homework10-solutions

homework10-solutions - ME 104 Homework 10 Solutions Prof...

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Unformatted text preview: ME 104: Homework 10 Solutions Prof. Karl Hedrick: University of California, Berkeley Chapter 17, Solution 3 FOR ANY DISK: Moment of inertia. Disk A: Disk B: m I ( r 2t ) 12 mr 2 1 tr 4 2 IA 1 2 b r4 IB 1 2 3n 4 (3b)(nr )4 1 2 IB br 4 3n 4 I A I total IA (1 3n 4 ) I A Work-energy. T1 T2 T1 U1 T2 : 0 M (4 ) 2 2 0 U1 1 I total 2 2 2 2 M M (4 rad) 2 1 (1 3n4 ) I A 2 2 2 8M (1 3n 4 ) I A 8 Mr 2 n2 IA 1 3n 4 For Point D on rim of disk B vD ( nr ) 2 2 or vD n2 r 2 2 2 Value of n for maximum final speed. For maximum vD : d n2 dn 1 3n 4 0 1 [n 2 (12n3 ) (1 3n 4 )(2n)] 0 (1 3n 4 )2 12n5 2n 6n5 2n(3n 4 1) n 0 and n 0 0 n 0.760 1 3 0.25 0.7598 Chapter 17, Solution 7 Work of external friction force on disk A. Only force doing work is F. Since its moment about A is M U1 2 rF , we have M rF r ( k mg ) Kinetic energy of disk A. Angular velocity becomes constant when 2 T1 T2 v r 0 1 I 2 2 2 11 2 mr 22 mv 2 4 v r 2 Principle of work and energy for disk A. T1 U1 2 T2 : 0 r k mg mv 2 4 v2 rad 4r k g v2 rev 8 r kg Angle change. Chapter 17, Solution 28 The point of contact with ground is the instantaneous center. Position 1. Point B is at the top. 1 vB 2r 1 vA r 1 T1 12121 mvB mv A I2 2 2 2 1 1 m(2r 1 ) 2 m( r 1 ) 2 2 2 3mr 2 12 1 (mr 2 ) 2 2 1 Position 2. Point B is at the bottom. 2 vB 0 vA r 2 T2 12121 2 mvB mv A I2 2 2 2 1 1 2 0 m( r 2 ) (mr 2 ) 2 2 2 mr 2 2 mr 2 (3 1 ) 2 9mr 2 2 1 2 2 Work. U1 2 mg ( h) 2mgr Principle of work and energy. T1 U1 2 T2 : 3mr 2 2 1 2mgr 2 1 9m g 3r 2 1 1 0.577 g r Chapter 17, Solution 55 Mass of disk B. mB rB rA 2 mA 2 125 mm 100 mm 4.6875 kg 3 kg Moment of inertia. I IA IB 1 (4.6875 kg)(0.125 m) 2 2 1 (3 kg)(0.1 m) 2 2 0.05162 kg m 2 Angular velocities. 1 200 rpm 800 rpm 2 60 2 60 20.944 rad/s 83.776 rad/s 2 Principle of impulse and momentum. Syst. Momenta1 Moments about B: Couple M. I 1 Syst. Ext. Imp.1 2 2 Syst. Momenta2 Mt M I I ( t 2 1) 0.05162 kg m 2 (83.776 rad/s 20.944 rad/s) 3s M 1.081 N m Chapter 17, Solution 79 Moment of inertia of yoke: Moment of inertia of disk: IC 0: I A 2 mkC 1.5 32.2 3 12 2 2.9115 lb s 2 ft 12 mr 4 1 2.5 4 32.2 12 mr 2 4 12 2 2.15666 lb s 2 ft 90 : I A 1 2.5 2 32.2 4 12 2 4.3133 lb s 2 ft Total moment of inertia about the x axis: 0: ( I x )1 90 : ( I x ) 2 IC IC IA 3 5.0682 10 IA lb s 2 ft lb s 2 ft 7.2248 10 3 Angular momentum about the x axis: 0: H1 90 : H 2 ( I x )1 ( I x )2 1 3 1 5.0682 10 2 7.2248 10 3 2 Conservation of angular momentum. H1 H 2 : 5.0682 10 2 3 1 1 7.2248 10 3 2 2 0.7015 (0.7015) (120 rpm) 84.2 rpm Chapter 17, Solution 85 Let Ω be the angular velocity of the cab and be the angular velocity of the blades relative to the cab. The absolute angular velocity of the blades is . 1 2 180 rpm 6 rad/s 240 rpm 8 rad/s Moments of inertia. Cab: Blades: IC IB 650 lb ft s 2 4 12 mL 3 (4) 1 3 55 (14) 2 32.2 446.38 lb ft s 2 The cab does not rotate. 1 2 0 2 Syst. Momenta1 Syst. Ext. Imp. 1 1 Syst. Momenta 2 2) 1) Moments about shaft: IB ( 1 1) IC Frt Frt IB ( IB ( 2 2 IC 2 Ft (446.38)(8 6) 2804.7 lb ft s Frt 2804.7 175.29 lb s r 16 mv2 Ft m 1250 32.2 Linear components: mv1 v2 Ft v1 175.29 55 (4) 32.2 v2 3.84 ft/s 14.61 lb 3.8398 ft/s (a) (b) Assume v1 Force. 0. F Ft t 175.29 12 F Chapter 17, Solution 88 Moment of inertia of ring. IR 1 mR R 2 2 Position 1 Position 1. vC 0. 0 90 (vC ) y vy R 2 Position 2 Position 2. Conservation of angular momentum about y axis for system. IR 1 mR R 2 2 mR R 2 1 IR 2 mC v y R 2 1 1 2 1 mR R 2 2 mC R 2 2 (mR 2mC ) R 2 2 mR mR 2mC 1 (1) Potential energy. Datum is the center of the ring. V1 mC gR V2 0 2 1 Kinetic energy: T1 T2 1 I R 12 2 1 mR R 2 4 1 2 IR 2 2 1 mR R 2 4 11 mR R 2 22 2 1 1 2 mC vx v 2 y 2 1 2 2 mC R 2 v2 2 2 1 mC v 2 y 2 PROBLEM 17.88 (Continued) Principle of conservation of energy: T1 V1 1 mR R 2 4 2 1 T2 V2 1 mR 4 2 kg 3 kg 0.25 m 35 rad/s mC gR mC mR R 1 1 mC R 2 2 2 2 1 mC v 2 y 2 (2) Data: (a) Angular velocity. From Eq. (1), 2 3 kg (35 rad/s) 3 kg 2(2 kg) 2 15.00 rad/s (b) Velocity of collar relative to ring. From Eq. (2), 1 (3 kg)(0.25 m) 2 (35 rad/s) 2 4 (2 kg)(9.81 m/s 2 )(0.25 m) 1 (3 kg) 4 57.422 4.905 24.609 v 2 y v2 y 37.716 1 2 2 (2 kg) (0.25 m) (15 rad/s) 2 1 2 (2 kg)v y 2 vy 6.14 m/s Chapter 17, Solution 98 mB 0.045 kg mP 9 kg IG 1 mP b2 6 1 (9)(0.200) 2 6 0.06 kg m 2 Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity ω vG b 2 . Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact. Syst. Momenta1 (a) Moments about A: + Syst. Ext. Imp. 1 2 Syst. Momenta2 (mB v0 cos 30 )h mB v0 sin 30 mB v0 h cos 30 b 2 0 IG IG mP vG b 2 b sin 30 2 1 mP b 2 4 0.100sin 30 ) (0.045)(400)(0.150 cos 30 1 (9)(0.2)2 4 21.588 rad/s 0.06 vB (0.100)(21.556) Ax ( t ) Ax (0.002) Ax 0.15 2.1556 m/s mP vG (9)(2.1556) 1920 N vG 2.16 m/s (b) Linear momentum: mB v0 cos 30 (0.045)(400 cos 30 ) AX 1920 N Linear momentum: mB v0 sin 30 (0.045)(400)sin 30 Ay 4500 N 4.892 kW Ay ( t ) Ay (0.002) Ay tan 0 0 4500 N 4500 1920 66.9 A 4892 N A 4.87 kN 66.9° Chapter 17, Solution 106 Moment of inertia. (a) First Impact at A. I 1 mL2 12 Syst. Momenta1 Syst. Ext. Imp. 1 e 1: ( v A )2 v1 2 Syst. Momenta 2 Condition of impact: Kinematics: Moments about A: mv1 L 2 v2 0 L 2 mv2 L 2 (v A ) 2 I 2 L 2 v1 m v2 ( vB ) 2 L 2 v1 v1 L 2 1 mL2 12 1 v1 2 2 2 3v1 L 1 v1 2 L 3v1 2L L v2 v1 2v1 (v A ) 2 3v1 (b) Impact at B. Syst. Momenta1 Syst. Ext. Imp . 1 e 1: ( v B )3 2 Syst. Momenta3 2v1 Condition of impact. Kinematics: L 2 v3 mv2 I 0 ( vB ) 2 mv3 L 2 L 2 I 3 2v1 L 2 Moments about B: 1 L v1 2 2 2 m 1 mL2 12 3v1 L 0 m 2v1 L 2 3 L 2 1 v1 2 1 mL2 12 3 3 3v1 L 1 v1 2 v3 (v A )3 2v1 L L 3v1 2L (vB )3 3v1 v3 v1 2v1 PROBLEM 17.106 (Continued) (c) Second Impact at A. Syst. Momenta3 Syst. Ext. Imp. 3 e 1: ( v A )4 4 Syst. Momenta 4 v1 4 Condition of impact. Kinematics: L 2 v4 (v A ) 4 L 2 L 2 I v1 L 2 4 Moments about A: 1 L v1 2 2 mv3 I 3 0 mv4 4 m 1 mL2 12 3v1 L 0 v4 m v1 v1 0 L 2 4 L 2 1 mL2 12 4 4 0 v1 v4 ...
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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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