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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 20—1* The radial—arm saw that has an operating speed of 1500 rev/min. The blade and motor armature have a combined weight of 3 lb and a centroidal radius of gyration of k6 = 1 in. When the saw is turned off, bearing friction and a magnetic brake
exert a constant braking torque T on the blade and motor. Determine the length of time that the blade rotates before coming to
rest if T = 0.015 lb°in. (bearing friction only). Determine the torque T necessary
to stop the blade in just 0.25 s. Solution The axle of the saw is a fixed axis
of rotation. The moment of inertia of
the blade and armature relative to this axis is I; [—32%] [—lz—JZ 6 2
646.998(10 ) slugft a. If the only torque about the axis of rotation
is bearing friction, then the angular momentum
principle (Eq. 20—9) gives (1500 rev/min = SON rad/s) —6 .
646.998(10 )(sou) — (3—2:; t H
O t = 81.3 s . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . .. Ans. b. If the blade stops rotating in just 0.25 s, then the angular momentum principle gives
6
646.998(10 )(SOﬂ)  T(0.25) = 0 T = 0.40652 lb‘ft = 4.88 lb'in. ... . . . . . . ........ . . . . . . .. Ans. ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 2018 The uniform slender rod AB (m = 3 kg, 3 = 800 mm) is resting on a frictionless horizontal surface when it is struck with an impulse P At = 20 N's as shown. If b = 650 mm, determine a. The angular velocity w of the rod
immediately after the impact. b. The velocity VA of end A immediately after the impact. Solution
The moment of inertia of the
rod relative to an axis through (a 1its mass center is 1 2
IG— 12 (3)(0.8>
uS
2 —.
= 0.16000 kg'm The principles of linear and angular momentum give
'4") + =
0 0 3VGX
+T + =
0 20 3VGY
C+ o + 0.25(20) = 0.160000
VGx = 0 m/s
= T
VGy 20/3 m/s
w = 31.3 rad/s C . . . . . . . . . . . . . . . . . . . . . . . . . ............. Ans.
h. Then, the relative velocity equation gives
_ = _ + 
vA VG vA/G
= mm 3 + (0.4)<31.2s><—3)
= —5.83 3 m/s = 5.83 m/s ‘L . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 2024* A 3kg uniform slender rod AB 800mm long hangs in a vertical
plane from a frictionless pivot when a 0.03—kg bullet strikes the rod
and becomes embedded in it. If the initial velocity of the bullet is v0 = 350 m/s, determine a. The rotation rate of the rod and bullet
immediately after the impact. b. The average force of contact between the
rod and the bullet for an impact duration
of At = 0.001 s. c. The average magnitude of the force exerted
on the rod by the frictionless pin at A for
an impact duration of At = 0.001 s. d. The total system energy lost in the impact.
e. The maximum angle through which the rod will
swing after the impact.
Solution
a. When the bullet and rod are
considered together, there is no angular impulse on the system about an axis through the fixed point A.
Therefore, the final angular momentum
about A is equal to the initial angular momentum about A C+ HA1. 0 + 0.500(o.o3)(350) = 5.2500 nms C+ HAf [—:'(3)(0.800)2+(0.03)(0.500)2]w f
wf = 8.11 rad/s C . . . . . . . . . . . . . . . .... b. For the bullet alone, the linear F:
momentum equation gives «—
+> 0.03(350) — F(0.001) = 0.03(0.500)(8.108) 0,03% F = 10,380 N = 10.38 kN .......... . . . . . . ....... . . . . . . .. Ans. (Problem 2024 continues ...) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 20—24 — cont.) c. For the rod alone the linear momentum equations
+9 0 + (Ax + F)(0.00l) 3(O.400)(8.108) O+A—W0.00l o
( y )( ) 648 N = 648 N 6— w= 3(9.8l) 29.43 N T V 6482 + 29.432 = 649 N initial and final energy of the system are 2
%(o.03)(350) = 1837.5 J 2 2 2
T i[—l—(3)(o.800) + 0.03(o.500) ](8.108) = 21.283 J f 2 3 and the percent loss of energy is Ti  Tf
is loss ——T(100)
i 1837.5 — 21.283
1837.5 e. Finally, use workenergy to determine the angle of swing of the rod.
21.283 + 3(9.8l)(—0.400) + 0.03(9.81)(O.500) + O
= O + 3(9.81)(—0.4OO cos 9) + 0.03(9.81)(0.500 cos 6) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 2029* The uniform rod falls horizontally and strikes the rigid corner with its right end. If the velocity of the
rod just before the impact is v0 and the
impact is perfectly elastic (e = 1), determine the angular velocity u and the velocity of the mass center VG immediately after the impact. Solution Just prior to the impact, the velocity
of the right end of the rod is the same
as that of the mass center, v0 downward.
The line of impact is the ydirection and the definition of the coefficient of restitution gives Also, the linear and angular momentum equations across the impact give +¢ . _ A = . V
mvo B t vaf (b)
I; 1 2
+ ——— A = ————
C+ C [2 (B t) 12 m6 (of (c)
Finally, using the relative velocity equation
a =   w ‘
VB VG + VB/G to relate va, va, and f gives
_ _ i w d
VBf ‘ vs: 2 f ( ) Combining Eqs. a through d gives mv — [imc‘Jw “(Lu VO] 0 6 f 2 f
(o = I, . . . . . . .
f 3vo/ C Ans
= ~L . . . . . .
VGf Vo/2 Ans ENGINEERING MECHANICS — Dynamics W.F. Ri16y & L.D. Sturges 2037 A slender uniform rod (C = 30 in., WAR = 4 lb) is released 0
from rest at an angle of 6 = 70 to the horizontal and strikes a hard horizontal surface as shown. If the initial height of the rod is h = 60 in. and the coefficient of restitution is e = 0.7, determine a. The angular velocity of the rod
immediately after the impact. b. The velocity of the mass center of
the rod immediately after the impact. If end B of the rod will strike the
surface as the rod rotates immediately
after the impact. Solution 3. Use work—energy to determine the speed of the bar just before it strikes the horizontal surface 60 1 4 2
0+4[12]+°‘ 2[32.2]VGi+0 = , i
vGi 17 94436 ft/s The line of impact is the yhdirection.
Then the definition of coefficient of restitution gives (uAf)  0
e = 0.7 =  ——Y—————
(17.94436) — o A G
(vAf)y = 12.56105 ft/s T “3A6
= (VGf)y — (15/12)(.‘.)ABf cos 70 A \709
= (va)y — 0.42753wABf and the linear and angular momentum equations
across the impact give 4
T: [W](—17.94436) + AM __;i__
32.2 (va)y 2
15 o 1 4 30
Ca' 0 [12 C°S 70] Am: 12 [32.2 12] wABf (Problem 2037 continues ..J EMHNEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 20—37  cont.) Rewriting these last three equations (ng)y  0.427536)“f = 12.56105
0.776f0wABf + 5.13030 AAt = 0
(va)y — 8.050 AAt = —17.94436
and solving gives
(vgf)y = 4.64 ft/s T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
wABf = 18.54 rad/s = 18.54 rad/s D . . . . . . . . . . . . . . . . . .. Ans. c. After the impact, the rod rotates at a
constant rate of 18.535 rad/s. Therefore, the rod will be vertical with B on the bottom A when
18.535t = 2.79253 rad (= 90° + 70°)
t = 0.15066 5 Also, after the impact the motion of the mass center is given by  A 2
a = 32.2 J ft/s
G
VG = (4.637 — 32.2t) j ft/s
—° 2 A
r6 = (1.25 sin 700 + 4.637t  16.1t ) J ft When t = 0.15066 5,
yG = 1.50773 ft = 18.09 in.
Therefore, end B will be 3.09 in. above the surface and it Will not hit the surface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 20—39* Bar AB of Fig. a is attached to a frictionless pin at A; bar CD is attached to a frictionless pin at E B and rests on a frictionless support at
F. Both AB and CD are uniform slender bars 36 in. long and weighing 5 lb. . 12 mar—12 ﬁnal
E F Both bars are initially at rest when
a slight disturbance causes bar AB to
fall to the right and strike bar CD as
shown in Fig. b. If the coefficient of restitution is e = 0.6, determine The angular velocities of both
bars immediately after the impact. The maximum angle of rebound of
bar AB after the impact. The average magnitude of the
support reaction at E for an
impact duration of 0.005 5. Solution Both bars move in fixed axis rotation. The moments
of inertia of the bars about these fixed axes are 2
1 5 36 2
IA 3 [ 32.2 ][ l2 ] — 0.46584 slug ft 2 2
1 5 36 5 6 2
IE 12 [ 32.2 ][ 12 ] + [ 32.2 ][ 12 J ’ 0‘15528 Slug'ft a. Use work—energy to determine the angular
velocity of bar AB before the impact 1 2 o
+ + = "‘—' . (O + '
0 5(18/12) 0 2 (0 46584) ABi 5(18/12) Sin 30 (.0 = .
A81 4 01248 rad/s 3 Bar AB rotates about a fixed axis through A = (18/12)(4.01248) = 6.01872 “ GABi en
CABi = (27/12)(4.01248) en = 9.02808
: (,0  . (1.)
(18/12) ABf( ) 1 5000 AB ) ( GABf <1 <1 <1 <1 —§
—§ 11
= — . to
(27/12) H 2 2500 A (.0
CABf ABf Bar CD rotates about a fixed axis through E w = 0
CD1 (Problem 20—39 continues ...) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges (Problem 2039 — cont.) VGCDJ'. = Vccm' = o
 = w A
vGCDf (6/12) CDf J ft/s
I = w _A
VCCDf (12/12) CDf( J) 00 A o A
= w — w '
CDf cos 3 en CDf 511'! 30 et ft/s The line of impact is the ndirection.
Then the definition of coefficient of
restitution gives  . w — w
( 2 2500 ABf) ( CDf 9.02808 — 0
while the angular momentum equations give CA: 0.46584(4.01248) + (27/12)Fnt = 0.465840 ABf
O
: + A = . to
CE 0 (12/12)F t cos 3o 0 15528 CDf
Solving these three equations gives wABf = 0.432 rad/s c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
w = . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . .. .
CDf 5 13 rad/s C Ans FAt = 0.92021 lb°s
b. Using work—energy again on bar AB after the impact gives
2
—:~(0.46584)(O.43211) + 5(18/12) sin 300 = 0 + 5(18/12) sin 6 9 = 30.380 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. c. Finally, the linear momentum equations for bar CD give —+: o + (F sin 30° + EX)At = o
o 5 6
= — A=——————.
T o + (Ey F cos 3o 1 t [ 32.2 ][ 12 ](5 13219)
EX = —92.02 lb = 92.02 lb e—
E = 239.1 lb T
y D
E = 256 lb 5 68.95 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ...
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