PROPRIETARY MATERIAL.

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PROLEM 4. 30- bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block which has a mass of 3 k.

1 PROLEM bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block which has a mass of 3 k. fter the impact, block slides on 30-k carrier C until it impacts the end of the carrier. Knowin the impact between and C is perfectly plastic and the coefficient of kinetic friction between and C is 0., determine (a) the velocity of the bullet and after the first impact, (b) the final velocity of the carrier. For convenience, label the bullet as particle of the system of three particles,, and C. (a) Impact between and : Use conservation of linear momentum of and. ssume that the time period is so short that any impulse due to the friction force between and C may be nelected. Σ mv +Σ Imp =Σmv (b) Components : mv0+ 0 = ( m + m) v v mv 0 (30 0 k)(450 m/s) = = = m + m (30 0 k + 3 k m/s v = 4.46 m/s Final velocity of the carrier: Particles,, and C have the same velocity v to the left. Use conservation of linear momentum of all three particles. The friction forces between and C are internal forces. Nelect friction at the wheels of the carrier. Σ mv +Σ Imp =Σmv 3 3 Components : ( m + m ) v + 0 = ( m + m + m ) v C ( m + m) v mv 0 v = = m + m + m m + m + m C C (30 0 k)(450 m/s) = = m/s 30 0 k + 3 k + 30 k v = m/s PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 843

2 PROLEM 4. For the system of particles of Problem 4., determine (a) the components v and v z of the velocity of particle for which the anular momentum H O of the system about O is parallel to the z ais, (b) the value of H O. PROLEM 4. system consists of three particles,, and C. We know that W = 5lb, W = 4lb, and W C = 3lb and that the velocities of the particles epressed in ft/s are, respectively, v = i + 3 j k, v = vi + vyj+ vzk, and vc = 3i j + k. Determine (a) the components v and v z of the velocity of particle for which the anular momentum H O of the system about O is parallel to the ais, (b) the value of H O. i j k H =Σ r mv =Σm y z O i i i i i i ( v ) ( v ) ( v ) i i y i z i j k i j k i j k = v v z = [5( 0 ) + 4(4 v z 6) + 3(6 0)] i + [5(8 0) + 4(3 v 4 vz ) + 3(0 8)] j + [5(0 0) + 4(8 4 v ) + 3( 6 + 8)] k HO = [(6 vz 6) i+ ( vz 6 vz + 6) j+ ( 6 v ) k ] () (a) For H to be parallel to the z ais, we must have H = H = 0: O (b) Substitutin into Eq. (): H H y y = 0: 6v 6 = 0 v = 7.5 ft/s z = 0: v 6(7.5) + 6 = 0 v = 8.33 ft/s HO = [ 6(8.33) ] 3. k H O = (4.5ft lb s) k z PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 859

PROLEM 4.3 In Problem 4.4, determine the enery lost as the bullet (a) passes throuh block, (b) becomes embedded in block. The masses are m for the bullet and m and m for the blocks.

3 PROLEM 4.3 In Problem 4.4, determine the enery lost as the bullet (a) passes throuh block, (b) becomes embedded in block. The masses are m for the bullet and m and m for the blocks. The bullet passes throuh block and embeds in block. Momentum is conserved. Initial momentum: mv0 + m (0) + m (0) = mv0 Final momentum: mv + mv + mv Equatin, mv0 = mv + mv+ mv mv + mv (6)(5) + (4.95)(9) m = = = lb v v The bullet passes throuh block. Momentum is conserved. Initial momentum: mv0 + m (0) = mv0 Final momentum: mv + mv Equatin, mv0 = mv + mv mv0 mv (0.0500)(500) (6)(5) v = = = 900 ft/s m The masses are: m = = lb s /ft 3. (a) 6 m = = lb s /ft m = = lb s /ft 3. ullet passes throuh block. Kinetic eneries: 0 (b) 3 efore: 0 0 ( T = mv = )(500) = ft lb fter: T = mv + mv = ( )(900) + (0.8633)(5) = 63. ft lb Lost: T0 T = = 5.7 ft lb enery lost = 6 ft lb ullet becomes embedded in block. Kinetic eneries: 3 efore: ( T = mv = )(900) = 68.9 ft lb fter: T3 = ( m+ m) v = (0.558)(9) = 6.9 ft lb Lost: T T3 = = 6.6 ft lb enery lost = 63 ft lb PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 886

4 PROLEM 4.38 Two hemispheres are held toether by a cord which maintains a sprin under compression (the sprin is not attached to the hemispheres). The potential enery of the compressed sprin is 0 J and the assembly has an initial velocity v 0 of manitude v 0 = 8m/s. Knowin that the cord is severed when θ = 30, causin the hemispheres to fly apart, determine the resultin velocity of each hemisphere. Use a frame of reference movin with the mass center. Conservation of momentum: Conservation of enery: v = + m 0 mv mv = v m V = m( v ) + m( v ) m = m v + m( v ) m m( m + m) ( v ) = m mv v = m ( m + m ) Data: m =.5 k m =.5 k V = 0 J ()(.5)(0) v = = 0 v = 0 m/s 30 (.5)(4.0) Velocities of and..5 v = (0) = 6 v = 6 m/s 30.5 v = [8 m/s ] + [6 m/s 30 ] v 4. m/s = v = [8 m/s ] + [0 m/s 30 ] v 7.39 m/s = PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 893

PROLEM 5. The brake drum is attached to a larer flywheel that is not shown. The motion of the brake drum is defined by the relation θ = 36t.6 t, where θ is epressed in radians and t in seconds.

5 PROLEM 5. The brake drum is attached to a larer flywheel that is not shown. The motion of the brake drum is defined by the relation θ = 36t.6 t, where θ is epressed in radians and t in seconds. Determine (a) the anular velocity at t = s, (b) the number of revolutions eecuted by the brake drum before comin to rest. Given: θ = 36t.6t radians Differentiate to obtain the anular velocity. dθ ω = = 36 3.t rad/s dt (a) t t = s, ω = 36 (3.)() ω = 9.6 rad/s (b) When the rotor stops, ω = 0. 0= 36 3.t t =.5 s θ = (36)(.5) (.6)(.5) = 0.5 radians In revolutions, 0.5 θ = θ = 3. rev π PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 009

6 PROLEM 5. In Problem 5.0, determine the velocity and acceleration of corner, assumin that the anular velocity is 9 rad/s and increases at the rate of 45 rad/s. PROLEM 5.0 The bent rod CDE rotates about a line joinin Points and E with a constant anular velocity of 9 rad/s. Knowin that the rotation is clockwise as viewed from E, determine the velocity and acceleration of corner C. E = E = 0.6 m r / = (0.5 m) j E = (0.4 m) i+ (0.4 m) j+ (0. m) k E 0.4i+ 0.4j+ 0.k λ E = = = ( i+ j+ k) E ω= ωeλe = (9 rad/s) ( i+ j+ k) 3 ω = (6 rad/s) i+ (6 rad/s) j+ (3 rad/s) k v = ω r = ( 6i+ 6j+ 3 k) ( 0.5) j=.5k i / α= α EλE = (45 rad/s ) ( i+ j+ k) 3 α = (30 rad/s ) i+ (30 rad/s ) j+ (5 rad/s ) k a = α r/ + ω ( ω r/ ) = α r/ + ω v i j k i j k a = = 3.75i+ 7.5k + 9 i+ (.5 + 9) j 4.5k v = (0.75 m/s) i+ (.5 m/s) k a = (.75 m/s ) i+ (.5 m/s ) j+ (3 m/s ) k PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 09

7 PROLEM 5.8 series of small machine components bein moved by a conveyor belt pass over a 0 mm radius idler pulley. t the instant shown, the velocity of Point is 300 mm/s to the left and its acceleration is 80 mm/s to the riht. Determine (a) the anular velocity and anular acceleration of the idler pulley, (b) the total acceleration of the machine component at. v = v = 300 mm/s r = 0 mm (a) v = ωr, (b) ( a ) = αr, t ( a ) = a = 80 mm/s t v 300 ω = = =.5 rad/s ω =.50 rad/s r 0 ( a) t 80 α = = =.5 rad/s r 0 ( a ) = r ω = (0)(.5) = 750 mm/s n α =.500 rad/s = ( ) t + ( ) n = (80) + (750) = 77 mm/s a a a 750 tan β =, β = a = 77 mm/s 76.5 PROPRIETRY MTERIL. 03 The McGraw-Hill Companies, Inc. ll rihts reserved. No part of this Manual may be displayed, 06

PROBLEM SOLUTION

PROBLEM SOLUTION PROLEM 13.119 35, Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which