# Ch16u7fd2u984cu89e3u7b54 - 163 The disk is originally...

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16–3.SOLUTIONAns.Ans.Ans.an=v2r;(aA)n=(11)2(2)=242 ft>s2at=ra;(aA)t=2(6)=12.0 ft>s2v=rv;vA=2(11)=22 ft>sv=8+6(0.5)=11 rad>sv=v0+actThe disk is originally rotating at If it issubjected to a constant angular acceleration of determine the magnitudes of the velocity and the nand tcomponents of acceleration of point Aat the instantt=0.5 s.a=6 rad>s2,v0=8 rad>s.2 ft1.5 ftBAV08 rad/s
16–10.SOLUTIONAngular Motion:The angular velocity of the blade at can be obtained byapplying Eq. 16–5.Motion of A and B:The magnitude of the velocity of points Aand Bon the bladecan be determined using Eq. 16–8.Ans.Ans.The tangential and normal components of the acceleration of points Aand Bcan bedetermined using Eqs. 16–11 and 16–12 respectively.The magnitude of the acceleration of points Aand BareAns.Ans.(a)B=2(at)2B+(an)2B=25.002+40.02=40.3 ft>s2(a)A=2(at)2A+(an)2A=210.02+80.02=80.6 ft>s2(an)B=v2rB=A2.002B(10)=40.0 ft>s2(at)B=arB=0.5(10)=5.00 ft>s2(an)A=v2rA=A2.002B(20)=80.0 ft>s2(at)A=arA=0.5(20)=10.0 ft>s2vB=vrB=2.00(10)=20.0 ft>svA=vrA=2.00(20)=40.0 ft>sv=v0+act=0+0.5(4)=2.00 rad>st=4 s20 ftBAac0.5 rad/s210 ftThe vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and begin to turn with a constant angular acceleration of ac= 0.5 rad>s2, determine the magnitude of the velocity and acceleration of points A and B on the blade when t=4 s.
16–14.The disk startsfrom rest and isgiven an angular accelerationwhere tisin seconds. Determine theangular velocity of the disk and itsangular displacementwhen t=4 s.a=(2t2) rad>s2,SOLUTIONWhen ,Ans.When ,Ans.u=16(4)4=42.7 radt=4 su=16t4Lu0du=Lt023t3dtv=23(4)3=42.7 rad>st=4 sv=23t3v=23t320tLv0dv=Lt02t2dta=dvdt=2t20.4 mP
16–19.SOLUTIONMotion of the Shaft:The angular velocity of the shaft can be determined fromWhen Motion of the Beater Brush:Since the brush is connected to the shaft by a non-slipbelt, thenAns.vB=¢rsrBvs=a0.251b(625)=156 rad>svBrB=vsrsvs=54=625 rad>st=4 svS=(t+14)t=vS1>4t2t0=vS1>42vs1Lt0dt=Lvs1dvS4vS3>4Ldt=LdvSaSASASThe vacuum cleaner’s armature shaft S rotates with an angular acceleration of a=4v3>4rad>s2, wherevis in rad>s.Determine the brush’s angular velocity when t=4 s, starting from v0=1 rad>s, at u =0.The radii of the shaft and the brush are 0.25 in. and 1 in., respectively. Neglect the thickness of the drive belt. – 1
AB125 mm200 mmvAvB16–33.The drivingbelt istwisted so that pulley Brotatesin theopposite direction to that of drive wheel A. If the angulardisplacement of Aisrad, where tisinseconds,determine the angular velocity and angularacceleration of Bwhen t=3 s.
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