homework9-solutions1

homework9-solutions1 - ME 104: Homework 9 Solutions Prof....

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Unformatted text preview: ME 104: Homework 9 Solutions Prof. Karl Hedrick: University of California, Berkeley Chapter 16, Solution 88 (a) Angular acceleration. Rod AB: I AB 1W 2 l AB 12 g 1 8 lb (2 ft) 2 12 32.2 0.8282 lb ft s 2 MB ( M B )eff : 3 (6 lb)(2 ft) (8 lb)(1 ft) 5 I AB I AB (0.8282 lb ft s 2 ) 0.8 lb ft 0.8 lb ft α 9.66 rad/s 2 (1) Entire assembly: Since AC is taut, assembly rotater about C as a rigid body. Kinematics: CB CGBC aBC a AB (0.75) 2 12 1 CB 2 0.625 ft 1.25 ft (0.625 ft) (0.75 ft) Kinetics: I BC 1 WBC (CB) 2 12 g 1 5 lb (1.25 ft) 2 12 32.2 0.0202 lb ft s 2 PROBLEM 16.88 (Continued) (b) Couple M. MC ( M C )eff : M M WBC (0.5 ft) (5 lb)(0.5 ft) mBC aBC (0.625 ft) I BC 5 lb (0.625 ft) 2 32.2 mAB a AB (1 ft) I AB 8 lb (0.75 ft) 2 32.2 I AB (0.0202 lb ft/s 2 ) 0.8 lb ft Substitute 9.66 rad/s 2 and from Eq. (1), IAB M M M 2.5 (0.06016)(9.66) (0.0202)(9.66) (0.1398)(9.66) 0.8 2.5 0.5860 0.1950 1.35 0.8 2.5 2.93 M 5.43 lb ft M 5.43 lb ft Chapter 16, Solution 100 Assume disk rolls: a r 8 ft 12 I mk 2 10 lb 6 ft 32.2 12 2 I MC ( M C )eff : (5 lb) 4 ft 12 0.07764 lb ft s 2 ( ma ) r I 2 1.6667 lb ft 1.6667 10 lb 8 ft 32.2 12 0.21566 0.07764 7.728 rad/s 2 a Fx ( Fx )eff : 7.73 rad/s 2 a 5.153 ft/s 2 r 8 ft 7.728 rad/s 2 12 F F 5 lb 5 lb ma 10 lb (5.153 ft/s 2 ) 32.2 F Fy ( Fy )eff : N 10 lb Fm sN 3.40 lb 0 N 10 lb 2.5 lb 0.25(10 lb) (a) Since F , Fm , Knowing that disk slides F kN Disk slides 0.20(10 lb) 2 lb PROBLEM 16.100 (Continued) (b) Angular acceleration. MG ( M G )eff : F (2 lb) 8 ft 12 8 ft 12 (5 lb) 4 ft 12 I (0.07764 lb ft s 2 ) 1.6667 lb ft 0.3333 0.07764 4.29 rad/s 2 4.29 rad/s 2 (c) Acceleration of G. Fx ( Fx )eff : F 5 lb ma 2 lb 5 lb 10 lb a 32.2 a 9.66 ft/s 2 a 9.66 ft/s 2 Chapter 16, Solution 111 The analysis will be restricted to very small rotation, i.e. position change is neglected. Kinematics: Assume From inside cover of text , a0 r OG a 4r 3 aO aG/O [a O [(OG ) r1 4 3 a [r 4r 3 Kinetics: m mass of half cylinder: IO IO 12 mr 2 I 12 mr 2 I m(OG )2 I mr 2 4r m 3 1 16 2 92 2 (a) Angular acceleration. MC ( M O )eff : Pr I (ma ) 1 4 3 r Pr Pr mr 2 mr 2 mr 2 1 2 1 2 3 2 16 92 16 92 8 3 mr 1 mr 2 4 3 8 1 3 1 16 92 4 r 3 Pr Pr mr 2 (0.65717) P 1.5357 mr 1.536 P mr PROBLEM 16.111 (Continued) (b) Minimum value of Fx s. ( Fx )eff : F ma F F Fy ( Fy )eff : N P mg N ( s ) min mr 1 4 3 mr (0.57559) P mr 0.8839 P mr (0.57559) 1.5357 0 mg F N P 0.8839 P mg P ( s ) min 0.884 P mg P Chapter 16, Solution 113 Kinematics: aA aB aB (a B ) x r aA r r , aB/A r + a B/ A r ( aB ) y r Kinetics: mh I 3mB mh r 2 3mB r 2 (a) Angular acceleration. MC ( M C )eff : WB r I mh a A r mB ( a B ) x r mB ( a B ) y r mB gr gr 3mB r 2 8r 2 (3mB )r 2 mB r 2 mB r 2 1g 8r (a B ) x (a B ) y 1 g 8 1 g 8 1 g 8 (b) Components of acceleration of B. (a B ) x r 1 g 8 (a B ) y r Chapter 16, Solution 119 Kinematics: aG/A 15 ft 12 1.25 a B/A 30 ft 12 2.5 aB aA a B/A : [aB ] [a A 30 ] [2.5 aA aB a ax 2.5 cos 30 2.387 30 (2.5 ) tan 30 aG [2.5 aA 1.443 30 ] [1.25 + aG/A ; a [2.887 [1.25 ] 1.443 1.25 ay [2.443 We have: ax 1.25 ; ay 1.443 I I 1 w2 L 12 g 1 8 lb (2.5 ft) 12 32.2 0.1294 lb ft s 2 PROBLEM 16.119 (Continued) (a) Angular acceleration. ME ( M E )eff : W (1.443 ft) I ma x (1.25 ft) ma y (1.443 ft) 8(1.443) 0.1294 8 (1.25) 2 37.2 8 (1.443) 2 32.2 11.544 1.035 11.154 rad/s 2 11.15 rad/s 2 (b) Reaction at B. MA ( M A )eff : B(2.5 ft) 2.58 I max (1.25 ft) (0.1294)(11.154) 8 (1.25)(11.154)(1.25) 32.2 2.58 1.443 4.330 8 1.155 lb B 1.155 lb ...
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This note was uploaded on 02/06/2010 for the course ME 104 taught by Professor Oreilly during the Fall '08 term at Berkeley.

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