Fall 2010 - Speliotopoulos - Final (solution)

Fall 2010 - Speliotopoulos - Final (solution) - AtVo =AsV-s...

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Unformatted text preview: AtVo =AsV-s --f l, v, tr) J .t h. *: vJ + jA! tP*,.* ) ) t-fv-'+ poo, e) d t At Pro"n ( 1) frorv'L tz) Se Vs= Z:Vc A. * -rvr' +J Ay = z8y A:I Ve.- +i *u-' lat/n?. , Tt'"<i lr 1s t"=ru V.t rr'. e- it *n-[ "s T?-l $r +L wa*er tc f*tt _, So )%1o S'- +l". ,{t.t*tt "e- r,! X= vo t:F) --- l 'tA' /- I V A'/A. ( r- nt/nl) w 2.JJ n7a3 -\ Physics 7A, UC Berkeley, Fall 2010, Prof. A. Speliotopoulos Grader: Aaron Alpert Force Considerations. The normal force is N = mg , and therefore, the maximum frictional force is fmax = µs N = µs mg . The rolling without slipping condition1 is v = rω , which implies a = rα. (Simply take the time deriva¯ ¯ tive to accomplish this.) That implies 1￿ r￿ 1￿ r￿ 5￿ F= τ→ ¯ F=2 2 τ= ¯ τ ¯ m I m 2mr 5 mr (1) The two relevant forces are friction and the spring force. When the friction is maximized, so is the spring force. At the maximal condition, fmax − Fs,max 5 = (−rfmax ) m 2mr → 7 Fs,max = fmax 2 (2) Final Exam Solution: Problem 4 As previously stated, the maximum frictional force is µs mg , which leads to Fs,max = − 7 µs mg . 2 Energy Considerations. Using Hooke’s Law, we can find the maximum displacement (denoted as δ ) which corresponds to the maximum force. Fs = −kx → 7 Fs,max = − µs mg = −k δ 2 → δ= 7µs mg 2k (3) The total energy stored in a spring stretched δ is ￿ ￿ 12 1 7µs mg 2 49µ2 m2 g 2 49(.8)2 (2)2 (9.8)2 Es = k δ = k = = = 15.06 J 2 2 2k 8k 8(100) (4) At the ball passes the equilibrium point, it no longer will have any potential energy, just kinetic energy from translation and rotation. 1 1 1 1 2 2 ￿ vmax ￿2 ¯ 7 Eeq = mvmax + I ω 2 = mvmax + ¯2 ¯2 mr = mvmax ¯2 (5) 2 2 2 25 r 10 Using the conservation of energy, Eeq = Es → 7 (2)¯max = 15.06 J v2 10 → v = 3.28 ¯ m s (6) Grading Rubric. 8 points for the force considerations (broken down into +2 for the rolling without slipping condition; +2 for F = ma; +2 for τ = I α; +2 for finding the max spring force). 12 points for the energy considerations (broken down into +2 for the maxium displacement; +4 for the energy in the spring; +4 for the kinetic and rotational energy; +2 for energy conservation). I am using the overbar, as in v , to indicate that the quantity refers to the center of mass. Also, I had to carefully ¯ choose that clockwise rotation was positive, or the condition would be v = −rω . ¯ 1 Problem 6 December 17, 2010 Nhoop = mω 2 R Ntable = mg F rhoop = µmω 2 R F rtable = µmg mr2 α = −µm(R2 ω 2 + gR) g α = −µ(ω 2 + ) R dω g = −µ(ω 2 + ) dt R dω = −µ g ω2 + R ￿ ω(t) ￿t dω = −µdt g ω2 + R ω0 0 ￿ ￿ ￿ R R R (tan−1 (ω ) − tan−1 (ω0 ) = −µt g g g ￿ ￿ R R π (tan−1 (ω ) − ) = −µt g g 4 ￿ ￿ R g π −1 tan (ω ) = −µt + g R 4 ￿ ￿ R π g ω = tan( − µt ) g 4 R ￿ ￿ g π g ω= tan( − µt ) R 4 R 1 mr2 α = τnet = −RF rhoop − RF rtable ω (T ) = 0 ￿ π g = µT 4 R ￿ π R T= 4µ g 2 1W onb vrry tvt '+e p--+> 7A" \/ 6isst-Lvn4 Lf v, mv $"'t Gr c*) ,.non& "* tfis ian t s t. (l) ynarnant"Ltv\: m G) ,nut) , {nvf = b\r1,)L+ tAnU'+ + Gt',*)"a' ( =g!L I #/flLt) za l,rJ ( r) v -+ N\IJ (3) a*g,rtnr yvr0ole'i'tuffi' l,rLVoL f (+) rurn (*) ^*J- , hr e' $" i' I -i='- \,.' LJ vnV" rnrff\ Frovn (3) ^^A tU) ,NtJinc'trs) r,o:dr)-,rnnL - arhi"l', si'^Plifie-s ao vn3 { t _-1$-(*)V" r;, [v,-v):tFL.) ?(; t'rrt* [ z) t w( h* \ r arnL rL\z T) l€ s,^bstit*le- (+) o-^tt[ L's ) '2 ;- fiVo + 4 t) si,,.3 \1 t :[m+M)= nT M'rru + /Vl '^=+ 3rn'M vvl -trL ! ..1 :ZmtM > M= Zvry of +hts uo.Lue-- ={v. , d= ho"Ye- "f mlrrl l'n t +) c !# , a^ J- (s ) , N& Jl* aL The- wtd-ss wilL fto"elbA art r" ta*lonct'L YYlotrola ut spe-e-oL V h"tf a p*.,.J "f -#*-t"' -{{'* clilst"noe- .t t so (ffi,)u. = zvo f S A = v,(*)',L r3T \ r -- *rcL t) ...
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This note was uploaded on 05/04/2011 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.

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