hw8 - al-ameri(aoa434 – Hwk08 – Stokes –(19102)1This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: al-ameri (aoa434) – Hwk08 – Stokes – (19102)1This print-out should have 18 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is Online Homework No. 8 (OH-08).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 thex-coordinate 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 they-coordinate 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 inter-ested only in the vertical coordinateZcmofthe center of mass, thusZcmsculpture=McubeMcube+MballZcmcube+MballMcube+MballZcmball.(2)al-ameri (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 accord-ing 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....
View Full Document

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.

Page1 / 6

hw8 - al-ameri(aoa434 – Hwk08 – Stokes –(19102)1This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online