kean (msk955) – Homework 07 – yao – (56725) 1This print-out should have 28 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsA small puck moves in a circle on a frictionlessairtable. The circular motion is enforced bystring tied to the puck and going through atiny hole in the middle of the table. Initially,the puck moves in a circle of radius R1atspeed v1. But l ater the string is pulled downthrough the hole forcing the puck to move ina smal ler circle of radius R2=12R1.RvWhat is the new speed of the puck?1. v2=12v12. v2=14v13. v2=1√2v14. v2= 4 v15. v2= v16. v2=√2 v17. v2= 08. v2= 2 v1correct9. v2= 2√2 v110. v2=12√2v1Explanation:Let the hole in the a irtable be the originof our coordinate system. Because the holeis tiny, the stri ng always pulls the puck inthe radial direction. Consequently, the stringtension force~T has zero torque (ab out theorigin). The other two forces on the puck —the weight~W and the normal force~N of thetable — cancel each other and each other’storques. Altogether, we have zero net torque,so the angular momentum of t he puck mustbe conserved:~L =~R × m~v = const.When the puck moves in a circle, the directionof the angular momentum is vertically up, andits magnitude is L = m v R . This is true bothbefore and after the string being pulled down,soL = m v1R1= m v2R2v2=R1R2v1= 2 v1.002 10.0 pointsThe system of point masses shown in t hefigure is rotating at an a ngular speed ofω = 1.87 rev/s. The masses (all equal ) areconnected by light, flexible spokes t hat can belengthened or shortened. At the beginning,each spoke is r = 0.918 m long.Note: An effect similar to that illustratedin this problem occurred in the early stages ofthe formation of our Galaxy. A s the massivecloud of dust and gas that was the source ofthe stars and planets contracted, an initiallysmall rotation increased wit h time.mmmmxyrrωWhat is the new angular speed if the spokesare shortened to 0.45 m?Correct answer: 7.78219 rev/s.Explanation:kean (msk955) – Homework 07 – yao – (56725) 2The initi al moment of inertia of the systemisIi=Xmiri2= 4 M r2i.The moment of inertia of the sy st em after thespokes are shortened isIf=Xmfrf2= 4 M r2f.We conserve angula r momentum asωf=IiIfωi=rirf2(1.87 rev/ s)= 7.78219 r ev/s .003 (part 1 of 2) 10.0 pointsA student sits on a rotating stool holding two5 kg masses. When his arms are extendedhorizontally, the masses are 0.77 m from theaxis of rotation, and he rotates with an angu-lar velocity of 1.2 rad/sec. The student thenpulls the weights horizontally to a shorter dis-tance 0.22 m from the rotation axis and hisangular velocity increases to ω2.ωiωfFor simplicity, assume the student himselfplus the stool he sits on have constant com-bined mo ment of inertia Is= 2.2 kg m2.Find the new angular velocity ω2of thestudent after he has pulled in the weights.Correct answer: 3.63443 rad/s.Explanation:Let : M = 5 kg ,R1= 0.77 m ,ω1= 1.2 rad/sec ,R2= 0.22 m , andω2= 3.63443 rad/sec ,As the student moves his arms, hi s momentof inertia changes fromI1= Is+ 2 m R21= (2.2 kg m2) + 2 (5 k g) (0.77 m)2= 8.129 kg m2toI2= (2.2 kg m2) + 2 (5 k g) (0.22 m)2= 2.684 kg m2,but his angular momentum is conserved,L = I1ω1= I2ω2.Consequently, his angular velocity i ncreasesfromω1= 1.2 rad/sec , toω2=I1I2ω1=(8.129 kg m2)(2.684 kg m2)(1.2 rad/sec )=3.63443 rad/sec .004 (part 2 of 2) 10.0 pointsWhen the st udent pulls the weights in, heperforms mechanical work — which increasesthe kinetic energy of the rotat ing system.Calculate the increase in the kinetic energy.Correct answer: 11.8737 J.Explanation:The rotational ki netic energy is given byK =I ω22=L22 I=L ω2.For the system in question — the studentand the weights — the angular moment umL stays constant while the moment of inertiakean (msk955) – Homework 07 – yao – (56725) 3decreases and the angular velocity increases;hence K increases by∆K =L ω22−L ω12=I1ω12[ω2− ω1]=(8.129 kg m2) (1.2 rad/sec)2× [(3.63443 rad/sec) − (1.2 rad/sec)]=11.8737 J .005 10.0 pointsA child of mass 49.4 kg sits on the edge ofa merry -go-round wi th radius 2.8 m and mo-ment of inertia 2 90.472 kg m2. The merry-go-round rotates with an angular velocity of2 rad/s. The child then walks towards thecenter of the merry-go-round and stops at adistance 1.20 4 m from the center. Now whatis the angular velocity of the merry-go-round?Correct answer: 3.74372 rad/s.Explanation:When the child moves inward, the moment ofinertia of the system iMGR+ichild(the merry-go-round plus the child) changes. Therefore,to conserve angular momentum, the angularvelocity of the system must change. Specifi-cally:Linit= Lfinal(ichild+ iMGR) ω = (ichild,2+ iMGR) ω2The mo ment of inertia o f the child is m r2.Thereforeω2=m r2+ iMGRm r22+ iMGRω .006 10.0 pointsWhen a spinning bike wheel is placed hor-izontally, hung from a pivot at one end, theaxis of rotation of the wheel will swing in ahorizontal circle.~TIn which direction does it turn (as viewedfrom above)?1.horizontally, CCWcorrect2.downward3.upward4.horizontally, CWExplanation:Precession is a direct demonstration ofX~τ =d~Ldt.~L changes in the direction of the torque,and the only torque about the pivot is dueto gr avity on the center of mass of the wheel,so the torque is CCW as viewed from above(along a horizontal plane).007 10.0 pointsA 3 kg rock is suspended by a masslessstring from one end of a 3 m measuring stick.0 1 2 33 kgWhat is the mass of the measuring stick ifit is balanced by a support force at the 1 mmark?Correct answer: 6 kg.Explanation:Let : m1= 3 kg ,x1= 1 m , andkean (msk955) – Homework 07 – yao – (56725) 4ℓ = 3 m .Because the stick is a uniform, symmetricbody, we can consider all of its weight to beconcentrated at t he center o f ma ss, soxCM=ℓ2and x2=ℓ2− 1 .0 1 2 3m1gx1x2m2gApplying torques,m2g x2= m1g x1m2=m1x1x2=(3 kg) ( 1 m)0.5 m=6 kg .008 10.0 pointsA wheel of radius R and negligible mass ismounted on a horizontal frictionless axle sothat the wheel is in a vertical plane. Threesmall objects having masses m, M , and 2 M,respectively, are mounted on the rim of thewheel, as shown.60◦RM2MmIf the system is in static equilibrium, whatis t he value of m in terms of M?1. m =5 M22. m = 2 M3. m =M24. m = M5. m =3 M2correctExplanation:Since the system is in static equilibrium,the net torque is zero:τ = m g R + M g R cos 60◦− (2 M) g R= m g R +12M g R − 2