© 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–1 z Determine the moment of inertia Iy for the slender rod The rod’s density r and cross-sectional area A are constant Express the result in terms of the rod’s total mass m y l SOLUTION A x Iy = LM x dm l = = L0 x (r A dx) r A l3 m = rAl Thus, Iy = m l2 Ans Ans: Iy = 791 m l2 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–2 z The solid cylinder has an outer radius R, height h, and is made from a material having a density that varies from its center as r = k + ar2, where k and a are constants Determine the mass of the cylinder and its moment of inertia about the z axis R h SOLUTION Consider a shell element of radius r and mass dm = r dV = r(2p r dr)h R m = (k + ar2)(2p r dr)h L0 m = 2p h( aR4 kR2 + ) m = p h R2(k + aR2 ) Ans dI = r2 dm = r2(r)(2p r dr)h R Iz = L0 r2(k + ar2)(2p r dr) h R Iz = 2ph L0 Iz = 2ph[ Iz = (k r3 + a r5) dr k R4 aR6 + ] aR2 p h R4 [k + ] Ans Ans: aR2 b p h R4 aR2 Iz = ck + d m = p h R2 ak + 792 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–3 y Determine the moment of inertia of the thin ring about the z axis The ring has a mass m R x SOLUTION 2p Iz = L0 r A(R du)R2 = 2p r A R3 2p m = L0 r A R du = 2p r A R Thus, I z = m R2 Ans Ans: Iz = mR2 793 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *17–4 y The paraboloid is formed by revolving the shaded area around the x axis Determine the radius of gyration kx The density of the material is r = Mg>m3 y2 50x 100 mm x SOLUTION 200 mm dm = r p y2 dx = r p (50x) dx Ix = y dm = L0 L2 = r pa 200 50 x {p r (50x)} dx 502 200 bc x d = rp a 502 b(200)3 200 m = L dm = L0 p r (50x) dx 200 = r p (50)c x2 d = rp a kx = 50 b (200)2 Ix 50 (200) = 57.7 mm = Am A3 Ans Ans: kx = 57.7 mm 794 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–5 y Determine the radius of gyration kx of the body The specific weight of the material is g = 380 lb>ft3 y3 = x in x in SOLUTION dm = r dV = rp y2 dx d Ix = 1 (dm) y2 = pry4 dx 2 Ix = prx4/3 dx = 86.17r L0 m = kx = L0 prx2/3 dx = 60.32r 86.17r Ix = = 1.20 in Am A 60.32r Ans Ans: kx = 1.20 in 795 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–6 y The sphere is formed by revolving the shaded area around the x axis Determine the moment of inertia Ix and express the result in terms of the total mass m of the sphere The material has a constant density r x2 + y2 = r2 x SOLUTION dIx = y2 dm dm = r dV = r(py2 dx) = r p(r2 - x2) dx dIx = r p(r2 - x2)2 dx r Ix = = 2 r p(r - x ) dx L- r pr r5 15 r m = = L- r 2 Ix = m r2 r p(r - x ) dx r p r3 Thus, Ans Ans: Ix = 796 mr2 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–7 y The frustum is formed by rotating the shaded area around the x axis Determine the moment of inertia Ix and express the result in terms of the total mass m of the frustum The frustum has a constant density r y –ba x b 2b b x z SOLUTION a dm = r dV = rpy2 dx = rp A 2b2 b2 x + b2 B dx x + a a2 dIx = 1 dmy2 = rpy4 dx 2 dIx = b4 b4 4b4 b4 rp A x4 + x3 + x2 + x + b4 B dx a a a a a Ix = L dIx = = 4b4 b4 b4 b4 x + b4 B dx rp A x4 + x3 + x2 + a L0 a a a 31 rpab4 10 a m = Ix = Lm dm = rp L0 A x2 + b2 a 2b2 x + b2 B dx = rpab2 a 93 mb 70 Ans Ans: Ix = 797 93 mb 70 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *17–8 y The hemisphere is formed by rotating the shaded area around the y axis Determine the moment of inertia Iy and express the result in terms of the total mass m of the hemisphere The material has a constant density r x2 y2 r2 x SOLUTION r m = LV r dV = r = rp cr y - Iy = = L0 r p x2 dy = rp L0 (r2 - y2)dy r y d = rp r3 3 r r r rp (dm) x2 = px4 dy = (r2 - y2)2 dy L0 L0 Lm rp y5 r 4rp c r y - r2 y3 + d = r 15 Thus, Iy = m r2 Ans Ans: Iy = 798 mr © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–9 Determine the moment of inertia of the homogeneous triangular prism with respect to the y axis Express the result in terms of the mass m of the prism Hint: For integration, use thin plate elements parallel to the x–y plane and having a thickness dz z –h (x – a) z = –– a h SOLUTION dV = bx dz = b(a)(1 - z ) dz h x b a y x dIy = dIy + (dm)[( )2 + z2] = x2 dm(x2) + dm( ) + dmz2 12 = dm( x2 + z2) = [b(a)(1 - a2 z z )dz](r)[ (1 - )2 + z2] h h k Iy = abr a3 h - z z ( ) + z2(1 - )]dz h h L0 = abr[ = [ 1 1 a2 (h4 - h4 + h4 - h4) + ( h4 - h4 )] h 3h3 abhr(a2 + h2) 12 m = rV = abhr Thus, Iy = m (a + h2) Ans Ans: Iy = 799 m ( a + h2 ) © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–10 The pendulum consists of a 4-kg circular plate and a 2-kg slender rod Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O O 2m Solution Using the parallel axis theorem by referring to Fig a, IO = Σ ( IG + md ) = c 1m 1 (2) ( 22 ) + ( 12 ) d + c (4) ( 0.52 ) + ( 2.52 ) d 12 = 28.17 kg # m2 Thus, the radius of gyration is kO = 28.17 IO = = 2.167 m = 2.17 m Am A4 + Ans Ans: kO = 2.17 m 800 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–106 The uniform bar of mass m and length L is balanced in the vertical position when the horizontal force P is applied to the roller at A Determine the bar’s initial angular acceleration and the acceleration of its top point B B L Solution + ΣFx = m(aG)x; P = maG d c + ΣMG = IGa; P = mLa a = 6P mL aG = P m P A L P a b = a mL2 ba 12 Ans aB = aG + aB>G - aB i = -P L i + m + ) (d aB = aB = - = P La m P L 6P - a b m mL 2P 2P = m m Ans Ans: 6P mL 2P aB = m a = 897 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–107 Solve Prob 17–106 if the roller is removed and the coefficient of kinetic friction at the ground is μk B L Solution + ΣFx = m(aG)x; d P - mkNA = maG P A L = a mL2 ba 12 NA - mg = ( P - mkNA ) c + ΣMG = IGa; + c ΣFy = m(aG)y; Solving, NA = mg aG = a = L a 6(P - mkmg) mL Ans aB = aG + aB>G +) a = (S B aB = aB = L L a + a La 2(P - mkmg) m Ans Ans: a = aB = 898 6(P - mk mg) mL 2(P - mk mg) m © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *17–108 The semicircular disk having a mass of 10 kg is rotating at v = rad>s at the instant u = 60° If the coefficient of static friction at A is ms = 0.5, determine if the disk slips at this instant v 0.4 m O G (0.4) m 3p u A SOLUTION Equations of Motion:The mass moment of inertia of the semicircular disk about its center of mass is given by IG = (10) A 0.4 B - 10 (0.16982) = 0.5118 kg # m2 From the geometry, rG>A = 20.16982 + 0.4 - 2(0.1698) (0.4) cos 60° = 0.3477 m Also, using sin u sin 60° law of sines, , u = 25.01° Applying Eq 17–16, we have = 0.1698 0.3477 a + ©MA = ©(Mk)A ; 10(9.81)(0.1698 sin 60°) = 0.5118a + 10(a G)x cos 25.01°(0.3477) + 10(a G)y sin 25.01°(0.3477) + ©F = m(a ) ; ; x G x + c Fy = m(a G)y ; (1) (2) Ff = 10(aG)x (3) N - 10(9.81) = - 10(aG)y Kinematics:Assume that the semicircular disk does not slip at A, then (a A)x = Here, rG>A = {- 0.3477 sin 25.01°i + 0.3477 cos 25.01°j} m = {- 0.1470i + 0.3151j} m Applying Eq 16–18, we have a G = a A + a * rG>A - v2rG>A -(aG)x i - (aG)y j = 6.40j + ak * ( -0.1470i + 0.3151j) - 42( -0.1470i + 0.3151j) -(a G)x i - (aG)y j = (2.3523 - 0.3151 a) i + (1.3581 - 0.1470a)j Equating i and j components, we have (a G)x = 0.3151a - 2.3523 (4) (a G)y = 0.1470a - 1.3581 (5) Solving Eqs (1), (2), (3), (4), and (5) yields: a = 13.85 rad>s2 (aG)x = 2.012 m>s2 Ff = 20.12 N (aG)y = 0.6779 m>s2 N = 91.32 N Since Ff (Ff)max = msN = 0.5(91.32) = 45.66 N, then the semicircular disk does not slip Ans Ans: Since Ff (Ff )max = ms N = 0.5(91.32) = 45.66 N, then the semicircular disk does not slip 899 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–109 The 500-kg concrete culvert has a mean radius of 0.5 m If the truck has an acceleration of m>s2, determine the culvert’s angular acceleration Assume that the culvert does not slip on the truck bed, and neglect its thickness m/s2 4m 0.5m SOLUTION Equations of Motion: The mass moment of inertia of the culvert about its mass center is IG = mr2 = 500 A 0.52 B = 125 kg # m2 Writing the moment equation of motion about point A using Fig a, a + ©MA = ©(Mk)A ; (1) = 125a - 500a G(0.5) Kinematics: Since the culvert does not slip at A, (aA)t = m>s2 Applying the relative acceleration equation and referring to Fig b, a G = a A + a * rG>A - v2rG>A a Gi = 3i + (a A)n j + (ak * 0.5j) - v2(0.5j) aGi = (3 - 0.5a)i + C (aA)n - 0.5v2 D j Equating the i components, (2) a G = - 0.5a Solving Eqs (1) and (2) yields aG = 1.5 m>s2 : a = rad>s2 Ans Ans: a = rad>s2 900 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–110 The 15-lb disk rests on the 5-lb plate A cord is wrapped around the periphery of the disk and attached to the wall at B If a torque M = 40 lb # ft is applied to the disk, determine the angular acceleration of the disk and the time needed for the end C of the plate to travel ft and strike the wall Assume the disk does not slip on the plate and the plate rests on the surface at D having a coefficient of kinetic friction of μk = 0.2 Neglect the mass of the cord B A M ϭ 40 lb и ft 1.25 ft D C ft Solution Disk: + S ΣFx = m(aG)x; T - FP = 15 a 32.2 G Plate: 15 b(1.25)2 d a - FP(1.25) + 40 - T(1.25) = c a 32.2 + ΣF = m(a ) ; S x G x FP - = a+ ΣMG = IGa; a 32.2 P aP = aG + aP>G +) (S aP = aG + a(1.25) aG = a(1.25) Thus aP = 2.5a Solving, FP = 9.65 lb aP = 36.367 ft>s2 a = 14.5 rad>s2 T = 18.1 lb +) (S s = s0 + v0 t + = + + t = 0.406 s Ans at c (36.367)t 2 Ans Ans: a = 14.5 rad>s2 t = 0.406 s 901 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–111 The semicircular disk having a mass of 10 kg is rotating at v = rad>s at the instant u = 60º If the coefficient of static friction at A is mV = 0.5, determine if the disk slips at this instant v 0.4 m O G (0.4) m 3p u A SOLUTION For roll A T(0.09) = 12 (8)(0.09)2 aA c + ©MA = IA a; (1) For roll B a + ©MO = ©(Mk)O; 8(9.81)(0.09) = 12 (8)(0.09)2 aB + 8aB (0.09) (2) + c ©Fy = m(aG)y ; T - 8(9.81) = - 8aB (3) Kinematics: aB = aO + (aB>O)t + (aB>O)n c aBd = c aOd + caB (0.09) d + [0] T T T aB = aO + 0.09aB A+TB (4) also, A+TB (5) aO = aA (0.09) Solving Eqs (1)–(5) yields: aA = 43.6 rad>s2 Ans aB = 43.6 rad>s2 Ans T = 15.7 N Ans aB = 7.85 m>s2 aO = 3.92 m>s2 Ans: The disk does not slip 902 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *17–112 v The circular concrete culvert rolls with an angular velocity of v = 0.5 rad>s when the man is at the position shown At this instant the center of gravity of the culvert and the man is located at point G, and the radius of gyration about G is kG = 3.5 ft Determine the angular acceleration of the culvert The combined weight of the culvert and the man is 500 lb.Assume that the culvert rolls without slipping, and the man does not move within the culvert ft O G 0.5 ft SOLUTIONS Equations of Motion: The mass moment of inertia of the system about its mass 500 (3.52) = 190.22 slug # ft2 Writing the moment equation of center is IG = mkG2= 32.2 motion about point A, Fig a, + ©MA = ©(Mk)A ; -500(0.5) = - 500 500 (a ) (4) (a ) (0.5) - 190.22a (1) 32.2 G x 32.2 G y Kinematics: Since the culvert rolls without slipping, a0 = ar = a(4) : Applying the relative acceleration equation and referrring to Fig b, aG = aO + a * rG>O - v2rG>A (aG)xi - (aG)y j = 4ai + (- ak) * (0.5i) - (0.52)(0.5i) (aG)xi - (aG)y j = (4a - 0.125)i - 0.5aj Equation the i and j components, (aG)x = 4a - 0.125 (2) (aG)y = 0.5a (3) Subtituting Eqs (2) and (3) into Eq (1), - 500(0.5) = - 500 500 (4a - 0.125)(4) (0.5a)(0.5) - 190.22a 32.2 32.2 a = 0.582 rad>s2 Ans Ans: a = 0.582 rad>s2 903 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–113 v0 The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor Determine the initial angular acceleration of the disk and the acceleration of its mass center The coefficient of kinetic friction between the disk and the floor is mk r SOLUTION Equations of Motion Since the disk slips, the frictional force is Ff = mkN The mass moment of inertia of the disk about its mass center is IG = m r2 We have + c ©Fy = m(aG)y; N - mg = N = mg + ©F = m(a ) ; ; x G x mk(mg) = maG aG = mkg ; Ans -mk(mg)r = a m r2 b a a = 2mkg b r Ans a+ ©MG = IGa; Ans: aG = mkg d 2mkg b a = r 904 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–114 v0 The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor Determine the time before it starts to roll without slipping What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is mk r SOLUTION Equations of Motion: Since the disk slips, the frictional force is Ff = mkN The mass moment of inertia of the disk about its mass center is IG = m r2 + c ©Fy = m(aG)y; N - mg = N = mg + ©F = m(a ) ; ; x G x mk(mg) = maG + ©MG = IGa; -mk(mg)r = - a m r2 b a aG = mkg a = 2mkg r Kinematics: At the instant when the disk rolls without slipping, vG = vr Thus, + B A; vG = (vG)0 + aGt vr = + mkgt t = vr mkg (1) and v = v0 + a t (a +) v = v0 + a - 2mkg bt r (2) Solving Eqs (1) and (2) yields v = v t = v0r 3mkg Ans Ans: v0 v0r t = 3mkg v = 905 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–115 A cord is wrapped around each of the two 10-kg disks If they are released from rest, determine the angular acceleration of each disk and the tension in the cord C Neglect the mass of the cord D 90 mm A SOLUTION C For A: c + ©MA = IA aA; T(0.09) = c (10)(0.09)2 d aA (1) T(0.09) = c (10)(0.09)2 d aB (2) 90 mm B For B: a + ©MB = IBaB; + T ©Fy = m(aB)y; (3) 10(9.81) - T = 10aB aB = aP + (aB>P)t + (aB>P)n (4) ( + T) aB = 0.09aA + 0.09aB + Solving, aB = 7.85 m>s2 aA = 43.6 rad>s2 Ans aB = 43.6 rad>s2 Ans Ans T = 19.6 N Ay = 10(9.81) + 19.62 = 118 N Ans: aA = 43.6 rad>s2b aB = 43.6 rad>s2d T = 19.6 N 906 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *17–116 The disk of mass m and radius r rolls without slipping on the circular path Determine the normal force which the path exerts on the disk and the disk’s angular acceleration if at the instant shown the disk has an angular velocity of V R SOLUTION r Equation of Motion: The mass moment of inertia of the disk about its center of mass is given by IG = mr2 Applying Eq 17–16, we have a + ©MA = ©(Mk)A ; ©Fn = m(aG)n ; mg sin u(r) = a mr2 b a + m(aG)t (r) [1] [2] N - mg cos u = m(aG)n Kinematics: Since the semicircular disk does not slip at A, then yG = vr and (aG)t = ar Substitute (aG)t = ar into Eq [1] yields mg sin u(r) = a mr2 b a + m(ar)(r) a = 2g sin u 3r Ans Also, the center of the mass for the disk moves around a circular path having a y2G v2r2 radius of r = R - r Thus, (aG)n = Substitute into Eq [2] yields = r R - r N - mg cos u = ma N = m v2r2 b R - r v2r2 + g cos u R - r Ans Ans: 2g sin u 3r v2r N = ma + g cos u b R - r a = 907 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–117 The uniform beam has a weight W If it is originally at rest while being supported at A and B by cables, determine the tension in cable A if cable B suddenly fails Assume the beam is a slender rod A B SOLUTION + c ©Fy = m(aG)y; c + ©MA = IAa; TA - W = Wa L –– L –– W L L W L b = c a b L2 d a + a b aa b g 4 12 g = Since aG = a a L –– W a g G L L a + ba g L b a = 12 g a b L TA = W TA = g W 12 L W L (a) a b = W a ba ba b g g L 4 W Ans Also, + c ©Fy = m(aG)y ; c + ©MG = IG a; Since a G = TA - W = TA a W a g G L W b = c a b L2 d a 12 g L a TA = W a b La g W L W a b La - W = - a b a g g a = 12 g a b L TA = g 12 W a bLa b a b g L TA = W Ans Ans: TA = 908 W © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–118 The 500-lb beam is supported at A and B when it is subjected to a force of 1000 lb as shown If the pin support at A suddenly fails, determine the beam’s initial angular acceleration and the force of the roller support on the beam For the calculation, assume that the beam is a slender rod so that its thickness can be neglected 1000 lb B ft A ft SOLUTION + ; a Fx = m(aG)x ; 500 1000 a b = (a ) 32.2 G x + T a Fy = m(aG)y ; 500 1000 a b + 500 - By = (a ) 32.2 G y a + a MB = a (Mk)B; 500 500 500(3) + 1000 a b (8) = (a ) (3) + c a b (10)2 d a 32.2 G y 12 32.2 aB = aG + aB>G - aBi = - (aG)x i - (aG)yj + a(3)j ( + T) (aG)y = a(3) a = 23.4 rad>s2 Ans By = 9.62 lb Ans By means that the beam stays in contact with the roller support Ans: a = 23.4 rad>s2 By = 9.62 lb 909 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17–119 The solid ball of radius r and mass m rolls without slipping down the 60° trough Determine its angular acceleration 30° SOLUTION d = r sin 30° = 30° r ©Ma - a = ©(Mk)a - a; 45° r r 2 mg sin 45°a b = c mr2 + m a b da a = 10g Ans 13 22 r Ans: a = 910 10g 1322 r © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *17–120 By pressing down with the finger at B, a thin ring having a mass m is given an initial velocity v0 and a backspin V when the finger is released If the coefficient of kinetic friction between the table and the ring is mk, determine the distance the ring travels forward before backspinning stops B ω0 v0 r SOLUTION + c ©Fy = 0; A NA - mg = NA = mg + ©F = m(a ) ; : x G x mk (mg) = m(aG) aG = mk g a + ©MG = IG a; mk (mg)r = mr a a = (c +) v = v0 + ac t = v0 - a t = + B A; mk g r mk g bt r v0 r mk g s = s0 + v0t + ac t2 s = + v0 a s = v0 r mk g v0 r v20 r2 b - a b (mk g) a 2 b mk g mk g v0 - v r Ans Ans: s = a 911 v0 r b a v0 - v0 r b mk g ... = r dV = r( 2p r dr)h R m = (k + ar2)(2p r dr)h L0 m = 2p h( aR4 kR2 + ) m = p h R2 (k + aR2 ) Ans dI = r2 dm = r2 (r) (2p r dr)h R Iz = L0 r2 (k + ar2)(2p r dr) h R Iz = 2ph L0 Iz = 2ph[ Iz = (k r3 ... Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by. .. Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form