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Unformatted text preview: Problem 18.38 The mass of the disk is 45 kg and its
radius is R = 0.3 m. The spring constant is k =f3IIIL'IN/m.
The disk is rolled to the left until the spring is com—
pressed 05 m and released from rest. (a) If you assume that the disk rolls, what is its angular
acceleration at the instant it is released? (b) What is the minimum coefﬁcient of static friction
for which the disk will not slip when it is released? Solution:
x0 = 70.5
k : 600 N/m
m : 45 kg
R : 03 m
10 : 5M3 : 2.025 N—mzt F3. : kx ZEY: _ FA _ : 1715104 ER: N — mg = Rolling implies am : —Roz
We have, at x : 70.5 m
* kx * f : mam
N 7 mg 2 U
— Rf : 100: not = 7R0:
Four eqns, four unknowns (£101, or, N, f) (4) Solving f = I00 N, N = 441.5 N
at : —l4.81 rad/s2 (clockwise) um : 4.44 m/s2
(b) for impending slip, : Mr N
m = f/N = 100/4415 m = 0.227 Problem 18.40 A 42lb sphere with radius R = 4 in
is placed on a horizontal surface with initial angular
velocity coo = 40 rad/s. The coefﬁcient of kinetic fric
tion between the sphere and the surface is uk 2 0.06.
What maximum velocity will the center of the sphere
attain, and how long does it take to reach that velocity? Strategy: The friction force exerted on the spinning
sphere by the surface will cause the sphere to acceler
ate to the right. The friction force will also cause the
sphere’s angular velocity to decrease. The center of the
sphere will accelerate until the sphere is rolling on the
surface instead of slipping relative to it. Use the relation
between the velocity of the center and the angular veloc—
ity of the sphere when it is rolling to determine when
the sphere begins rolling. Solution: Given
W = 42 1b, g = 32.2 ft/sz, m = W/g, R = 4/12 a, 11k = 0.06
We have ZFX : ,ukN 2 ma ZEviN—ngO 2 9
ZMG tut/NR = ngYt Solving we ﬁnd _ SUM: a = 14.49 rad/$2, a = 111. g = 1.932 ft/s2 From kinematics we learn that a! = 14.49 rad/32, a) = (14.49 rad/32): — (40 rad/s) a = 1.932 ft/sz, v = (1.932 mg):
when we reach a steady motion we have
u = —Ra) => (1.932 131/52): = —(0.33 ft)[(l4.49 rad/52y — (40 rad/5)] Solving for the time we ﬁnd r=l.97s => v=3.81ft/s Problem 18.45 The l8kg ladder is released from rest
in the position shown. Model it as a slender bar and
neglect friction. At the instant of release, determine
(a) the angular acceleration of the ladder and (b) the
normal force exerted on the ladder by the ﬂoor. (See
Active Example 18.3.) Solution: The vector location of the center of mass is rG =
(L/2) sin 30°i + (L/2) cos 300j : li + 1.732j (m). Denote the normal
forces at the top and bottom of the ladder by P and N. The vector
locations of A and B are rA = L sin 30°i = 2i (In). r3 = L cos 30°j = The VﬁClOKS I'A/G = I'A — l‘G = ll— I'B/G =
r3 7 1‘0 = 7 li + 1.732j (m). The moment about the center of mass is M=l'B/G XPil‘A/G XN. 1 J k 1 J k
M: —l 1732 0 + l —l.732 0
P 0 0 O N 0 : (4.73219 + N)k (Nm). From the equation of angular motion: (1) —l.732 P + N : Ia. From
Newton‘s second law: (2) P = mrrr, (3) N — mg = max. where ax,
a". are the accelerations of the center of mass. From kinematics: ac; : aA + 05 x rG/A — wer/A. The angular veloc
ity is zero since the system was released from rest, 1 j k
ac = rrAi—l— 0 O a = (mi—1.732ai—aj
fl 1.732 0 : (aA — l.732n!)i — Otj (m/s2). from which a, = 70:.
Similarly,
I j k
ao=ag+o£><rG/g.a(;=33+ 0 0 a
l —l.732 0 = (sz + l.732ai + otj. from which at : l.7326¥. Substitute into (1), (2) and (3) to obtain three
equations in three unknowns: 71.732P + N = Id. P = m(l.732)cr,
N — mg 2 —ma. Solve: (a) Ct = 1.84 rad/52, P = 57.3 N, (b) N =
143.47 N 30° mg 4m Problem 18.48 The masses of the bar and disk are
14 kg and 9 kg, respectively. The system is released
from rest with the bar horizontal. Determine the bar’s
angular acceleration at that instant if (a) the bar and disk are welded together at A,
(b) the bar and disk are connected by a smooth pin
at A. Strategy: In part (b), draw individual freebody
diagrams of the bar and disk. Solution: L=l.2m R=O.3m
mg:l4 kg 171;):9kg 0 is a ﬁxed point
For the bar 1 7. l w 7
‘ i 7 7 _ 7 
[0—1 mBL — 12(14)“ ) —l68Nm L 2
[OH = [C +1713 3 103 z 6.72 N—m2
For the disk:
:A : §mDR3 = grows)? = 0.405 N_m2 [00 21,; + mng 213.37 N—m2 The total moment of inertia of the welded disk and bar about
0 is ['1‘ = 103 + 10;) = Nm2 ZR: 0X = 0 = mam 2F}: 0" — mgg — mpg : (1713 + 1711))(16‘. L
2M0: — mgg — meg : [Tu We can solve the last equation for or without ﬁnding the location
and acceleration of the center of mass. G. Solving, at = —9.38 rad/s2 (clockwise) (b) In this case, only the moment of inertia changes. Since the disk
is on a smooth pin, it does not rotate. It acts only as a point mass
at a distance L from point 0.
In this case, [gm 2 mle and I} = 103 + [gm 2 19.68 Nm2
We now have , I" I /
ZMU. — (7)1ngg—meg=frcr Solving oz’ : 79.57 rade2 (clockwise) Problem 18.65 Bars 0Q and PQ each weigh 6 lb.
The weight of the collar P and friction between the
collar and the horizontal bar are negligible. If the system
is released from rest with 6 = 45°, what are the angular
accelerations of the two bars? Solution: Let (mg and OIPQ be the clockwise angular acceleration
of bar 0Q and the counterclockwise angular acceleration of bar PQ.
The acceleration of Q is i k
HQ =ag+a0Q XFQm= 0 0 —0£0Q
2 cos 45° 2 sin 45° 0 : ZUQQ sin 450i — 20t0Q COS 45Dj. The acceleration of P is 2110 = aQ + (3po X l‘p/Q i k
(“xi = 20100 sin45°i * ZOEOQ C05450j + 0 0 org
2 cos 45° 2 sin 45° 0 Equatiﬂg i and j components,
(1]) = 20mg sin 45° — Zapg sin 45° (1) O = —2(xOQ cos 45° + 20!}3Q c0545° (2). The acceleration of the center of mass of bar PQ is ac = 3Q +£2po X l'g/Q = 200g sin 450i i j k
—2achos45°j+ 0 0 OtPQ
cos 45° — sin 45° 0 Hence,
((0‘ : ZOIOQ Sin 45D  DIPQ Sin 450 (3): no). : 72mm) cos 45° + apQ cos 45° (4). From the diagrams: The equation of angular motion of bar 0Q is 2 M0 = 100mg: Q,(2sin 450) — Q_v(2cos45°) + 6cos45° : §(6/32.2)(2)2a09 The equations of motion of bar PQ are
2 Fir : — Qt : (6/32.2)aci (6)
2 F)» : N — Q3» — 6 : (6/32.2)ac_v (7) 1 7
2M = (N + Qy + Q,.r)(cos45°) = ﬁ(6/32.2)(2)'CEPQ (8). Solving Equations (l)—(8), we obtain 0mg : apQ : 6‘83 radJ’s2 (5). 18.124 should be solved
for the information to solve 18.125 Problem 18.124 Each bar is l m in length and has
a mass of 4 kg. The inclined surface is smooth. If the
system is released from rest in the position shown, what
are the angular accelerations of the bars at that instant? 30° Solution: For convenience, denote H = 45°. r: = 303 and L =
1 m. The acceleration of point A is i k
at = don >< Parr) = 0 0 arm Leosr‘) LsinH U at = 0,0,, {—iL sin!) + jL cos in (M53). ThwlrﬂanJlnf'n' . . . e e . .
6 race 6 1m) 0 IS 1 so 3“: b} The equations of motion tor the bars: lor the pin supported lelt hill‘: 3.41 = 31; + urib’ X Page L
(5) AIVL cost! — Axis siné! — mg (;) C056 = {0,1961A.
i j k
8,11: :13 + 0' 0 (My . “IL: 4 a
—LcosH Lsinr) 0 where 10,; = ( 3 )= E kgm‘.
n  . 3
3A = 38 — 10M 8 L “"9 ‘JQ’A‘llL €059 ("VS—l The equations of motion for the right bar: From the constraint on the motion at B‘ Equate the expressions for . . . . — AI — 3 sin = mang ,
the acceleration ol A to obtain the two equations: ‘ ‘6 ' . . (7 —/l‘—HI +Bcos =mu;, ..
(l) — 00,; L 5111“ = in; (305 ff — 0,13]. .N'ln lot ) " g ﬂ ( 1B" I. i'. . I. .
and (’2) arm.me = (if; sinﬁ — «mi. e039. (8) A). (g) cos(J + A; (g) $me + B (;)51n{}CUNﬁ The acceleration of the center of mass of AB is
L .
— B (3039 smﬁ = f(‘.\mx’,13, “(Lt3 =1l.t + «an X roami l l w
i j k where 10,“; = (l—j) oth = (g) kgm:
= 31 + 0 0 “at; _ h
’ L cos o L sin ti . . . . . .
a — A} These eight equations In eight unknowns are solved by iteration: Ar =
' ' —9.27 N, A}. = 1.15 N. mm = 0.425 radisz. (1,”; = —.59 radi’sg,
a = 45.43 N, mm]. = —0.8610 111153, am ,1," = —0.2001 m3
LCM}; _. . €3sz _ . 3 '
30,1, 3 = 31.11 + $111 0 + $035 U" [mils }.
from which
. Loco; . 1 (3)130131 = —o'0,iL smb’ + sInH (mIS'l. [Xle (4) “0,11” : 09,; L cost‘i + cos 6. Problem 18.125 Each bar is l m in length and has
a mass of 4 kg. The inclined surface is smooth. If the
system is released from rest in the position shown, what
is the magnitude of the force exerted on bar 0A by the
support at O at that instant? Solution: The acceleration of the center of mass of the bar 0A is i j k
0 0 0'0“ a;.=ot, I‘;',=a‘. ‘ (01 “1X (’1 ‘+ LeosH Lsth
2 2
L sin 9 _ L cos 6' _ 1
ﬂow = — 7 (Xe11+ a (mu (mist).
The equations 0f mono“: Use the solution to Problem 18.140: 9 = 45°. and = 0.425 radt'sl. A. = —l9.27 N, m = 4 kg, from which [3 = 18.6? N. F) =
Fx + Ax = Mann Fr + A."  me = mam 33.00 N. from which F = VH1? + F3 = 42.00 N ...
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This note was uploaded on 03/29/2009 for the course MECHENG 240 taught by Professor Perkins during the Spring '09 term at University of Michigan.
 Spring '09
 PERKINS

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