Fowles08 - 1 CHAPTER 8 MECHANICS OF RIGID BODIES PLANAR...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
1 CHAPTER 8 MECHANICS OF RIGID BODIES: PLANAR MOTION 8.1 (a) For each portion of the wire having a mass 3 m and centered at , 2 2 b b , ( ) 0,0 , and , 2 2 b b y x b b 1 0 0 2 3 2 3 cm b m b m x m   = + +     = 1 0 2 3 2 3 cm b m b m b y m   3 = + + =     (b) ( ) 1 2 2 2 ds xdy b y dy = = ( ) 1 2 2 2 0 1 b cm y y b y m ρ = dy ( ) ( ) 1 2 2 2 2 2 0 2 2 1 4 y b y cm b y d b y y b ρ π ρ = = = 4 y 3 cm b π = From symmetry, 4 x 3 cm b π = (c) The center of mass is on the y-axis. ( ) 1 2 2 2 ds xdy by dy = = ( ) ( ) 3 1 2 2 0 0 1 1 2 2 0 0 2 2 b b cm b b y by dy y dy y by dy y dy ρ ρ = = 3 y = 5 cm b (d) The center of mass is on the z-axis. ( ) 2 2 2 x y dz bzdz π + = dv r dz π π = = 1
Image of page 1

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

View Full Document Right Arrow Icon
2 2 0 0 0 0 b b cm b b z bzdz z dz z bzdz zdz ρ π ρπ = = 2 3 cm z b = (e) The center of mass is on the z-axis. α is the half-angle of the apex of the cone. is the radius of the base at z = 0 and is radius of a circle at some arbitrary z in a plane parallel to the base. r D r tan r r b b z α = = D , a constant ( ) 2 2 2 tan dv r dz b z dz π π α = = 2 3 1 1 tan 3 3 b b 2 m r ρ π πρ = = D α ( ) ( ) 2 2 2 2 3 0 3 0 3 2 tan 3 2 1 tan 3 b b cm z b z dz z b b b ρ π α πρ α = = z bz z dz + 4 cm b z = 8.2 2 0 0 b cm b cx dx xdx x dx cxdx ρ ρ = = 2 3 cm b x = 8.3 The center of mass is on the z-axis. Consider the sphere with the cavity to be made of a (i) solid sphere of radius and mass a s M , with its center of mass at , and (ii) a solid sphere the size of the cavity, with mass 0 z = c M and center of mass at 2 a z = − . The actual sphere with the cavity has a mass s c M m M = and center of mass . cm z 1 0 s 2 c cm a M m z = + M 3 4 3 s M a π ρ = , 3 4 3 2 c a M π ρ = 3 3 3 3 1 0 2 2 2 cm a a a a z a   = +     2
Image of page 2
3 14 cm a z = 8.4 (a) 2 2 2 0 3 2 2 z i i i m b b R = = + + I m 2 6 z mb I = (b) ds rd dr θ = , sin R r θ = 2 z I R ds ρ = 2 2 4 0 4 sin r b z r I r rdr d π θ π θ ρ θ θ = = = =− = 4 2 4 4 sin 4 z b I d π π ρ θ θ = 2 sin 2 sin 2 4 d θ θ θ θ = 4 1 4 4 2 z b I ρ π = 2 1 4 m b ρπ = ( ) 2 2 4 z mb I π π = (c) 2 x ds hdx b dx b = = Where the parabola intersects the line , y b = ( ) 1 2 x by b = = ± 2 4 2 2 b b y b b x x I x b dx bx dx b b ρ ρ = = 4 4 15 y I b ρ = 2 2 4 3 b b x m b dx b b ρ ρ = = 2 1 5 y I mb = (d) dv 2 RhdR π = h b z = 3
Image of page 3

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

View Full Document Right Arrow Icon
4 ( ) ( ) 1 1 2 2 2 2 R x y bz = + = 1 2 1 2 b dR dz z = ( ) ( ) 1 1 2 2 2 0 1 2 2 b z b I R dv bz bz
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern