Hw11 - al-ameri(aoa434 – Hwk10 – Stokes –(19102)1This print-out should have 16 questions.Multiple-choice questions may continue onthe next

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Unformatted text preview: al-ameri (aoa434) – Hwk10 – Stokes – (19102)1This print-out should have 16 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is Online Homework No 10. It is duebefore 6.00 am on Tuesday 21 April, AbuDhabi Time. (Monday 9.00 pm in Texas)00110.0 pointsA homogeneous cylinder of radius 20 cm andmass 50 kg is rolling without slipping along ahorizontal floor at 2 m/s.How much work was required to give it thismotion?Correct answer: 0.15 kJ.Explanation:Let :r= 20 cm,m= 50 kg,andv= 2 m/s.Because the cylinder rolls without slipping,the translational speed isv=r ωand the total kinetic energy of the cylinder isKtotal=12mv2+12I ω2=12mv2+12parenleftbigg12mr2parenrightbiggparenleftBigvrparenrightBig2=34mv2=34(50 kg) (2 m/s)2·1 kJ1000 J=.15 kJ,which is also the work required to give thecylinder this motion.keywords:002(part 1 of 2) 10.0 pointsA skater spins with an angular speed of13.8 rad/s with his arms outstretched. Helowers his arms, decreasing his moment of in-ertia from 47 kg·m2to 43 kg·m2.a) Calculate his initial rotational kineticenergy.Correct answer: 4475.34 J.Explanation:Basic Concepts:L=IωKE=12Iω2Given:Ii= 47 kg·m2If= 43 kg·m2ωi= 13.8 rad/sSolution:KEi=12Iiω2i=12(47 kg·m2)(13.8 rad/s)2= 4475.34 J003(part 2 of 2) 10.0 pointsb) Calculate his final rotational kinetic en-ergy.Correct answer: 4891.65 J.Explanation:Solution:Angular momentum of the system is con-served, soIfωf=Iiωiωf=IiωiIfKEf=12IfparenleftbiggIiωiIfparenrightbigg2=12I2iω2iIf=12(47 kg·m2)2(13.8 rad/s)243 kg·m2= 4891.65 J004(part 1 of 2) 10.0 pointsA cylinder with moment of inertia 26.3 kg m2rotates with angular velocity 5.93 rad/s on aal-ameri (aoa434) – Hwk10 – Stokes – (19102)2frictionless vertical axle. A second cylinder,with moment of inertia 25.2 kg m2, initiallynot rotating, drops onto the first cylinder andremains in contact.Since the surfaces arerough, the two eventually reach the same an-gular velocity.I2I1ωBeforeωAfterCalculate the final angular velocity.Correct answer: 3.02833 rad/s.Explanation:From conservation of angular momentum(I1+I2)ω=I1ω,orω=I1I1+I2ω=(26.3 kg m2)(26.3 kg m2) + (25.2 kg m2)(5.93 rad/s)=3.02833 rad/s.005(part 2 of 2) 10.0 pointsShow that energy is lost in this situation bycalculating the ratio of the final to the initialkinetic energy.Correct answer: 0.51068.Explanation:Ki=12I1ω2andKf=12(I1+I2)ω2,ThereforeKf=12(I1+I2)ω2=12(I1+I2)parenleftbiggI1I1+I2parenrightbigg2ω2=parenleftbiggI1I1+I2parenrightbigg12I1ω2=parenleftbiggI1I1+I2parenrightbiggKi.SoKfKi=I1I1+I2=(26.3 kg m2)(26.3 kg m2) + (25.2 kg m2)=.51068<1.00610.0 pointsIn a circus performance, a large 4.2 kg hoopof radius 3.3 m rolls without slipping....
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This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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Hw11 - al-ameri(aoa434 – Hwk10 – Stokes –(19102)1This print-out should have 16 questions.Multiple-choice questions may continue onthe next

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