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Unformatted text preview: For pendulum, can: 3/! in vacuum .. 0.5 Hz =. 11‘ rut/sec 1: $4,: = 9vsl/n'z = 0.9940 m (44 = w“ 1.. T: ;n viscous medium = 0,4,5 Hz = 0.9“. mJ/Sa:
2.. 101 1—130?! “"9
T__°’nzd =77 , >=l.3m
0"
‘5': %o.4$§7
1 u o .
Ezua‘bfan of moéu‘on: m9 9 + Ct 9 + M72 9 0 Cd: = 2(m11) ("In ’
‘ 0.4}f7 2.706 J
S'mce T = f_?___ .3 , Ce W Nm—Aec/ra’ . (a) Viscous d3mping, (b) Coulomb damping. (iii) (a) rd = 0.2 sec, fd = 5 Hz, wd = 31.416 rad/sec.
(b) Tn = 0.2 sec, I‘n = 5 Hz, w, = 31.410 rad/sec. xi (a) (a) = e‘ “n ’*
xiel
xi 2 1r g
1n = In 2 = 0.0931 =
[Km 1 ._ {3
or 39.9590 (1 = 0.4804 or g = 0.1090
Since wd =wn V 1 — ,weﬁnd
«0 31.410
wn = = = 31.0005 rad/sec
1 _ (2 V0.9s798
_ 2 _ _5_9_9_ 2 __ 4
k —m w _ 9 81 (31.0005) —5.0910 (10 )N/m
0
s‘ = — s:
cc 2 m wn Hence 0 = 2 m can 5 = 2 (93:11) (31.0005) (0.1090) = 353.1104 N—s/m (b) From Eq. (2.116):
_5_0(_)_
9.81
Using N = W = 500 N,
I”: 0.002 I: a (0.002) (5.0304 (104))
4 w 4 (500) k = m w: = (31.410)2 = 5.0304 (10‘) N/m = 0.0503 Assume that the bicycle and the be]; {all as a rigid body by 5 cm at point A. Thus
the mass will be subjected to an initial downward displacement of 5 cm (t =
0 assumed at point A): xo=0.05m,5:o =0 a)n == VE= = 24.7614 rad /sec me.l 800 100 4038.5566 »
wd = 0),, V 1 — g” = 24.7014 1 — 0.24702 = 23.9905 rad/sec Response of the system:
x(t) =Xe—W‘t sin (cud t +¢)
2 l
, 2
xo + 3’ OJ. xo
“:1 where X = x3 + l 2 '2'
= (0.05), + 0.2476) (24.7614) 0.05)] =0.051607m 23.9905
‘ . x0 we! 0.05 (23.9905)
d = ‘1 —— = t —1 —_—————— = . o
g 3” 4’ tan *0 + g wn x0 . an [0.2470 (24.7614) (0.05) 75 6645
359 _' _ Thus the displacement of the boy (positive downward) in vertical direction is given by
x(t) = 0.051007 e" “3" t sin (23.0905 t + 75.0045°) m Consider a small angular displacement of the bar 0 about its static equilibri,
position. Newton’s second law gives: 3  3! 3! 6’ 6’ t’ t’
$M=J°9=‘k[”T TJ‘°[9:HI]‘“‘ 9:][7] I
. ~ cl 3  I 1
1e, JoQ+—ls 0+:kt’20 0 ‘ V. I 104 7
where Jo = 2; m (’2. The undamped natural frequency of torsional vibration is given by: ___ 3U” = 36k
w“ 410 V 7111 Let x = displacement of mass and P = tension in rope on the left of mass.
Equations of motion: 2F=mi€=~k1§.P or P=.mSE—kx (1)
2M=J09=Pr2—c(9r1) (2)
Using Eq. (1) in (2), we obtain
(mii+kx)r2cl9r1=Jo.0' (3)
With x = 9 r2, Eq. (3) can be written as: _
(Jo+mr;)'é+crfb+k:;9=o (4) For given data, Eq. (4) becomes
.. 2 .
[5 + 10 (0.25)’) 9 + c (0.1) 9 + k (0.25)2 9 = 0 or 5.025 2} + 0.12c ('9 + 0.0625 1: 9 = 0 (5)
Since amplitude is reduced by 80% in 10 cycl, X! 1.0 cm c w. r.l —=—=5= 1“ 0.2 1‘1
lnT=ln5=L6094=10§wan (0)
11 Since the natural frequency (assumed to be undamped torsional vibration frequency) is 5 Hz, wn = 2 7r (5) = 31.416 rad/sec. Also
1 2 1r 2 1r 0.2
 (7) “‘4 aTaanl—e VIe Eq. (6) gives 10‘ 0.2 32.832 g‘ 1.6094 = 10 g (31.410) = _
i.e., 'V 1 — g2 = 62‘s” g = 39.0400 g 1.6094 ' Le. 0.02561
c ( r' 9) Thus we obtain: ’ g C
0‘ I 0.0625 C c c g=0.02561 = —— =___=_____
cc 2 0),I 2 (5.625) (31.410) Friction force = p N: 0.2 (5) = 1 N. k = 62756 = 250 N/m. Reduction in amplitude in each cycle = 4 u N = gill = 0.016 m. Number of halfcycles k 250
executed before the motion ceases (r):
_. _”‘ N
r) X" k = 0.1—0.004 a 12
— 2 u N 0.008
k Thus after 6 cycles, the mass stops at a distance of 0.1  6 (0.016) = 0.004 m from
the unstressed position of the spring. can =’\/_—l£.= \ l = 22.1472 rad/sec
In T W
“ wn = 2— = 0.2837 sec Thus total time of vibration = 8 1', : 1.7022 sec. ...
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This note was uploaded on 09/08/2010 for the course MAE 315 taught by Professor Wu during the Summer '08 term at N.C. State.
 Summer '08
 WU

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