10/23/2003 PHY Lecture 14R 1


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1 Announcements. Remember  Tuesday, Oct. 8 th, 9:30 AM Second exam (coverng Chapters 94 of HRW) Brng the followng: a) equaton sheet b) Calculator c) Pencl d) Clear head e) Note: If you have kept up wth your HW, you may drop your lowest exam grade. Today Thursday, Oct. 3th, 4 PM Physcs Colloquum by Professor Bernd Schüttler, Dept. of Physcs, U. Ga wll dscuss the analyss of bologcal systems n terms of a physcal and mathematcal model 3. Today s lecture revew Chapters 94, problem solvng technques 0/3/003 PHY 3  Lecture 4R
2 Gravtatonal forces and energy r m Potental energy : U () r GMm r v 0 Energy needed to escape Earth s gravtatonal feld, assumng an ntal velocty v 0 : K mv + U v 0 0 K E f GMm R GM R E + U 0 f 5000 km/h 0/3/003 PHY 3  Lecture 4R
3 Energy needed to go from one stable crcular orbt to another: R R E K + U From Newton's law for crular orbt v m r GMm mv r GMm E r GMm r : E E GMm R + GMm R 0/3/003 PHY 3  Lecture 4R 3
4 Energy needed to go from one stable crcular orbt to another  Example: R R How much energy s needed to take a satellte of mass m00kg from the nternatonal space staton (R R E +390 km) to ts usual orbt (R R E +600 km)? E E GMm R J R 0/3/003 PHY 3  Lecture 4R 4
5 τ r F L r p F gravty Gm m r ˆr Problem solvng sklls Math sklls Equaton Sheet Advce:. Keep basc concepts and equatons at the top of your head.. Practce problem solvng and math sklls 3. Develop an equaton sheet that you can consult. 0/3/003 PHY 3  Lecture 4R 5
6 Problem solvng steps. Vsualze problem labelng varables. Determne whch basc physcal prncple apples 3. Wrte down the approprate equatons usng the varables defned n step. 4. Check whether you have the correct amount of nformaton to solve the problem (same number of knowns and unknowns. 5. Solve the equatons. 6. Check whether your answer makes sense (unts, order of magntude, etc.). 0/3/003 PHY 3  Lecture 4R 6
7 ŷ of mass r r j xˆ r COM m r m 0/3/003 PHY 3  Lecture 4R 7
8 Poston of the center of mass: r com m r m Velocty of the center of mass: v com m v m Acceleraton of the center of mass: a com m a m 0/3/003 PHY 3  Lecture 4R 8
9 Physcs of composte systems: F m a dm v dp Centerofmass velocty: Note that: F F total M v d com v com m m v m M v 0/3/003 PHY 3  Lecture 4R 9
10 A new way to look at Newton s second law: F ma m dv d( mv) dp Defne lnear momentum p mv Consequences:. If F 0 dp 0 p constant. For system of partcles: If F 0/3/003 PHY 3  Lecture 4R 0 dp F 0 0 p dp constant
11 Statement of conservaton of momentum: m v 0 m v m v f f cosθ + m v snθ m v f f snφ cosφ If mechancal (knetc) energy s conserved, then: f m v m v + m v f 0/3/003 PHY 3  Lecture 4R
12 Snapshot of a collson: P Impulse: P f dp F ( t ) dp F( t) t t dp t t F( t) J 0/3/003 PHY 3  Lecture 4R
13 Angular moton s angular dsplacement θ(t) dθ angular velocty ω (t) angular acceleraton dω α (t) natural unt radan Relaton to lnear varables: s θ r (θ f θ ) v θ r ω a θ r α 0/3/003 PHY 3  Lecture 4R 3
14 v r ω r ω r v r ω Specal case of constant angular acceleraton: α α 0 : ω(t) ω + α 0 t θ(t) θ + ω t + ½ α 0 t ( ω(t)) ω + α 0 (θ(t)  θ ) 0/3/003 PHY 3  Lecture 4R 4
15 Newton s second law appled to centerofmass moton dv F m Ftotal M dv CM Newton s second law appled to rotatonal moton τ v dv F m r F r m τ τ r total F m ω r r I d dω ( ω r ) Iα dm 0/3/003 PHY 3  Lecture 4R 5 dv (for rotatng about prncpal axs) r I F m d
16 Object rotatng wth constant angular velocty (α 0) ω R vrω v0 Knetc energy assocated wth rotaton: K where : I m v m r m r moment of nerta 0/3/003 PHY 3  Lecture 4R 6 ω Iω ;
17 Knetc energy assocated wth rollng wthout slppng: K rot Iω I mr Dstance to axs of rotaton Rollng: K K + tot com K rot If there s K tot no slppng : + M I MR v com v com Rω 0/3/003 PHY 3  Lecture 4R 7
18 Torque and angular momentum Defne angular momentum: L r For composte object: L Iω p Newton s law for torque: dω dl τ total I If τ total 0 then L constant In the absence of a net torque on a system, angular momentum s conserved. 0/3/003 PHY 3  Lecture 4R 8
19 Centerofmass r CM m r m Torque on an extended object due to gravty (near surface of the earth) s the same as the torque on a pont mass M located at the center of mass. m τ r m g( j) r r CM { } r { ( j) } Mg CM 0/3/003 PHY 3  Lecture 4R 9
20 Noton of equlbrum: F τ 0 Noton of stablty: 0 θ r T Fma T mg cos θ 0 mg sn θ ma θ τi α r mg sn θ mr α mra θ mg(j) Example of stable equlbrum. 0/3/003 PHY 3  Lecture 4R 0
21 Analyss of stablty: F τ 0 0 0/3/003 PHY 3  Lecture 4R
22 0/3/003 PHY 3  Lecture 4R
23 0/3/003 PHY 3  Lecture 4R 3
24 0/3/003 PHY 3  Lecture 4R 4
25 0/3/003 PHY 3  Lecture 4R 5
26 0/3/003 PHY 3  Lecture 4R 6
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