kean (msk955) – Homework 06 – yao – (56725) 1This print-out should have 21 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsThe speed of a moving bullet can be deter-mined by allowing the bullet to pass throughtwo rotating paper disks mounted a distance95 cm apart on the same axle. From theangular displacement 23.3◦of the two bul-let holes in the disks and the rotational speed368 rev/min of the disks, we can determinethe speed of the bullet.23.3◦v368 rev/min95 cmWhat is the speed of t he bullet?Correct answer: 90.0258 m/s.Explanation:Let : ω = 368 rev/mi n ,d = 95 cm , andθ = 23.3◦.θ = ω tt =θω,so the speed of the bullet i sv =dt=d ωθ=(95 cm) (368 rev/min)23.3◦×360◦1 rev1 m100 cm1 min60 s= 90.0258 m/s .keywords:002 (part 1 of 2) 10.0 pointsA dentist’s drill starts from rest. After 7.28 sof constant angular acceleration it turns at arate o f 10740 rev/min.Find the drill’s angula r accelerati on.Correct answer: 154.491 rad/s2.Explanation:Let : t = 7.28 s andωf= 10740 rev/mi n .Since ω0= 0,α =ωf− ω0t=ωft=10740 rev/min7.28 s·2 πrev·1 min60 s=154.491 rad/s2.003 (part 2 of 2) 10.0 pointsThroughout what angle does the drill rotateduring this period?Correct answer: 4093.87 rad.Explanation:θ = ω0t +12α t2= 0 +12(154.491 rad/s2) (7.28 s)2=4093.87 rad .004 10.0 pointsAn electric motor rotating a workshop grind-ing wheel at a rate of 21 0 rev/min is switchedoff. A ssume constant angular deceleration ofmagnitude 2.88 rad/s2.Through how many revolutions does thewheel turn before it finally comes to rest?kean (msk955) – Homework 06 – yao – (56725) 2Correct answer: 13.3627 rev.Explanation:Let : ω0= 210 rev/min ,α = −2.88 rad/s2, andωf= 0 .From kinematics,ω2f= ω20+ 2 α θθ =ω2f− ω202 α=−ω202 α=−(210 rev/min)22 (−2.88 rad/s2)·1 min60 s2×2 π rad1 rev=13.3627 rev .005 10.0 pointsIn the spin cycl e of a washing ma chine, thetub o f radius 0.474 m develops a speed of603 rpm.What is the max imum linear speed withwhich water leaves the machine?Correct answer: 29.9313 m/s.Explanation:Let : r = 0.474 m andω = 603 rpm .Angular and linear velocity are related byv = r ω= (0.474 m)603revmin2 πrev1 min60 s=29.9313 m/s .keywords:006 10.0 pointsA small wheel o f r adius 1.8 cm drives a largewheel of radius 1 4.3 cm by having their cir-cumferences pressed together.If the small wheel turns at 441 rad/s, howfast does the larger one turn?Correct answer: 55.5105 rad/s.Explanation:Let : r1= 1.8 cm ,r2= 14.3 cm , andω1= 441 rad/s .The rims of the wheels are forced to travelat the same linear speed, sov = r1ω1= r2ω2,ω2=r1ω1r2=(1.8 cm) (441 rad/s)14.3 cm=55.5105 rad/s .007 10.0 pointsThe dri ver of a car traveling at 40.3 m/s ap-plies the brakes and undergoes a constantdeceleration of 1.54 m/s2.How many revolutions does each tire makebefore the car comes to a stop, assuming thatthe car does not skid and that the tires haveradii of 0.33 m?Correct answer: 254.311 rev.Explanation:Let : v0= 40.3 m/s ,a = −1.54 m/s2,vf= 0 , andr = 0.33 m .a = α r and v0= r ω0, soω2= ω20+ 2 α θ = 0θ =−ω202 α=v0r22ar=−v202 a rkean (msk955) – Homework 06 – yao – (56725) 3=−(40.3 m/s)22 (−1.54 m/s2) (0.33 m)·1 rev2 π rad=254.311 rev .008 10.0 pointsThe angular vel ocity in rad/s of the minutehand o f a watch i s:1. π/602. π/1800 correct3. 1800/π4. 60/π5. πExplanation:The minute hand of a watch complet es onerevolution in one hour.So, we haveω =1 revolution1 hr2πradiansrevolution1 hr3600 s=π1800rads009 10.0 pointsConsider t hree objects of equal masses butdifferent shapes: a solid disk, a thin ring, anda thin hollow square. The ring and the squareare hollow and their perimeters carry all themass, but the disk i s solid and has uniformmass density over its whole area.diskRringRsquare2 RCompare the t hree objects’ moments of in-ertia w hen rotated around their respectivecenters of mass.1. Icmring> Icmsquare> Icmdisk2. Icmsquare> Icmring> Icmdiskcorrect3. Icmdisk> Icmring> Icmsquare4. Icmdisk> Icmsquare> Icmring5. Icmsquare> Icmdisk> Icmring6. Icmring> Icmdisk> IcmsquareExplanation:The mom ent of inertia I =Zr2dm is I =M ×r2, where M is the object’s net mass andr is the average distance of massive points inthe object from the rotational axis. For ahoop o r a thin ring, all massive points are atthe same distance R from the axis, sor = RFor the square the distance from the centerranges from R to R√2 , sor > RandIcmdisk< M R2Icmring= M R2Icmsquare> M R2.ThusIcmsquare> Icmring> Icmdisk010 10.0 pointsCalculate the moment of inertia for the sys-tem. The rods of length L are massless.bbbbbAxis4m 3mm2mLL LLL1. 32 m L2correct2. 30 m L2kean (msk955) – Homework 06 – yao – (56725) 43. 16 m L24. 52 m L25. 41 m L2Explanation:The moment of inertia i sI = m L2+ 2 m (2 L)2+ 3 m(2 L )2+ L2 +4 m (L2+ L2)= m L2+ 8 m L2+ 15 m L2+ 8 m L2= (1 + 8 + 15 + 8) m L2= 32 m L2.011 10.0 pointsA 0.25 m power sawblade has a mass of 0.5 kgdistributed uniformly as in a disc.What is the rotational kinetic energy at3600 rpm (revolutions per minute)?1. 9.1 × 102J2. 2.6 × 103J3. 8.9 × 101J4. 3.6 × 101J5. 1.1 × 103J correctExplanation:Let : r = 0.25 m ,m = 0.5 kg , andω = 3600 rpm .The moment of inertia of a disk of uniformdensity isI =12m r2.The rotational kinetic energy of the saw-blade is thenKrot=12I ω2=1212m r2ω2=14(0.5 kg )(0.25 m)2(3600 rpm)2×2πrev21 min60 s2=1110.33 J .012 (part 1 of 2) 10.0 pointsA uniform solid disk and a uniform ring areplace side by side at the top of a rough inclineof height h.If they are released from rest and roll with-out slipping, determine the velocity vringofthe ring when it reaches the bottom.1. vring=pg h correct2. vring= sin θpg h3. vring=r4 g h34. vring=p2 g h5. vring= sin θrg h3Explanation:Total kinetic energy consists of transla-tional and rotational kinetic energies. Noslipping means v = r ω and for the ringI = M R2.The ring has kinetic energy due to transla-tional and rotational velocity. If we denotethe mass of the ring with m and total kineticenergy with K, we haveK =12m v2+12Iω2=12m v2+12IvR2=12m +IR2v2.By conserva tion of energyKf+ Uf= Ki+ Ui12m +IR2v2+ 0 = 0 + m g h .kean (msk955) – Homework 06 – yao – (56725) 5v2=2 g h1 +Im R2= g h .(1)vring=pg h .013 (part 2 of 2) 10.0 pointsWhich object reaches the bottom first?1.