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# GEN PHY PHY 302K

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This 6 page Class Notes was uploaded by Malvina Orn on Monday September 7, 2015. The Class Notes belongs to PHY 302K at University of Texas at Austin taught by Zhen Yao in Fall. Since its upload, it has received 31 views. For similar materials see /class/181821/phy-302k-university-of-texas-at-austin in Physics 2 at University of Texas at Austin.

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Date Created: 09/07/15

hampton idh225 7 Homework 08 7 yao 7 56775 1 This print out should have 12 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering 001 100 points A small puck moves in a circle on a frictionless airtable The circular motion is enforced by string tied to the puck and going through a tiny hole in the middle of the table Initially the puck moves in a circle of radius R1 at speed m But later the string is pulled down through the hole forcing the puck to move in a smaller circle of radius R2 7 R1 What is the new speed 112 of the puck 1 112 2 vl 2 112 111 3 112 i111 4 112 401 5 112 2111 correct 1 6 7 7 02 43901 7 UQ V501 1 8 7 7 02 201 9112 0 1 10 UQ 701 2M Explanation Basic Principle Angular momentum conservation x mx const Let the hole in the airtable be the origin of our coordinate system Because the hole is tiny the string always pulls the puck in the radial direction Con sequently the string tension force T has zero torque about the origin The other two forces on the ppck 7 the weight W and the normal force N of the table 7 cancel each other and each others torques Altogether we have zero net torque and therefore the angular momentum of the puck must be con served R gtlt mx const When the puck moves in a circle the direction of the angular momentum is vertically up and its magnitude is L my B This is true both before and after the string being pulled down hence L mlel m UQRQ and therefore R1 112 7X01 2111 R2 002 100 points The system of point masses shown in the gure is rotating at an angular speed of w 171 revs The masses all equal are connected by light exible spokes that can be lengthened or shortened At the beginning each spoke is 7 103 m long Note An effect similar to that illustrated in this problem occurred in the early stages of the formation of our Galaxy As the massive cloud of dust and gas that was the source of the stars and planets contracted an initially small rotation increased with time hampton idh225 7 Homework 08 7 yao 7 56775 2 What is the new angular speed if the spokes are shortened to 0664 m Correct answer 411466 revs Explanation The initial moment of inertia of the system is Ii 2mm The moment of inertia of the system after the spokes are shortened is If 27724er 4M7 We conserve angular momentum as 4 i If 22171reVS 75f 411466 revs wf w 003 100 points A student holds two lead weights each ofmass 54 kg When the students7 arms are extended horizontally the lead weights are 086 m from the axis of rotation and the student rotates with an angular speed of 16 radsec The moment of inertia of student plus stool is 44 kg m2 and is assumed to be constant ie the students arms are masslessl Then the student pulls the lead weights horizontally to a radius 015 m from the axis of rotation 1 30 0ng a b Calculate the nal angular speed of the system Correct answer 426885 rads Explanation Basic Concepts 2E 7 o 1 Km 5 1w Solution The initial moment of inertia of the system is I I 2 m R2 44 kg m2 2 54 kg 086 m2 123877 kg mg The nal moment of inertia of the system is If S 2 m r2 44 kg mg 2 54 kg 015 m2 4643 kg m2 From conservation of the angular momentum it follows that Ii w wf Therefore Ii 123877 kng 16 radSec 4643 kg m2 426885 rads 004 100 points A child of mass 496 kg sits on the edge of a merry go round with radius 26 m and mo ment of inertia 264884 kg mg The merry go round rotates with an angular velocity of 15 rads The child then walks towards the center of the merry go round and stops at a distance 0936 m from the center Now what is the angular velocity of the merry go round Correct answer 291975 rads Explanation When the child moves inward the moment of inertia of the system iMGR Hem1d the merry go round plus the child changes Therefore to conserve angular momentum the angular hampton idh225 7 Homework 08 7 yao 7 56775 3 velocity of the system must change Speci cally Limit Lfinal ismId more ichild iMGR W The moment of inertia of the child is mfg Therefore m7quot iMGR d2 2 mm more 005 100 points Three 6 kg masses are located at points in the any plane A I I 54 cm I I quotlgO 47 cm What is the magnitude of the resultant force caused by the other two masses on the mass at the origin The universal graVita tional constant is 66726 gtlt 10 11 N m2kg2 Correct answer 136423 gtlt 10 8 N Explanation Let m 6kg 47cm047m y54cm054m and G 66726 gtlt 10711 Nm2kg2 The force from the mass on the right points in the x direction and has magnitude G 2 F1Gimn m 2 66726 gtlt 10711 N mQkg 6 kg 047 m2 108743 gtlt 10 8 N The other force points in the y direction and has magnitude 66726 gtlt 10711 N m2kg2 6 kg 054 m 823778 gtlt 1079 N F2 O I F 646 The magnitude of the resultant force is F I F12 F22 108743 gtlt 108 N2 12 823778 gtlt 1079 NP 136423 gtlt 10 8 N 006 100 points The planet Krypton has a mass of 68 gtlt 1023 kg and radius of 28 gtlt 106 m What is the acceleration of an object in free fall near the surface of Krypton The graVita tional constant is 66726 gtlt 10 11 N m2kg2 Correct answer 578746 ms2 Explanation Let M 68 gtlt1023 kg R 28 gtlt106 m and G 66726 gtlt 10711 N 1112ng Near the surface of Krypton7 the graVita tion force on an object of mass m is Mm FG R2 hampton idh225 7 Homework 08 7 yao 7 56775 so the acceleration a of a free fall object is F a E gKrypton m M 0 66726 gtlt 10711 N mQkgg X 68 gtlt 1023 kg 28 gtlt 106 m 578746 ms2 007 100 points In another solar system is planet Drii39f7 Which has 5 times the mass of the earth and also 5 times the radius How does the gravitational acceleration on the surface of Driff compare to the gravita tional acceleration on the surface of the earth l 1 Its 7th as much 25 1 2 Its 5th as much correct 3 Its the same7 10 ms2 4 There is no gravity on Driff because 5 times 4000 miles the radius of the earth7 is 20000 miles7 far beyond the pull of gravity 5 Its 25 times as great 6 Rs 5 times as much Explanation Let MD 5Me and RD 5 Re Gravitational force is M F mg G 2m 7 G M M 7 olt 7 so mD 9D 7123 MDTE 5merz fl 9e E me 7123 T me5 re 5 762 l 9DEge 008 100 points Consider a satellite moving near the Earth surface7 the radius of its orbit 7 is approxi mately the radius of the Earth R The period T of the satellite is 9 1T2 7 VP 2 T 5 9 9 3 T 1 I 7 R 4 T 277 5 correct 9 Explanation gw2Rso w1 and 2 Tilso w 009 part 1 0f 2 100 points In order for a satellite to move in a stable circular orbit of radius 6686 km at a constant speed7 its centripetal acceleration must be inversely proportional to the square of the radius 7 of the orbit What is the speed of the satellite The universal gravitational constant is hampton idh225 7 Homework 08 7 yao 7 56775 5 667259 gtlt 10 11 N mQkg2 and the mass if the earth is 598 gtlt 1024 kg Correct answer 772529 ms Explanation Let r 6686 km 6686 gtlt106 m M 598 gtlt 1024 kg and G 667259 gtlt 10 11 N m2kg2 The centripetal force is provided by the universal force from the earth Mm my r2 7 Where M is the mass of the earth Thus GM 7 7 667259 x 1011 N rn2kg2 6686 gtlt 106 m gtlt V598 gtlt1024 kg 772529 ms 010 part 2 0f 2 100 points Find the time required to complete one orbit 1 Correct answer 151053 h Explanation The time to complete one orbit is T27r7 U i 27r6686 gtlt106 m 1h 7 772529 ms 3600 s 011 100 points Two satellites A and B7 Where B has twice the mass of A7 orbit the earth in circular orbits The distance of satellite B from the earth7s center is tWice the distance of satellite A from the earth7s center What is the ratio of the orbital period of satellite B to that of satellite A TB 1 7 2 TA f TB 2 7 1 2 TA TB 3 7 1 4 TA TB 4 7 1 8 TA TB 5 7 8 TA TB 6 7 iis TA 7 T73 8 correct TA TB 8 7 i12 TA 913 1 TA TB 10 7 2 TA Explanation According to Kepler7s third law7 the square of the orbital period is proportional to the cube of the orbital radius Therefore7 T31 TE Bi 7 Bid Therefore7 TB RB 32 32 7 7 2 8 TA RA f 012 100 points The planet Saturn has a mass 952 times that of the earth and a radius 947 times that of the earth Find the escape speed for objects on the surface of Saturn Correct answer 355109 kms Explanation Let MS 952ME and RS 947RE hampton idh225 7 Homework 08 7 yao 7 56775 The escape speed from Saturn is 2GM v53 R35 1 and the escape speed from Earth is 2 G ME UeE RE Dividing7 ves RE MS 1517 RS ME HE Ms UeS 7 6E 3 a S E 1 52 i 1 T 112 kms 355109 kms

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