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Unformatted text preview: alameri (aoa434) – Hwk08 – Stokes – (19102)1This printout should have 18 questions.Multiplechoice questions may continue onthe next column or page – find all choicesbefore answering.This is Online Homework No. 8 (OH08).All answers should be submitted online before0600 on Tuesday 31 Mar (UAE time).001(part 1 of 2) 10.0 pointsThree balls with masses of 3 g, 2 g, and 4 g,respectively, are connected by massless rods.The balls are located (in meter intervals) asin the figure.3 g2 g4 g0 1 2 3 4 5 6 7 8 9123456789x[m]y[m]What is thexcoordinate of the center ofmass?Correct answer: 3.88889 m.Explanation:Let :(x1, y1) = (1 m,8 m),(x2, y2) = (8 m,5 m),(x3, c3) = (4 m,2 m),m1= 3 g,m2= 2 g,andm3= 4 g.The center of mass of a 3–body system islocated atvectorRcm=m1vectorr1+m2vectorr2+m3vectorr3m1+m2+m3.Thusxcm=m1x1+m2x2+m3x3m1+m2+m3=(3 g)(1 m) + (2 g)(8 m) + (4 g)(4 m)(3 g) + (2 g) + (4 g)=3.88889 m.002(part 2 of 2) 10.0 pointsWhat is theycoordinate of the center ofmass?Correct answer: 4.66667 m.Explanation:ycm=m1y1+m2y2+m3y3m1+m2+m3=(3 g)(8 m) + (2 g)(5 m) + (4 g)(2 m)(3 g) + (2 g) + (4 g)=4.66667 m.00310.0 pointsAn abstract sculpture consists of a ball (radiusR= 51 cm) resting on top of a cube (each sideL= 140 cm long). The ball and the cube aremade of the same material of uniform density;there are no hollow spaces inside them. Thebottom face of the cube rests on a horizontalfloor.How high is the sculpture’s center of massabove the floor?Correct answer: 90.3759 cm.Explanation:Basic Principle:If a body comprisesseveral parts and we know the location of eachpart’s center of mass, the the whole body’scenter of mass is located atvectorRcmwhole=summationdisplaypartsMpartMwholevectorRcmpart.(1)For the problem at hand, we have two parts— the cube and the ball — and we are interested only in the vertical coordinateZcmofthe center of mass, thusZcmsculpture=McubeMcube+MballZcmcube+MballMcube+MballZcmball.(2)alameri (aoa434) – Hwk08 – Stokes – (19102)2Each part here has its center of mass in itsgeometric center, hence for the cube of sizeL×L×Lresting on the floorZcmcube=L2(3)while for the ball of radiusRresting on top ofthe cubeZcmball=L+R.(4)Next, the masses follow from volumes according toMcube=Vcubeρ=L3ρ,(5)Mball=Vballρ=4π3R3ρ,(6)henceMcubeMcube+Mball=3L33L3+ 4πR3,(7)MballMcube+Mball=4πR33L3+ 4πR3,(8)and the densityρof the scuplture’s materialcancels out of these formulae.Finally, we substitute eqs.(3–4) and (7–8)into eq. (2) and obtainZcmsculpture=3L33L3+ 4πR3×L2+4πR33L3+ 4πR3×(L+R)= 90.3759 cm.(9)00410.0 pointsA 6 kg steel ball strikes a wall with a speedof 10.4 m/s at an angle of 52.1◦with thenormal to the wall. It bounces off with thesame speed and angle, as shown in the figure....
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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.
 Spring '09
 NAJAFZADEH
 mechanics, Work

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