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Unformatted text preview: £50 I 16100111111 “ll“ . PROBLEM 17.11 The double pulley shown has a mass of 14 kg and a centroidal radius of
gyration of 165 mm. Cylinder A and block B are attached to cords that are
wrapped on the pulleys as shown. The coefﬁcient of kinetic friction
between block B and the surface is 0.25. Knowing that the system is
released from rest in the position shown, determine (a) the velocity of
cylinder A as it strikes the ground, (1)) the total distance that block B moves before coming to rest. 25!) mm SOLUTION Let VA = speed of block A, VB 2 speed of block B, a) = angular speed of pulley. Kinematics. VA = rAa) = 0.250a) v3 = rBa) = 0.15060 SA = rAB = 0.2506 s3 = r36 = 0.1506 (a) Cylinder A falls to ground. SA = 0.900 m s3 = M(0.900) = 0.540 m
0.250
Work ofweightA: UHZ = mAgsA = (11.5)(9.81)(0.900) = 101.534 J
Normal contact force acting on block B: N = mg = (9)(9.81) = 88.29 N
Friction force on block B: Ff = ,ukN = (0.25)(88.29) = 22.0725 N
Work offriction force: UH2 = —FfsB = —(22.0725)(0.540) = —11.919 J
Total work: UH2 =101.534 — 11.919 = 89.615 J
. . 1 2 1 2 1 2
K1net1c energy: T1 = 0; T2 = —mAvA + —Ia) + —vaB
2 2 2
T2 = lmArja)2 + lmckza)2 + lmgrgw2
2 2 2
= %[(11.5)(0.250)2 + (14)(0.165)2 + (9)(0.150)2]a)2 = 0.6512502
Principle of Work and Energy. T1 + UH2 = T:
0 + 89.615 = 0.6512w2 a) = 11.7309 rad/s
Velocity of cylinderA: VA = (0.250)(11.7309) vA =2.93m/s§< PROBLEM 17.11 CONTINUED (b) Block B comes to rest. For blockB and pulley C. T3 = élwz +ém3v123; T4 = 0 :
1 2 2 1 2 2 1 2 2 2
T3 = 3ka a) +EmBrBa) = 5 (14)(0.165) +(9)(0.150) ](11.7309)
= 40.159 J
Work of friction force: U3<)4 = —Ffsj9 = —22.07255};
Principle of Work and Energy. T3 + U3_,4 2 T4: 40.159 — 22.0725s;g = o s}, = 1.819 In Total distance for block B. d = s3 + SE: d =0.540+1.819 SOLUTION PROBLEM 17.15 A slender 6kg rod can rotate in a vertical plane about a pivot at B. A
spring of constant K = 600 N/m and an unstretched length of 225 mm is
attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated
through 90°. Position I . BG = 0.270 m CD = 0.5252 + 0.1802 = 0.555 m
x1 = CD — 10 = 0.555 — 0.225 = 0.330 m 12 = %(600)(0.330)2 = 32.67 J h1 = BG = 0.270 m (Vg)1 = Wh1 = mgh = (6)(9.81)(—O.270) = —15.892 J Kinetic energy: 7] = 0
Position 2. CD = 0.525 — 0.180 = 0.345 m
Spring: x2 = CD — [0
x2 = 0.345 — 0.225 = 0.120m 1 )2 = 5(600)(0.120)2 = 432 J Gravity: Kinetic energy: = 1—12—(6)(0.900)2 = 0.405 kgm2 172 = (30% = 0.27002 T2 =%m1722 + gig); = %(6)(0.270a)2)2 + %(0.405)a)22 = 0.4212a;22 Conservation of Energy. T1 + VI = T2 + V2: 0 + 32.67 — 15.892 = 0.4212022 + 0 + 4.32 (02 = 5.44 002 = 5.44 rad/s) 4 SOLUTION PROBLEM 17.29 The mechanism shown is one of two identical mechanisms attached to the
two sides of a ZOOlb uniform rectangular door. Edge ABC of the door is
guided by wheels of negligible mass that roll in horizontal and vertical
tracks. A spring of constant k = 40 lb/ft is attached to wheel B. Knowing
that the door is released from rest in the position 0 = 30° with the spring
unstretched, determine the velocity of wheel A just as the door reaches the vertical position. Kinematics. Locate the instantaneous center at point 1. VA = (5 sin 6) a)
Moment of inertia. T = LEOOY = 51.7601bsZft
12 32.2 Kinetic energy. T = lﬂ(5cosl9)2 02 + l(51.760)a)2
2 32.2 2 = (77.640cos2 0 + 25.880)a)2 Potential energy. Gravity. Datum at levelA. Vg = Wh = (200)(—55in6)  ~1000sin€ ftlb Two springs. Ve = 2(é—ke2j = k(5$in(9 ~ Ssin30°)2 = 25k(sin6 — 0.5)2
Position 1. (9 = 30°
= —500 ft~1b
Position 2. 0 = 90° 0) (02 T2 = (77.640cos2 90° + 25.880)a)2 = 25.88002 ‘,~\_ PROBLE
V2 = M 17.29 CONTINUED (V )2 + (Vg)2 = (25)(40)(sin90° — 0.5)2 — 500sin90° 8 —750 ftlb Conservation of energy. T1 + Vl = T2 + V2: Kinematics. 0 — 500 = 25.880w2 — 750 a) = 3.1080 rad/s
VA = (55in90°)(3.1080) VA = 15.54 ft/s m4 PROBLEM 17.62 A drum of 4—in. radius is attached to a disk of 8in. radius. The disk and
drum have a combined weight of 10 lb and a combined radius of gyration
of 6 in. A cord is attached to the drum at A and pulled with a constant
force 1’ of magnitude 5 lb. Knowing that the disk rolls without sliding and
that its initial angular velocity is 10 rad/s clockwise, determine (a) the
angular velocity of the disk after 5 s, (b) the corresponding total impulse
of the friction force exerted on the disk at B. SOLUTION V = VG = rBa)
‘2
Moment of inertia. I = Wk
g
Kinetics.
Syst Momenta1 + Syst Ext Imp1 _,2 = Syst Momenta2
x. . — W — 
/ moments about B. [(01 + Ever + WtrB srnﬂ — Pt(rB — rA)
= 7002 + [7er
g
£0? + r§)(a)2 — ml) 2 [WrB sinﬂ — P(rB  rA)]t
g PROBLEM 17.62 CONTINUED WrB sinﬂ — P(rB — rA) (02 = w1+ W(/?2 + r13) gt W=lOlb, rA =4in.=%ft, rB =8in.=§ft i=6in.=%ft P=51b t=55 ml 2 —10rad/s ,8 = 30° (10)(§)sin30° — (5)(§ — 4) ———2———2—(32.2)(5)
mm) + (4) ] (1.66667)(32.2)(5)
6.9444 602 : —10+ = —lO + = 28.64 rad/s (02 = 28.6 rad/s ‘34 +/ components parallel to incline: E17] + Wtsinﬂ — Pt — det = 14:72
g g 1th = (Wsinﬂ — 10%ng _ 7,) = (Wsinﬂ  P)t = (105in30° — 5)(5) — = —8.001bs
det = 8.001bs 730% /\ PROBLEM 17.68 A sphere of radius r and mass m is placed on a horizontal ﬂoor with no
linear velocity but with a clockwise angular velocity wo. Denoting by
,uk the coefﬁcient of kinetic friction between the sphere and the ﬂoor,
determine (a) the time z‘1 at which the sphere will start rolling without
sliding, (b) the linear and angular velocities of the sphere at time t1. SOLUTION
Moment of inertia. Solid sphere. I— = gmr2
rim
KN / 1w: \\
‘i‘ 1; +  2 —
/ _ G * 2 m4};
"N "'1’” x”;
. , “we!"
Ft
N’L
Syst Momenta1 Syst Ext Imp1 —>2 Syst Momenta2
+Ty components: Ntl — th = 0 N = W = mg (l)
“31.x components: Ft1 = mv2 (2)
+3 moments about G: lwo — Ftlr = 7602 (3)
Since F = ,ukN = ykmg, Equation (2) gives
ykmgt1 = mv2 or v2 2 ,uk gt1 (4)
Using the value for l in Equation (3),
Emrza)0 — ,ukmgtlr = Zera)2 or (02 = (00 — i’ukgtl (5)
5 5 2 r
(a) Time II at which sliding stops.
From kinematics, v2 = r0)
5 2 m;
#kgt1 = “00 — ‘ﬂkgﬁ t1 = —_‘£ ‘
2 7 #kg
(b) Linear and angular velocities.
. 2
From Equation (4), v2 = ,ukggﬂ v2 = —ra)0 > {
7 #kg 7
From Equation (6), wz = 2 = Zwo 002 = 12— 0) 4 r 7 7 /"\ PROBLEM 17.81 A 1.8kg collar A and a 0.7kg collar B can slide without friction on a 7
frame, consisting of the horizontal rod 0E and the vertical rod CD, which
is free to rotate about its vertical axis of symmetry. The two collars are
connected by a cord running over a pulley that is attached to the frame at
0. At the instant shown, the velocity VA of collar A has a magnitude of
2.1 m/s and a stop prevents collar B from moving. The stop is suddenly
removed and collar A moves toward E. As it reaches a distance of 0.12 m
from 0, the magnitude of its velocity is observed to be 2.5 m/s.
Determine at that instant the magnitude of the angular velocity of the
frame and the moment of inertia of the frame and pulley system
about CD. ‘
__
1— ——1
SOLUTION
Components of velocity of collar A. vi, = (VA )3 + (VA )2 (1)
Constraint of rod 0E. (VA )0 = rAa) (2)
Constraint of cable AB. ArA = AyB, (VA )r = v3 (3)
Position 1. (NA) = 0.1m, [(vAH1 = 0, (VA)1 = 2.1m/s
2
From Equation (1), (2.1)2 = 0 + [(vA )01 [(VA)6:L = 2.1m/s A
From Equation (2), (2.1) = 0.1601 (01 = 21 rad/s
From Equation (3), v3 = 0
Potential energy. Take position 1 as datum. V1 = 0 (4)
Angular momentum. (H0)1 = 1021 + mA [(VA)r]1(rA)1:
(H0)1 = [(21) + (1.8)(2.1)(0.1) (H0)1 = 211 + 0.378 (5)
. . 1 2 1 2 1 2
Kinetic energy. 71 = —Ia)1 + —mAvA + —vaB:
2 2 2
1 2 l 2
T1 = E[(21) + E(1.8)(2.1) T1 = 220.51 + 3.969 (6)
Position 2. (rA)2 2 0.12m, (VA)2 = 2.5 m/s 0) = wz
From Equation (2), [(VA )012 = 0.12602
. 2 2 2 2 2 2
From Equation (1), [(VA),:2 = (VA)2 — [(vA)0]2 = (2.5) — (0.12) (02 : 6.25 — 0.014460% L A A‘ PROBLEM 17.81 CONTINUED From Equation (3), v; = 6.25 — 0.0144022
Change in radial position. ArA = (rA)2 — (0)1 = 0.02 m
From Equation (3), AyB : 0.02 m
Potential energy. V2 = mBg(AyB) = (0.7)(9.81)(0.02)
V2 = 0.137341 (7)
Angular momentum. (H0)2 = 1602 + mA [(VAMZ (rA)2:
(H0)2 = [(02 + (1.8)(0.12a)2)(0.12) (H0)2 = (1 + 0.02592)a)2 (8)
Kinetic energy. T2 = %1w§ + émAvj + évag:
T2 2%103 + %(1.8)(2.5)2 + %(0.7)(6.25 — 0.0144022)
12 = (0.51 — 0.00504)a)22 + 7.8125 (9)
Conservation of angular momentum. (HO)1 = (H0)2: 211 + 0.378 = (I + 0.02592)a)2 Solving for (02, (02 = % = % (10)
+ . Conservation of energy. T1 + VI = T2 + V2: 220.51 + 3.969 = (0.51 — 0.00504)a)22 + 7.8125 + 0.13734 2
220.51 — (0.51 — 0.00504)% — 3.98084 = 0 220.51132 — 0.51N2 + 0.00504N2 — 3.98084D2 = 0 220.51(12 + 0.051841 + 0.0006718464) — 051(44112 + 15.8761 + 0.142884) + 0.00504(44112 + 15.8761 + 0.142884) — (3.98084)(12 + 0.051841 + 0.0006718464) = 0 013 +1.7345212 — 0.049651671 — 0.001954378 = 0 Solving the quadratic equation for I, _ 0.04965167 i 0.126590 1 = 0.050804 and —0.022179 3.46904 PROBLEM 17.81 CONTINUED Reject the negative root. _ (21)(0.050804) + 0.378 From Equation (10), Q2 — 0 050804 + 0 02592 a) = 18.83 rad/s { I = 0.0508 kgm2 4 PROBLEM 17.98 The slender rod AB of length L forms an angle ﬂ with the vertical axis as
it strikes the frictionless surface shown with a vertical velocity V1 and no
angular velocity. Assuming that the impact is perfectly elastic, derive an
expression for the angular velocity of the rod immediately after
the impact. SOLUTION Moment of inertia. Perfectly elastic impact. Kinetics. M Syst Momenta1 Syst Ext Imp1 _,2 Syst Momentaz _+> horizontal components: 0 + 0 = m17x Kinematics. v0 2 VA + vGM [Vy l] = [v1 l l+[(vA)x w» ]+[3w‘§ ,6] Velocity components T : )moments about A: mvl 55in ﬂ + 0 = —mVy gsin ﬂ + fa) mvl 55inﬂ = m[§a)sinﬂ — vljésinﬂ + émLza) (émL2 + imL2 sin2 ,ij = mvlL sinﬂ 12'
“—81% £4
l+3s1n ﬂL SOLUTION Moment of inertia. 3 Wall}? Syst Momenta1 Sphere A: +T components: +T linear components: 3:. linear components: PROBLEM 17.109 The 9kg rigid body BD consists of two identical 60mmradius spheres
and the rod which connects them and has a centroidal radius of gyration
of 250 mm. The body is at rest on a horizontal frictionless surface when it
is struck by the 3kg sphere A which has a radius of 60 mm and is moving
as shown with a velocity VI of magnitude 4 m/s. Assuming a perfectly
plastic impact, determine immediately after the impact (a) the angular
velocity of body BD, (b) the velocity of point G. ll + Syst Ext Imsz = 0+0=mA(vA) (VA) =0
y y Let (VA )x be named vA and note that Q = — = —. Then, 4;) moments about B: Kinematics. a): 0+0=mA(vA)y+mVy Vy=0
mAvI +0 2 mA(vA)x + mix 17x = Inmihl —(vA)x]
3 1
m 9 3
__ 1
Vx=3‘(V1‘VA) 0 + 0 = I_a) — (bsin60°)mi7x mb sin 60° _ b sin 60°
—————v — (vI m]? X‘ 31:2 W) VB 2 VG +vB/G = [Vx M ]+[ba) TI 30°]  2
(v3)x = l(v1 — VA) + bcos30°$ 3 (V1 — VA) Impact condition. For perfectly plastic impact, (a) (b) PROBLEM 17.109 CONTINUED 1 3b2 (v3)x = §[l + Ejbﬁ — VA)
_ 1 H M _ 3 (4)(0.250)2
(v3)x — VA = e(v1— 0) 0.69333(v1— VA) — VA = ev1
(v1— VA) = 0.69333(v] — VA) 1.69333vA = (0.69333 — e)v1
e = 0
1.69333vA = 0.69333v1 VA = 0.40945v1 v1 — VA = 0.59055v1 = (0.59055)(4) = 2.3622 m/s (0.300)sin 60° w = (3)(0.250)2 (2.3622) 1
_ =— 2.3622
v 3( > < co = 3.27 rad/s )4 = 0.787 m/s —~ 4 ...
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 Summer '06
 

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