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Unformatted text preview: Problem 20.13 Solve Problem 20.12, assuming that
the secondary reference frame is ﬁxed with respect to
the vertical shaft. Solution: wha.‘ = 2j + 4k(rad/s), so
(a) The angular velocity of the disk relative to the secondary refer— i J k
ence frame is VC 2 v0 + what x ram 2 0 + 0 2 4
0.3 O 0
wrcl = wdi + wk 2 6i + 4k (rad/s).
= 1.2' — . .
(b) The angular velocity of the reference frame is J 0 6k (m/s)
The velocity of P is
$2 = woj = 2j (rad/s),
thd'k’ l l“t' i‘ik
so e IS sanguarveoclyls vpva+wer/C : l.2j—0.6k+ 6 2 4
0 0.2 0 a) = $2 + a... = 6i + 2j + 4k (rad/s). Let 0 be the origin and C the center of the disk. The angular velocity = ‘08] + ['2'] + 0'6k (m/s). of the horizontal bar is Problem 20.27 The airplane’s inertia matrix in terms
of the bodyﬁxed coordinate system is shown. At
the present instant, the airplane’s angular velocity
and angular acceleration are (0 = 0.13i + 0.24j — ‘
0.08k (rad/s) and a = 0.139i + 0.108j (rad/s2). What is
the total moment about the center of mass due to the
forces and couples acting on the airplane? Solution:
Mx 46.000 0 0 0139
My = 0 113,000 0 0.108 kgmZ/sz
Mz 0 0 134,000 0
0 0.08 0.24 46,000 0 0
+ —0.08 0 —0.13 0 113,000 0
—0.24 0.13 0 0 0 134,000
0.13
x 0.24 kgmZ/s2
—o.03 Solving we ﬁnd M = (M11 + Myj + Mzk) = (5.99i + l3.lj + 2.0911) kNm Problem 20.33 In terms of the coordinate system y
shown, the inertia matrix of the 6kg slender bar is Ixx _Ixy _Ixz
_I)’x IY)’ —I)’Z "‘sz Izy Izz 1
0.500 0.667 0
= [0.667 2.667 0 ]kg—m2.
0 0 3.167 The bar is stationary relative to an inertial reference
frame when the force F 2 12k (N) is applied at the right
end of the bar. No other forces or couples act on the bar.
Determine (a) the bar’s angular acceleration relative to the inertial
reference frame and (b) the acceleration of the right end of the bar relative
to the inertial reference frame at the instant the
force is applied. Solution: (a) In terms of the primed reference frame shown, the coordinates of
the center of mass are ‘ /. ’ 1m
_ 1 2
x, _ x/lm, +x’2m2 _ (0)§<6)+(1)§(6)
m1+m2 1 2
_. 6 ._
3( )+ 3(6)
=0.667 m, l 2
yl‘ml +y/2m2 (0.5)§(6)+ (0)5(6) m1+m2 I 1 2
3(6) + 3(6) 2 0.167 m. The moment of F about the center of mass is
M = (1.333i— 0.167j) x 12k = —2i — l6j (Nm).
From Eq. (20.19) with a) = 52 = 0,
—2 0.5 0.667 0 dwx/dt
16 = 0.667 2.667 0 dwy/dt .
0 0 0 3.167 dwz/dt
Solving, we obtain a = 6.0” —— 7.50j (rad/$2).
(b) From Newton’s second law, 2F = 12k = (6)ao, the accelera tion of the center of mass is a0 = 2k (m/sz). The acceleration of
pt A is aA=ao+axrA/o+wX(le‘A/o)
i j k 6.01 —7.50 0
1.333 —0.167 0 =2k+ +0 =11.0k(m/s2). Problem 20.40 The 5—kg triangular plate is connected
to a balland—socket support at 0. If the plate is released
from rest in the horizontal position, what are the com—
ponents of its angular acceleration at that instant? ‘3
Solution: From the appendix to Chapter 20 on moments of inertia,
the inertia matrix in terms of the reference frame shown is m l m l
— ~hh3 ~——  22 0
A(12 ) A<8bh
m l m l
= —— 4;th — — 3 0
“1 A(s ) A(4hb m l 3 l3
0 0 A(12bh+ hb) 4
' 0.300 —0.675 0
= —0.675 2.025 0 kg—mz. 0 ' 0 2.325 From Eq. (20.13) with w = ﬂ = 0,
The moment exerted by the weight about the ﬁxed pt. 0 is —9.8] 0.300 —0.675 0 dwx/dt
2_ I . .. 29.43 = —0,675 2.025 0 day/21; .
Zmo = (3171+ 5/1.!) X (—mgk) 0 0 0 2.325 zlwz/dt = —9.8i + 29.43j (NV'Im). Solving, we obtain at = 14.5 j (rad/s2). ...
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 Spring '08
 N/A
 Dynamics

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