Problem Set 5: Universal Law of Gravitation; Circular Planetary Orbits

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Warren Dennis
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1 Poblem Set 5: Univesal Law of Gavitation; Cicula Planetay Obits Design Engineeing Callenge: Te Big Dig.007 Contest Evaluation of Scoing Concepts: Spinne vs. Plowe PROMBLEM 1: Daw a fee-body-diagam of a spee in a ole, like is used to old te sot-puts and te balls. Figue 1: Diagam of a spee in a ole F N mg Nv Figue : FBD of spee in a ole. Assume no fiction ove line contact.

2 PROMBLEM : Te foce condition fo making a ball o sot-put to just ise up out of te ole. Summing te foces in te -diection fist wit igt as te positive diection: F = + 0 = F => F = N N (1) Summing te foces in te y-diection wit up as te positive diection: + Fy 0 mg N v () = = + => Nv = mg Lastly summing moments aound te contact point: M contact _ point = 0 = F ( ) + mg (3) => F = mg Note fo 0<<, te above moment am fo N v was obtained fom Figue 3, wic esults in equation (4). - Figue 3: Dimensions of spee. + = ( ) => = (4)

3 PROBLEM 3: Te foce condition fo making a ball sot-put to leave te ole and stat olling acoss te platte. Tis poblem is just like poblem ecept tee is an acceleation tem. Summing te foces in te -diection fist wit igt as te positive diection: F = + ma = F N => F = ma + N Summing te foces in te y-diection wit up as te positive diection: (5) + Fy 0 mg N v (6) = = + => Nv = mg Lastly summing moments aound te contact point: M = 0 = F ( ) + mg contact _ point => F = ma + mg wee te moment am fo N v was obtained fom Figue 3. (7) Note te diffeence between Equations (3) and (9) is te latte as an additional tem of ma.

4 PROBLEM 4: Angula velocity of te platte must be acieved in ode to meet te foce conditions in (3)? Fo foce conditions in poblem (3), we need te centifugal foce. v = = = (8) F Fcentifigual maadial m R but te linea velocity is equal to te angula velocity times te distance between te cente of platte to cente of ball wic is denoted by R, efe to Figue 4. Tus equation (8) becomes: ( ωr) Fcentifigual = m R F m R => = ω (9) Substituting fo F found in poblem 3 into equation (9) poduces: a + g = ω R a => ω = + g R R ( ) (10) R Figue 4: Distance between cente of platte to cente of sot-put.

5 PROBLEM 5: How ad would a lasso ave to pull (o a blade to pus) in ode to meet te foce conditions in (3)? Te amount of foce equied would be te same as equation (7): F = ma + mg (11) PROBLEM 6: Wat is a bette concept fo libeating te sot-puts o ockey balls, spinning te platte o pulling o pusing tem off? Compaing equations (3), (7), and (11), te diection acceleation needs to be consideed fo libeating te ball wile pusing o pulling it off wee wen otating, te moment of inetia needs to be known. Tus pusing te ball equies less foce, but as te platte is spun wit te pope angula velocity, moe balls fly off fo tat amount of foce. PROBLEM 7: How do consideations of macine design compleity and feasibility affect te oveall best concept? Spinning te Platte: a. Can te motos of te ca andle te amount of toque necessay to get te angula velocity needed? If sot-put balls can not be spun off, instead could te ockey balls be? Ten pus te sot-put off te platte? b. Wic is quicke wen emoving te balls fom te platte, spin o pus/pull? Pulling/pusing: a. Can te motos andle pusing te sot-put off te platte plus te ockey balls in one effot o ae multiple effots needed?

6 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Pysics Depatment Pysics 8.01T Fall Tem 003 Ball Coming Out of Socket Hee s ow solve poblem of a ball coming out of its socket on a otating table witout using non-intetial coodinate systems o knowing anyting about moments. Te figue below sows a ball just about to come out, in te sense tat it makes contact wit te table at only one point on te cicumfeence of te socket. Te ball as adius a and te socket as adius. a N N tan θ mg Te figue sows a fee body diagam wit all te foces acting on te ball sown in ed. Newton s nd law fo te vetical ais is N sin θ + N tan cos θ = mg wee θ = cos 1 ( a ) Te oizontal foce components can poduce a maimum centipetal foce F c = N cos θ N tan sin θ If R is te adius of te ball s cicula pat, te ball stats to come out wen mrω mrω ma = N cos θ N tan sin θ If tee is no fiction, N tan = 0 and we obtain ω ma = Te ole of fiction is discussed on te net page. g R tan θ J. D. Litste 1 Octobe 1, 003

7 Te ole of fiction can be undestood if we tink in a little moe detail about ow te ball escapes wen tee is no fiction. As soon as ω > ω ma te ball stats to slide up te edge of te socket. Tat inceases θ, and teefoe educes te maimum centipetal foce N cos θ. Tis is an unstable situation so te ball will escape. Because te ball slides up te edge of te socket as it escapes, any fictional foce must act in te opposite diection to N tan as dawn in te figue. Teefoe N (sin θ µ cos θ) = mg Rωma = N (cos θ + µ sin θ) m cos θ + µ sin θ = sin θ µ cos θ g = 1 + µ tan θ tan θ µ g d dµ (Rω ma) = 1 + tan θ (tan θ µ) g So you can see tat adding some fiction makes it ade to get te ball out, in te sense tat ω ma inceases. Tat is consistent wit ou intuitive epectation. Howeve, tee is anote possibility: tat te ball will oll ate tan slide ove te edge of te socket. In tat case, te value fo ω c is te same as fo te fictionless case. Tis means tat only effect of fiction is to make te ball oll out ate tan slide out of te socket. It does not cange te citical otation speed. If you ae not sue you believe tis, I found te easiest way to sow it was to go into te otating coodinate system and calculate te toques of te weigt and of te centifugal foce about te contact point at te edge of te socket. J. D. Litste Octobe 1, 003

Physics 111 Lecture 5 Circular Motion

Physics 111 Lecture 5 Circular Motion Physics 111 Lectue 5 Cicula Motion D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Multiple Objects q A block of mass m1 on a ough, hoizontal suface is connected to a ball of mass m by a lightweight