{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

p09fl8 - Lecture 8 Constraints(21 Sep 09 A Falling body 1...

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

View Full Document Right Arrow Icon
Lecture 8: Constraints (21 Sep 09) A. Falling body 1. Vector directions using spherical polar coordinates and assuming spher- ical earth ˆ r, ˆ ϑ, ˆ ϕ , = ω ˆ z . Cross products: ˆ r × ˆ ϑ = ˆ ϕ ; ˆ ϑ × ˆ ϕ = ˆ r ; ˆ ϕ × ˆ r = ˆ ϑ cyclic 2. For the angular velocity = ω ˆ z , use ˆ z = cos ϑ ˆ r - sin ϑ ˆ ϑ and then ˆ z × ˆ r = sin ϑ ˆ ϕ 3. Hence the centrifugal term is (correct typo) × ( × ˆ r ) = - ω 2 [cos ϑ sin ϑ ˆ ϑ + sin 2 ϑ ˆ r ] and the effective acceleration at earth’s surface is ~g = - g 0 ˆ r - × ( × R e ˆ r ) = - [ g 0 - ω 2 R e sin 2 ϑ r + 1 2 ω 2 R e sin 2 ϑ ˆ ϑ 4. Drop the subscript for body-frame time derivatives, drop the centrifugal term and approximate the gravitational acceleration as ~g = - g ˆ r ; hence m d 2 r dt 2 = m~g - 2 m~ω × d r dt 5. Solve in successive approximations: r ( t ) = r 0 ( t ) + r 1 ( t ) + ... (order in powers of ω ) d 2 r 0 dt 2 = - g ˆ r d 2 r 1 dt 2 = - 2 × d r 0 dt 6. Initial condition is static: r 0 (0) = ( R e + h 0 r and ˙ r 0 (0) = 0: r 0 ( t ) = [ R e + h 0 - 1 2 gt 2 r ; the time of fall is t f = q 2 h 0 /g . 1
Image of page 1

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

View Full Document Right Arrow Icon
7. The equation for the first order term [initial conditions r 1 (0) = 0 and ˙ r 1 (0) = 0] is d 2 r 1 dt 2 = 2 gt~ω × ˆ r = 2 gωt sin ϑ ˆ ϕ r 1 ( t ) = 1 3 gωt 3 sin ϑ ˆ ϕ B. Constraints 1. FW Sec. 13; G Sec. 1.3 2. Start from a description in Cartesians in 3D: vector positions
Image of page 2
Image of page 3
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