ) + m 4 L = 0.003 kg-m. Part () I 1 = m 1R = 4.9 10 6 kg-m, 4 m = ρwlh = 0.31 kg, I = (1/1)m (L + w ) + m (L/) The composite: I L0 = I 1 + I I 3 + I 4 = 0.00367 kg-m = 0.00108 kg-m. Part (3) m 3 = m 1 = 0.049 kg, I 3 = ( ) 4R ( m + I 1 + m 3 L 4R ) 3π 3π = 0.00041179 kg-m. 54

Problem 18.107 Determine the moment of inertia of the object in Problem 18.106 about the ais through the center of mass of the object parallel to L 0. The center of mass is located relative to L 0 is given b = ( m 1 4R ) ( + m (0.05) m 3 0.1 4R ) + m 4 (0.1) 3π 3π m 1 + m m 3 + m 4 = 0.066 m, I c = m + I Lo = 0.0065 + 0.00367 = 0.0010 kg-m Problem 18.108 The thick plate consists of steel of densit ρ = 15 slug/ft 3. Determine the moment of inertia of the plate about the z ais. Divide the object into three parts: Part (1) the rectangle 8 in b 16 in, Parts () & (3) the clindrical cut outs. Part (1): m 1 = ρ8(16)(4) = 4.444 slugs. 4 in 4 in in in 4 in 8 in 4 in z 4 in I 1 = (1/1)m 1 (16 + 8 ) = 118.5 slug-in. ( ) 1 Part (): m = ρπ( )(4) 1 3 = 0.4363 slug, I = m ( ) + m (4 ) = 7.854 slug-in. Part (3): m 3 = m = 0.4363 slugs, The composite: I 3 = I = 7.854 slug-in. I z-ais = I 1 I = 10.81 slug-in I z-ais = 0.715 slug-ft Problem 18.109 Determine the moment of inertia of the object in Problem 18.108 about the ais. Use the same divisions of the object as in Problem 18.108. ( ) 1 Part (1) : I 1-ais = m 1 (8 + 4 ) = 9.63 slug-in, 1 Part () : I -ais = (1/1)m (3( ) + 4 ) = 1.018 slug-in. The composite: I -ais = I 1-ais I -ais = 7.59 slug in = 0.1916 slug ft 543

Problem 18.110 The airplane is at the beginning of its takeoff run. Its weight is 1000 lb. and the initial thrust T eerted b its engine is 300 lb. ssume that the thrust is horizontal, and neglect the tangential forces eerted on its wheels. (a) (b) If the acceleration of the airplane remains constant, how long will it take to reach its takeoff speed of 80 mi/hr? Determine the normal force eerted on the forward landing gear at the beginning of the takeoff run. T 6 in 1 ft 7 ft The acceleration under constant thrust is a = T m = 300(3.) 1000 = 9.66 ft/s. The time required to reach 80 mph = 117.33 ft/s is T W t = v a = 117.33 9.66 = 1.1 s The sum of the vertical forces: F = R + F W = 0. The sum of the moments: M = 7F 0.5T 1R = 0. Solve: R = 856.5 lb, F = 143.75 lb R F Problem 18.111 The pulles can turn freel on their pin supports. Their moments of inertia are I = 0.00 kg-m, I B = 0.036 kg-m, and I C = 0.03 kg-m. The are initiall stationar, and at t = 0 a constant M = N-m is applied at pulle. What is the angular velocit of pulle C and how man revolutions has it turned at t = s? 100 mm 100 mm 00 mm B C 00 mm Denote the upper and lower belts b the subscripts U and L. Denote the difference in the tangential component of the tension in the belts b T = T L T U, T B = T LB T UB. R B1 T UB T U R R C B C T L T LB R B From the equation of angular motion: M + R T = I α, R B1 T + R B T B = I B α B, R C T B = I C α C. From kinematics, R α = R B1 α B, R B α B = R C α C, from which α = R B1R C R R B α C = (0.)(0.) (0.1)(0.1) α C = 4α C, α B = R C R B α C = 0. 0.1 α C = α C. Substitute and solve: α C = 38.5 rad/s, from which ω C = α C t = 76.9 rad/s ( ) 1 N = θ = α C π 4π ( ) = 1. revolutions 544

Problem 18.11 -kg bo is subjected to a 40-N horizontal force. Neglect friction. 40 N (a) (b) If the bo remains on the floor, what is its acceleration? Determine the range of values of c for which the bo will remain on the floor when the force is applied. c B 100 mm 100 mm (a) From Newton s second law, 40 = ()a, from which 40 N C a = 40 = 0 m/s. mg (b) The sum of forces: F = + B mg = 0. The sum of the moments about the center of mass: M = 0.1B 0.1 40c = 0. Substitute the value of B from the first equation into the second equation and solve for c: 100 mm 100 mm B c = (0.1)mg (0.) 40 The bo leg at will leave the floor as 0, from which c (0.1)()(9.81) 40 0.0491 m for values of 0. 545

Problem 18.113 The slender, -slug bar B is 3 ft long. It is pinned to the cart at and leans against it at B. B a (a) (b) If the acceleration of the cart is a = 0 ft/s, what normal force is eerted on the bar b the cart at B? What is the largest acceleration a for which the bar will remain in contact with the surface at B? 60 Newton s second law applied to the center of mass of the bar ields B B + = ma G, W = ma G, ( ) L cos θ + (B + ) ( L sin θ ) = I G α, W where a G, a G kinematics, are the accelerations of the center of mass. From a G = a + α r G/ ω B r G/ = 0i ft/s where α = 0, ω B = 0 so long as the bar is resting on the cart at B and is pinned at. Substitute the kinematic relations to obtain three equations in three unknowns: B + = ma, W = 0, ( ) L cos θ + (B + ) ( L sin θ ) = 0. Solve: B = W cot θ ma. For W = mg = 64.34 lb, θ = 60, m = slug, and a = 0 ft/s, B = 1.43 lb, from which the bar has moved awa from the cart at point B. (b) The acceleration that produces a zero normal force is a = g cot θ = 18.57 ft/s. 546

Problem 18.114 To determine a 4.5-kg tire s moment of inertia, an engineer lets the tire roll down an inclined surface. If it takes the tire 3.5 s to start from rest and roll 3 m down the surface, what is the tire s moment of inertia about its center of mass? 330 mm 15 motion, From Newton s second law and the angular equation of R α mg sin 15 f = ma, Rf = Iα. From these equations and the relation a = Rα, we obtain f mg a a = mg sin 15 m + I/R. (1) N We can determine the acceleration from s = 1 at : 3 = 1 a(3.5), obtaining a = 0.490 m/s. Then from Eq. (1) we obtain I =.05 kg-m. 547

Problem 18.115 Pulle weighs 4 lb, I = 0.060 slug-ft, and I B = 0.014 slug-ft. If the sstem is released from rest, what distance does the 16-lb weight fall in 0.5 s? B 8 in 1 in 16 lb 8 lb The strateg is to appl Newton s second law and the equation of angular motion to the free bod diagrams. Denote the rightmost weight b W R = 16 lb, the mass b m R = 0.4974 slug, and the leftmost weight b W L = 4 + 8 = 1 lb, and the mass b m L = 0.3730 slug. R B = 8 in. is the radius of pulle B, I B = 0.014 slug-ft, and R = 1 in. is the radius of pulle, and I = 0.060 slug-ft. Choose a coordinate sstem with the ais positive upward. T 3 T T T 1 T 1 The 16 lb. weight: (1) T 1 W R = m R a R. Pulle B: The center of the pulle is constrained against motion, and the acceleration of the rope is equal (ecept for direction) on each side of the pulle. () R B T 1 + R B T = I B α B. From kinematics, (3) a R = R B α B. Combine (1), () and (3) and reduce: (4) T = W R + ( I B RB + m R ) a R Pulle : (5) T + T 3 W L = m L a, where a is the acceleration of the center of the pulle. (6) R T 3 + R T = I α. From the kinematics of pulle, the acceleration of the left side of the pulle is zero, so that the acceleration of the right side relative to the left side is a right = a R j = a left + α (R i) i j k = 0 0 α = 0 + R α j, R 0 0 W L W R The total sstem: Equate (4) and (9) (the two epressions for T ) and solve: a R = ( ( ) WL W R + I B RB + m R + I 4R + m L 4 Substitute numerical values: a R = 15.7 ft/s. The distance that the 16 lb. weight will fall in one-half second is s = a R t = 15.7 = 1.96 ft 8 ). from which (7) a R = R α, where the change in direction of the acceleration of the 16 lb. weight across pulle B is taken into account. Similarl, the acceleration of the right side relative to the acceleration of the center of the pulle is a right = a R j = a + α (R i) = a + R α j, from which (8) a = a R. Combine (5), (6), (7) and (8) and reduce ( ) to obtain (9) T = W I 4R + m a. 4 548

Problem 18.116 Model the ecavator s arm BC as a single rigid bod. Its mass is 100 kg, and the moment of inertia about its center of mass is I = 3600 kg-m.if point is stationar, the angular velocit of the arm is zero, and the angular acceleration is 1.0 rad/s counterclockwise, what force does the vertical hdraulic clinder eert on the arm at B? B.4 m 3.0 m C 1.7 m 1.7 m The distance from to the center of mass is d = (3.4) + (3) = 4.53 m. The moment of inertia about is B mg I = I + d m = 8,70 kg-m. From the equation of angular motion: 1.7B 3.4mg = I α. Substitute α = 1.0 rad/s, to obtain B = 40,170 N. Problem 18.117 Model the ecavator s arm BC as a single rigid bod. Its mass is 100 kg, and the moment of inertia about its center of mass is I = 3600 kg-m. The angular velocit of the arm is rad/s counterclockwise and its angular acceleration is 1 rad/s counterclockwise. What are the components of the force eerted on the arm at? The acceleration of the center of mass is i j k a G = α r G/ ω r G/B = 0 0 α ω (3.4i + 3j) 3.4 3 0 = 16.6i 8.6j m/s. From Newton s second law: = ma G = 19,900 N, + B mg = ma G. From the solution to Problem 18.13, B = 40,170 N, from which = 38,70 N 549

Problem 18.118 To decrease the angle of elevation of the stationar 00-kg ladder, the gears that raised it are disengaged, and a fraction of a second later a second set of gears that lower it are engaged. t the instant the gears that raised the ladder are disengaged, what is the ladder s angular acceleration and what are the components of force eerted on the ladder b its support at O? The moment of inertia of the ladder about O is I 0 = 14,000 kg-m, and the coordinates of its center of mass at the instant the gears are disengaged are = 3m, = 4m. O α = (00)(9.81)(3) 14,000 The moment about O, mg = I o α, from which = 0.40 rad/s. The acceleration of the center of mass is i j k a G = α r G/O ω r G/O = 0 0 α = 4αi + 3αj 3 4 0 F F (, ) mg a G = 1.68i 1.6j (m/s ). From Newton s second law: F = ma G = 336 N, F mg = ma G, from which F = 1710 N 550

Problem 18.119 The slender bars each weigh 4 lb and are 10 in. long. The homogenous plate weighs 10 lb. If the sstem is released from rest in the position shown, what is the angular acceleration of the bars at that instant? 45 8 in From geometr, the sstem is a parallelogram, so that the plate translates without rotating, so that the acceleration of ever point on the plate is the same. Newton s second law and the equation of angular motion applied to the plate: F F B = m p a PG, F + F B W p = m p a PG. The motion about the center of mass: ( ) ( ) ( ) 0 4 4 F + F + F B 1 1 1 ( ) 0 + F B = I p α = 0. 1 40 in F 4 lb. F B F F F B B 4 lb. F B F B F B Newton s second law for the bars: F + W B = m B a BG, F + = m B a BG. F B + B W B = m B a BG. F B + B = m B a BG. The angular acceleration about the center of mass: ( ) ( ) ( ) 5 5 5 F cos θ + F sin θ 1 1 1 cos θ + ( 5 1 ) sin θ = I B α, ( ) ( ) ( ) 5 5 5 F B cos θ + F B sin θ B 1 1 1 cos θ + B ( 5 1 ) sin θ = I B α. From kinematics: the acceleration of the center of mass of the bars in terms of the acceleration at point is i j k a BG = α r G/ ω r G/ = 0 0 α 5 1 cos θ 5 1 sin θ 0 = 5 5 sin θαi 1 1 cos αj (ft/s ). From which ( ) ( ) 5 5 a BG = sin θα, a BG = cos θα, 1 1 since ω = 0 upon release. The acceleration of the plate: i j k a P = α r P/ ω r P/ = 0 0 α 10 10 cos θ 1 1 sin θ 0 Substitute to obtain the nine equations in nine unknowns: (1) F F B = () F + F B W p = ( ) 10 m p sin θα, 1 ( ) 10 m p cos θα, 1 (3) 0F + 4F + 0F B + 4F B = 0, (4) ( ) 5 F + W B = m B cos θα, 1 (5) ( ) 5 F + = m B sin θα, 1 (6) ( ) 1 F sin θ + F cos θ sin θ + cos θ = I B α, 5 (7) ( ) 5 F B + B = m B sin θα, 1 (8) ( ) 5 F B + B W B = m B cos θα, 1 (9) ( ) 1 F B cos θ + F B sin θ B cos θ + B sin θ = I B α. 5 The number of equations and number of unknowns can be reduced b combining equations, but here the choice is to solve the sstem b iteration using TK Solver Plus. The results: F =.1 lb, F = 1.68 lb, F B = 3.3 lb, = 3.3 lb, = 4.58 lb, B = 4.4 lb, B = 5.68 lb. α = 30.17 rad/s. = 10 10 sin θαi 1 1 cos θαj (ft/s ). From which a P = ( ) ( ) 10 10 sin θα,a P = cos θα. 1 1 551

Problem 18.10 slender bar of mass m is released from rest in the position shown. The static and kinetic friction coefficients of friction at the floor and the wall have the same value µ. If the bar slips, what is its angular acceleration at the instant of release? Choose a coordinate sstem with the origin at the intersection of wall and floor, with the ais parallel to the floor. Denote the points of contact at wall and floor b P and N respectivel, and the center of mass of the bar b G. The vector locations are l r N = il sin θ,r P = jl cos θ,r G = L (i sin θ + j cos θ). From Newton s second law: θ P µn = ma G,N + µp mg = ma G, where a G, a G are the accelerations of the center of mass. The moment about the center of mass is M G = r P/G (P i + µp j) + r N/G (Nj µ Ni) : M G = PL M G = i j k sin θ cos θ 0 1 µ 0 + NL ( ) PL (cos θ + µ sin θ)k + From the equation of angular motion, ( ) PL (cos θ + µ sin θ) + ( NL ( NL i j k sin θ cos θ 0. µ 1 0 ) (sin θ µ cos θ)k ) (sin θ µ cos θ) = I B α From kinematics: ssume that at the instant of slip the angular velocit ω = 0. The acceleration of the center of mass in terms of the acceleration at point N is a G = a N + α r G/N ω r G/N i j k = a N i + 0 0 α L sin θ L cos θ 0 ( a G = a N from which ) ( αl cos θ i + αlsin θ ) j, a G = L sin θ α. Substitute to obtain the three equations in three unknowns, ml cos θ (1) P µn = α, () µp + N = ml sin θ α + mg. (3) PL NL (cos θ + µ sin θ) + (sin θ µ cos θ) = I Bα. Solve the first two equations for P and N: P = ml µmg (1 + µ (cos θ µ sin θ)α + ) (1 + µ ). N = ml mg (1 + µ (sin θ + µ cos θ)α + ) (1 + µ ). Substitute the first two equations into the third, and reduce to obtain α [I B + ml 4 ( 1 µ 1 + µ )] = mgl ( ) µ sin θ mgl 1 + µ cos θ. ( 1 µ ) 1 + µ Substitute I B = ( 1 1 ) ml, reduce, and solve: α = (3(1 µ ) sin θ 6µ cos θ)g ( µ )L The acceleration of the center of mass in terms of the acceleration at point P is a G = a P + α r G/P. P µp a G = a P + α r G/P ω r G/P = a P j + ( αl cos θ a G = i j k 0 0 α L cos θ, 0 L sin θ ) ( i + a P + ) αl sin θ j, mg θ N µn from which a G = L cos θ α. 55

Problem 18.11 Each of the go-cart s front wheels weighs 5 lb and has a moment of inertia of 0.01 slug-ft. The two rear wheels and rear ale form a single rigid bod weighing 40 lb and having a moment of inertia of 0.1 slug-ft. The total weight of the go-cart and driver is 40 lb. (The location of the center of mass of the gocart and driver, not including the front wheels or the rear wheels and rear ale, is shown.) If the engine eerts a torque of 1 ft-lb on the rear ale, what is the go-cart s acceleration? 15 in 6 in 4 in 16 in 60 in B Let a be the cart s acceleration and α and α B the wheels angular accelerations. Note that a = (6/1)α, (1) 1 ft-lb B a = (4/1)α B. () 1 ft-lb 40 50 lb B Front wheel: F = B + f B = (10/3.)a, (3) 40 lb B 10 lb B f f B F = B + N B 10 = 0, (4) N N B M = fb (4/1) = (0.0)α B. (5) Rear Wheel: F = + f = (40/3.)a, (6) F = + N 40 = 0, (7) M = 1 f (6/1) = (0.1)α. (8) Cart: F = B = (190/3.)a, (9) F = B 190 = 0, (10) M = B [(15 4)/1] + B [(60 16)/1] + [(15 6)/1] (16/1) 1 = 0. (11) Solving Eqs. (1) (11), we obtain a =.99 ft/s. 553

Problem 18.1 Bar B rotates with a constant angular velocit of 10 rad/s in the counterclockwise direction. The masses of the slender bars BC and CDE are kg and 3.6 kg, respectivel. The ais points upward. Determine the components of the forces eerted on bar BC b the pins at B and C at the instant shown. B 400 mm 10 rad/s C D E 700 mm 400 mm 700 mm The velocit of point B is i j k v B = ω B r B = 0 0 10 0 0.4 0 B B C = 0.4(10)i = 4i (m/s). W BC C C The velocit of point C is i j k v C = v B + ω BC r C/B = 4i + 0 0 ω BC 0.7 0.4 0 = 4i + 0.4ω BC i + 0.7ω BC j (m/s). From the constraint on the motion at point C, v C = v C j. Equate components: 0 = 4 + 0.4ω BC, v C = 0.7ω BC, from which ω BC = 10 rad/s, v C = 7 m/s. The velocit at C in terms of the angular velocit ω CDE, v C = v D + ω CDE r C/D i j k = 0 + 0 0 ω CDE = 0.4ω CDE j, 0.4 0 0 from which ω CDE = 7 = 17.5 rad/s. 0.4 The acceleration of point B is a B = ω B r B = (10 )(0.4)j = 40j (m/s ). The acceleration at point C is a C = a B + α BC r C/B ω BC r C/B. i j k a C = 40j + 0 0 α BC ωbc (0.7i 0.4j) (m/s ). 0.7 0.4 0 a C =+(0.4α BC 0.7ω BC )i + ( 40 + 0.7α BC + 0.4ω BC )j (m/s ). The acceleration in terms of the acceleration at D is i j k a C = 0 0 α CDE ωcde ( 0.4i) 0.4 0 0 C W CE The acceleration of the center of mass of BC is a G = 40j + i j k 0 0 α BC 0.35 0. 0 from which a G = 61.5i + 148.44j (m/s ) ωbc (0.35i 0.j), The equations of motion: B + C = m BC a G, B + C m BC g = m BC a G, where the accelerations a G, a G are known. The moment equation, 0.35C + 0.C 0.B 0.35B = I BC α BC, where α BC, is known, and ( ) 1 I BC = m BC L BC 1 = 0.1083 kg-m,0.4c 0.15m CE g = I D α CE, ( ) 1 where I D = m CE L CE 1 + (0.15) m CE = 0.444 kg-m, is the moment of inertia about the pivot point D, and 0.15 m is the distance between the point D and the center of mass of bar CDE. Solve these four equations in four unknowns b iteration: B = 1959 N, B = 138 N, C = 081 N, C = 9 N. = 0.4α CDE j + 0.4ω CDE i. Equate components and solve: α BC = 481.5 rad/s,α CDE = 84.19 rad/s. 554

Problem 18.13 t the instant shown, the arms of the robotic manipulator have the constant counterclockwise angular velocities ω B = 0.5 rad/s, ω BC = rad/s, and ω CD = 4 rad/s. The mass of arm CD is 10 kg, and the center of mass is at its midpoint. t this instant, what force and couple are eerted on arm CD at C? 300 mm 30 B 50 mm 0 C 50 mm D The relative vector locations of B, C, and D are C r B/ = 0.3(i cos 30 + j sin 30 ) = 0.598i + 0.150j (m), C C 15 mm mg D r C/B = 0.5(i cos 0 j sin 0 ) = 0.349i 0.08551j (m), r D/C = 0.5i (m). The acceleration of point B is a B = ω B r B/ = (0.5 )(0.3 cos 30 i + 0.3 sin 30 j), a B = 0.065i 0.0375j (m/s ). The acceleration at point C is a C = a B ωbc r C/B = a B ωbc (0.349i 0.08551j). a C = 1.005i + 0.3045j (m/s ). The acceleration of the center of mass of CD is a G = a C ω CD (0.15i) (m/s ), from which a G = 3.005i + 0.3045j (m/s ). For the arm CD the three equations of motion in three unknowns are C m CD g = m CD a G,C = m CD a G,M 0.15C = 0, which have the direct solution: C = 101.15 N, C = 30.05 N. M = 1.64 N-m, where the negative sign means a direction opposite to that shown in the free bod diagram. 555

Problem 18.14 Each bar is 1 m in length and has a mass of 4 kg. The inclined surface is smooth. If the sstem is released from rest in the position shown, what are the angular accelerations of the bars at that instant? O 45 B 30 For convenience, denote θ = 45, β = 30, and L = 1 m. The acceleration of point is i j k a = α O r /O = 0 0 α O. L cos θ Lsin θ 0 a = α O ( il sin θ + jl cos θ) (m/s ). mg mg 30 B The acceleration of is also given b a = a B + α B r /B. i j k a = a B + 0 0 α B. L cos θ Lsin θ 0 a = a B iα B L sin θ jα B L cos θ(m/s ). From the constraint on the motion at B, Equate the epressions for the acceleration of to obtain the two equations: (1) α O L sin θ = a B cos β α B L sin θ, and () α O L cos θ = a B sin β α B L cos θ. The acceleration of the center of mass of B is a GB = a + α B r GB/ = a + i j k 0 0 α B L sin θ, 0 L cos θ a GB = a + Lα B sin θi + α BL cos θj (m/s ), The equations of motion for the bars: for the pin supported left bar: (5) L cos θ L sin θ mg ( ml ) where I O = = 4 3 3 kg-m. The equations of motion for the right bar: (6) B sin β = ma GB, (7) mg + B cos β = ma GB, (8) ( L B ( ) L cos θ = I O α O, ) ( ) ( ) L L cos θ + sin θ + B sin θ cos β ( ) L cos θ sin β = I CB α B, ( ) 1 where I GB = ml = 1 ( ) 1 kg-m. 3 These eight equations in eight unknowns are solved b iteration: = 19.7 N, = 1.15 N, α O = 0.45 rad/s, α B = 1.59 rad/s, B = 45.43 N, a GB = 0.8610 m/s, a GB = 0.601 m/s from which (3) a GB = α O L sin θ + Lα B sin θ(m/s ), (4) a GB = α O L cos θ + Lα B cos θ. 556

Problem 18.15 Each bar is 1 m in length and has a mass of 4 kg. The inclined surface is smooth. If the sstem is released from rest in the position shown, what is the magnitude of the force eerted on bar O b the support at O at that instant? The acceleration of the center of mass of the bar O is a GO = α O r G/ = a + i j k 0 0 α O L sin θ, 0 L cos θ F mg a GO = L sin θ The equations of motion: α O i + L cos θ α O j (m/s ). F + = ma GO,F + mg = ma GO. F Use the solution to Problem 18.140: θ = 45, α G = 0.45 rad/s, = 19.7 N, m = 4 kg, from which F = 18.67 N, F = 38.69 N, from which F = F + F = 4.96 N Problem 18.16 The fied ring gear lies in the horizontal plane. The hub and planet gears are bonded together. The mass and moment of inertia of the combined hub and planet gears are m HP = 130 kg and I HP = 130 kg-m. The moment of inertia of the sun gear is I s = 60 kg-m. The mass of the connecting rod is 5 kg, and it can be modeled as a slender bar. If a 1 knm counterclockwise couple is applied to the sun gear, what is the resulting angular acceleration of the bonded hub and planet gears? Planet gear Hub gear Connecting rod Sun gear 140 mm 340 mm 40 mm 70 mm The moment equation for the sun gear is Ring gear (1) M 0.4F = I s α s. For the hub and planet gears: () (0.48)α HP = 0.4α s, (3) F Q R = m HP (0.14)( α HP ), (4) (0.14)Q + 0.34F I HP ( α HP ). For the connecting rod: F Hub & Planet Gears M Sun Gear F Q R R Connecting Rod (5) (0.58)R = I CR α CR, where I CR = ( ) 1 m GR (0.58 ) = 0.561 kg-m. 3 (6) (0.58)α CR = (0.14)α HP. These si equations in si unknowns are solved b iteration: F = 148.7 N,α s = 10.74 rad/s, α HP = 5.37 rad/s,q= 1383.7 N, R = 1.5 N,α CR = 1.96 rad/s. 557