Since the cylinder rolls without slipping, the point of contact with the ground is the instantaneous center. r Ë Á 1 2ˆ = = = r

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1 PROBEM 7.7 A 0-kg uniform cylindrical roller, initially at rest, is acted upon by a 90-N force as shown. Knowing that the body rolls without slipping, determine (a) the velocity of its center G after it has moved.5 m, (b) the friction force required to prevent slipping. SOUTION Since the cylinder rolls without slipping, the point of contact with the ground is the instantaneous center. Kinematics: v = rw Position. At rest. T = 0 vg Position. s= 5. m vg = v w = r T = mv + Iw vg = mvg + Ê mr Ë Á ˆ Ê ˆ Ë Á r 3 3 = = = mv G ( 0) 5 v G v G Work: U Æ = Ps = ( 90)(. 5) = 35 J. F f does no work. (a) Principle of work and energy. T + U Æ = T : = 5v G v G = 9 v G = 300. m/s Æ (b) Since the forces are constant, a = a = constant + a G G vg = s 9 = ( )(. 5) = 3 m/s SF = ma: P- F = ma x f Ff = P-ma = 90 -( 0)( 3) F f = N PROPRIETARY MATERIA. 00 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced 736

PROBEM 7.3 The -kg rod AB is attached to a collar of negligible mass at A and to a flywheel at B. The flywheel has a mass of 6 kg and a radius of gyration of 80 mm.

2 PROBEM 7.3 The -kg rod AB is attached to a collar of negligible mass at A and to a flywheel at B. The flywheel has a mass of 6 kg and a radius of gyration of 80 mm. Knowing that in the position shown the angular velocity of the flywheel is 60 rpm clockwise, determine the velocity of the flywheel when Point B is directly below C. SOUTION Moments of inertia. Rod AB: Flywheel: IAB mabab ( kg)(0.7 m) 0.78 kg m IC mk (6 kg)(0.8 m) 0.58 kg m Position. As shown. 0. sin h (0.7)cos m V WABh ()(9.8)(0.339) J Kinematics. vb r 0. Bar AB is in translation. 0, v v AB B T mabv IABAB IC ()(0. ) 0 (0.58) 0.37 PROPRIETARY MATERIA. 03 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 767

3 PROBEM 7.3 (Continued) Position. Point B is directly below C. h AB r (0.7) m V WABh ()(9.8)(0.).7088 J Kinematics. vb r 0. vb AB v vb 0. T mabv IABAB IC ()(0. ) (0.78)( ) (0.58) Conservation of energy. T V T V : () Angular speed data: 60 rpm rad/s Solving Equation () for, rad/s 8.7 rpm PROPRIETARY MATERIA. 03 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 768

4 PROBEM 7.97 A bullet weighing 0 gm is fired with a horizontal velocity of 550 m/s into the lower end of a slender 7.5-kg bar of length = 800 mm. Knowing that h = 300 mm and that the bar is initially at rest, determine (a) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at C, assuming that the bullet becomes embedded in 0.00 s. SOUTION Bar: = 800mm = 08. m m= 75. kg I = m = ( 75. )( 08. ) = 0. kg m Bullet: m 0 = 0 g= 0. 0 kg Support location: h = 300 mm = 0. 3 m Kinematics. v = ( - h) w = ( ) w = 05. w Kinetics. v B G Ê ˆ = - h Ë Á w = ( ) w = 0. w Syst. Momenta + Syst. Ext. Imp. Æ = Syst. Momenta Ê ˆ Moments about C: mv 0 0( - h) = mv 0 B( - h) + mvg - h I Ë Á + w ( 00. )( 550)( 05. ) = ( 00. )( 05. w)( 05. ) + ( 75. )( 0. w)( 0. ) + ( 0. w) PROPRIETARY MATERIA. 00 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced 859

5 PROBEM 7.97 (Continued) (a) = 0. 0w w + 0. w = 0. 85w or w =. 680 =. 7 rad/s + Horizontal components: v v B G = ( 0. 5)(. 680) =. 30 m/s = ( 0. )(. 680) =. 680 m/s - mv + C( t) = -mv - mv : C( t) = m( v -v )-mv B G 0 0 B 0 C( t) = ( 0. 0)( ) - ( 7. 5)(. 680) = N s (b) C = C t t = C = 50 N Æ PROPRIETARY MATERIA. 00 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced 860

6 PROBEM 7.7 Member ABC has a mass of. kg and is attached to a pin support at B. An 800-g sphere strikes the end of member ABC with a vertical velocity v of 3 m/s. Knowing that 750 mm and that the coefficient of restitution between the sphere and member ABC is 0.5, determine immediately after the impact (a) the angular velocity of member ABC, (b) the velocity of the sphere. SOUTION et Point G be the mass center of member ABC. m kg m m m. kg AC IG mac (.)(0.750) 0.5 kg m Kinematics after impact. Conservation of momentum., v G, v A Moments about B: mv 0 mv IGmACvG mv mv IG ma (0.800)(3)(0.875) (0.800)(0.875) v [0.5 (.)(0.875) ] v () PROPRIETARY MATERIA. 03 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 9

7 PROBEM 7.7 (Continued) Coefficient of restitution. v va v ev ( v ) A Solving Eqs. () and () simultaneously. (a) Angular velocity. 3 v (0.5)(3 0) () 3.00 rad/s (b) Velocity of. v v m/s PROPRIETARY MATERIA. 03 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using 95

8 PROBEM 7.6 A slender rod CE of length and mass m is attached to a pin support at its midpoint. A second and identical rod AB is rotating about a pin support at A with an angular velocity w when its end B strikes end C of rod CE. enoting by e the coefficient of restitution between the rods, determine the angular velocity of each rod immediately after the impact. SOUTION Rod AB. Kinematics. w = w w = w Principle of impulse and momentum. ( ) = Ø v AB w ( v ) = w Ø AB Syst. Momenta + Syst. Ext. Imp. Æ = Syst. Momenta Moments about A: Iw + m( vab ) -( Ft) = I( wab ) + m( vab ) + m w + m w F t m w m AB wab Ë Á ˆ -( ) = ( ) + ( ) m w -( Ft) = m ( wab ) 3 3 F t = m [ w -( wab ) ] 3 () PROPRIETARY MATERIA. 00 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced 953

9 PROBEM 7.6 (Continued) Rod CE. Principle of impulse and momentum. Syst. Momenta + Syst. Ext. Imp. Æ = Syst. Momenta + Moments about : ( Ft) = I ( wce ) ( Ft) = m ( wce ) Substitute for ( F t) from () m [ w -( wab ) ] = m ( wce ) 3 w -( w ) = ( w ) Condition of impact. e =coefficient of restitution. AB CE () ( v ) e= ( v ) -( v ) B C B w e= ( wce ) -( wab ) È From Eq. () w - w -w w Î Í ( CE ) e = ( CE ) ( wab ) = ( wce ) - we (3) w ( + e) = ( wce ) ( ) = w ( + e) w CE From Eq. (3) ( wab ) - w( + e) - we= w + we- we ( w AB ) = w ( -e) PROPRIETARY MATERIA. 00 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced 95

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: Plane Motion of Rigid Bodies: Energy and Momentum Methods. Tenth Edition CHAPTER

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: Plane Motion of Rigid Bodies: Energy and Momentum Methods. Tenth Edition CHAPTER Tenth E CHAPTER 7 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P. Self California State Polytechnic University Plane Motion