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Unformatted text preview: PROBLEM 13.180 A 300g collar A is released from rest, slides down a frictionless rod, and
strikes a 900—g collar B which is at rest and supported by a spring of
constant 500 N/m. Knowing that the coefﬁcient of restitution between the
two collars is 0.9, determine (a) the maximum distance collar A moves up
the rod after impact, (17) the maximum distance collar B moves down the rod after impact. SOLUTION After impact Velocity of A just before impact, v0 v0 = ,/2g = 2(9.81m/52)(1.2 m)sin30° = (9.81)(1.2)(0.5) = 3.431m/s Conservation of momentum
t4“. mAvo = vaB — mAvA: 0.3v0 = 0.9vB — 0.3vA (1)
Restitution
(vA + v3) = e(v0 + 0) = 0.9v0
Substituting for v3 from (2) in ( 1)
0.3v0 = 0.9(0.9v0 — vA) — 0.3vA, 1.2vA = 0.51v0 vA = 1.4582 m/s, v3 = 1.6297 m/s (a) A moves up the distance d where; %MAvﬁ = MAgdsin30°; %(1.4582 m/s)2 = (9.81m/s2)d(o.5) A), 61,, = 0.21675 m = 217 mm 4 “3
ID", ,‘f (D (b) Static deﬂection = x0, B moves down All)? @ / I Aﬁ’x‘, Conservation of energy (1) to (2) dc . " “35" : Position (1) — spring deﬂected, x0 TI+V1=T2+V2; Tl=ém3v123$ T2=0 PROBLEM 13.180 CONTINUED
VI = Ve + Vg = ékxé + mBgstin30°
, , x + 1
V2 = Ve + Vg = [00 dBlocdx = 314d; + 2de0 + x3) 1 2 ékxg + mgdg Sin30° + EvaB = %k(d§ + Zde0 + x3) + 0 + 0 m; = vaé; 500d; = 09(16297)2,d3 = 00691 m dB = 69.1mm4 éll) 111111 T__ SOLUTION PROBLEM 13.185 A 9.13kg sphere A of radius 90 mm moving with a velocity of
magnitude v0 = 2 m/s strikes a 830g sphere B of radius 40 mm which is
hanging from an inextensible cord and is initially at rest. Sphere B swings
to a maximum height h after the impact. Determine the range of values of
h for values of the coefﬁcient of restitution 6 between zero and one. Angle of impulse force from geometry of A and B 8 30mm somm‘:Zf
6 8 6 = cos—(ﬂ) = 22.62° :zomm Momentum consideration Ball A: Ball B: Restitution M41539
+ = vama
0*“ C mmrﬂ’x TRAt FAt
EC) l O :. 0—4 mBVizsz 23h (1) my 0?» Approach Separation
v’ 0056 — v’ c036 + ' Sing
6: B (A)x (VA)y VA =V0=2m/s
VA c059
e _ v33 — (v’A)x + (12:4)y tan0 _ v2; — (v;)x + (v'A)y(%) VA 2 PROBLEM 13.185 CONTINUED
A: mAvA sin19 = mA(v:4)x(sint9) + mA(v'A)y(cosé?)
10 = 5(v;,)x +12(v;,)y (2) . __ I r
A +B. mAvA — mA(vA)x + vaB 2 = (v'A)x + % (3)
From the Equation for e
I I 7 5
e = 0; VB — (VAX + (ml13) = 0 (4)
I I I 5
e=l; vB—(vA)x+(vA)y[Ti]=2 (5) Simultaneous solution of Equations (2), (3) and (4) for e = 0 and Equations (2), (3) and (5) for e = 1 yeilds
e = 0: (v'A)x = 1.8357 m/s, (my = 0.0685 m/s, v2; = 1.8072 m/s e =1: (V'A)x =1.6714m/s, (v2) = 0.1369 m/s, v53 = 3.6144m/s y , 2
h = (VB) = 0.1665 m, 0.6658m 166.5 5 h S 666mm‘ PROBLEM 15.6 When the power to an electric motor is turned on the motor reaches its
rated speed of 2400 rpm in 4 s, and when the power is turned off the
motor coasts to rest in 40 5. Assuming uniformly accelerated motion,
determine the number of revolutions that the motor executes (a) in
reaching its rated speed, (b) in coasting to rest. 2400 2
ml = 2400 rpm 2 (——6())(ﬂ = 807r rad/s, (00 = 0, ll: 45
(a) col = 010 + at = at, a = % 2 gig— = 207: rad/s2
.91: wot + laﬁ = 0 + l(207:)(42) =160mad = £5 = 80rev
2 2 27:
61 = 80 rev 4
(b) 601 = 807: rad/s, wz = 0, 1‘2 — t1 = 40 s
(02 =wl+a(t2—tl), a: wz—w‘ 20—80” =—27Irad/s2
t2 — r1 40
92 — .91 = mtg _ q) + 50502 _ t1)2 = (807:)(40) + %(—27r)(40)2
= 1600” radians = 1600” = 800 rev 192 — 61 = 800 rev 4 2n PROBLEM 15.13 The bent rod ABCDE rotates about a line joining points A and E with a
constant angular velocity of 12 rad/s. Knowing that the rotatiOn is
clockwise as viewed from E, determine the velocity of comer C . 'l SOLUTION
rA/E =—l6i+16j+8kin. [AE =\/162+162+32 =24in_
Angular velocity.
(0 12
co=——r =— —l6i+l6'+8k
[AE A/E 24( J )
co = —(8 rad/s)i + (8 rad/s)j + (4 rad/s)k
r05 = —(16 in.)i + (6 in.)j
Velocity ofC.
i j k
vC =coer/E = —8 8 4 = —24i—64j+80k
—16 6 0
v6 = —(24.0 in./s)i — (64.0 in./s) j + (80.0in./s)k4
Angular Acceleration. (1 = 0 Acceleration of C. i j k
aC=aer/E+o)xvc=0+ —8 8 4
—24 —64 80 896i + 544j + 704k ac = (896 in./sz)i + (544 in./sz) j + (704in./s2)k 4 PROBLEM 15.18 The belt shown moves over two pulleys without slipping. Pulley A starts
from rest with a clockwise angular acceleration deﬁned by the relation
05 = 120 — 0.002602, Where a is expressed in rad/s2 and a) is expressed
in rad/s. Determine, after onehalf revolution of pulley A, (a) the
magnitude of the acceleration of point B on the belt, (b) the acceleration
of point P on pulley C. SOLUTION ~1—rev = nradians, rA =160mm = 0.160 m, rC 2100mm = 0.100m 2 a=120_0,002w2 zwi‘lﬂ, Jim—2.2M?=dg
d6 120 — 0.00%) 60000 — 0) Integrating and applying initial condition a) = 0 at 6 = 0 and noting that :9 = 7r radians at the ﬁnal state, 1:; 500 male) 60000 2 = —2501n(60000  wz : Ede = 7r
" a) _ 2
—250[h1(60000 — 02) — 11160000] 2 —250 111%— = 7r — 02 = (BAH/250
60000 002 = 60000[1— (gr/25°] = 749.26 radz/sz
a) = 27.373 rad/s) a = 120 — 0.002 of = 120 — (0.002)(749.26) = 118.50 rad/s) (a) Tangential velocity and acceleration of point B on the belt. v3 = v, = 73,60 = (0.160)(27.373) = 4.3797 m/s
a3 = aA = rAa = (0.160)(118.50) = 18.96 m/s2 a3 = 18.96 m/s2 4 (b) Acceleration of point P on pulley C. p = 0.100 m
C VP = v3 = 4.3797 m/s i _ 4.37972
0.100 = 191.82 m/szw PROBLEM 15.18 CONTINUED (aP), = a3 = 18.96 m/s2 1 up = (191.82)2 + (18.96)2 = 192.8 m/s2 18.96 , = 5.64°
191.82 ﬂ tanﬂ= aP = 192.8 m/s2 7 5.64°4 ...
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 Summer '06
 

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