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Unformatted text preview: PROBLEM 1T.4E A uniform 16811: cube is attached to a uniform lSﬁib circular shaft as
shown and a couple M of constant magnitude 13!] 1121'111. is applied to
the shaft. Knowing that r = 4 in. and L = [2 in... determine the time
required for the angular velocity.r of the systern to increase from moo rpm
to 21101} rpm. SOLUTION 2 2
Moments of inertia. Cube: imp} + L2) = LEHE] + [E] ] = [1.86951 llznsE ﬁ 12 12 32.2 12 12 1 1155 4 L. l:
‘3' 12 2
J = {1.26915 lhszft Total: F 1.13s? [basin Angular speeds. all = Itlﬂt'} rpm = 113;}?! rada's (5:12 2 EDGE! rpm = 29:3 rada's Applied couple. M = 181} loin. = 15.0 113111 Syst Momental + ﬁrst Ext lmpHg = ﬁrst Moments:3 Moments about cylinder axis: fail + M! = in); — M {LIBSTHEUU * rung
_ M _ 15.13 I PROBLEM 17.53
Each of the gears A and B weighs LS lb and has a radius of gyration of i=3; I ._H [.6 in.. while gear C 1weighs 8 lb and has a radius of gyration of 4 in.
_ .. 5. ﬂ” .: Assume that kinetic friction in the bearings of gears A, B and C." produces
"v ﬂl‘r'?f_______.. ;" couples of constant magnitude 0.1 lhft, 0.] lbﬁ. and 0.2 lbﬂ,
.~ respectively. Knowing that the initial angular velocity of gear C is
I: xfﬁlu ' 2000 rpm, determine the time required for the system to come to rest.
ﬁin ._F
‘32. $2.. ”'13 ASLIi“.
'.« .,*_!,""\": .i' .. ;: aoLUTIoN
_ _ 3 1.5 4 3
Moments ofinertia. I2 = In. = ”210' = {—M = 322.15 x104" lbszft
g 22.2
_ 22.22. {a} a}: . .
if = C '1 =—'* =21605x10" lbsﬂ
g 32.2
Kinematics. {art )1 = 2000 rpm = 209.44 radfs {4.224 = £34223 = .F'Lﬂ‘rt [rsAL = {2223}. = ${mcjl = $909.44} = 523.60 Iadx's ’.i Georgi:
I202
‘i' = @ O
52.. a
536st "i"IIi~mei:Ita1 + Syst Ext ImpH 1 = Syst Momentaz
1'tjrnolnents about A: L (“J4L + if4 ff'gro'r — (MrL! = 0
_: ,, 2.4 [322.15 2.10 ~}[52s.ea} + E" [Emctr — an
or 0.1.t — 0.2ij4ﬁdt = 0.43362 {1}
Gear 3: By a similar analysis 0.1: — 0.2 [FMcf! = 0.43362 [2} Let x = java: + [are and add Equations [1} and {2}. a2: — 0.2.: = [1.36224 {3} Syst Mnmentu. Syst Ext Imp: ﬂ = ﬁrst P'l'IIlnwmmI Jmoments about C'.’ IE [arch — [MLLi — r5 [Fxfcdr + r(.ij~d.' = I]
(21505x 103)[209.44}— 0.2: — ﬁﬂacm + [Eggs]
[LEI + 415:: = 5.7'316 Solving Equations. {3] and {4} simultaneously, r = 11.368, 1: = 1:12 [bag PROBLEM 1 15? Show that, when a rigid slah rotates about a ﬁxed axis through 0 IV.
perpendicular to the slab, the system of the momenta of its particles is
equivalent to a single vector of magnitude mFaJ, perpendicular to the line 0:}, and applied to a poi_nt P on this line, called the center of
percussion, at a distance GP = .t 2 I? from the mass center oi" the slab. SOL UTIDN Kinematics. Point 0 is ﬁxed. System momenta. Components parallel to Hi: Moments about G: ...
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 Summer '06
 

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