Lab 9 Circular Motion Part 2

Lab 9 Circular - Lab 9 Circular Motion Part 2 Lab Report(100 pts A t PHYS 1114 201 Day and Time Station Check Part I Table 1 Position and Maximum

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Unformatted text preview: Lab 9: Circular Motion Part 2 Lab Report (100 pts) A, t PHYS 1114 201 Day and Time: Station Check: Part I: Table 1. Position and Maximum An Exercises and Questions for Part I 1. Use the position and maximum angular speed data recorded in Table l to compute the quantities listed in Table 2 and record the results in Table 2 below. Show work for the 50 gram mass only. The moment of inertia of the main platter is Ip = 7.5 x 10'3 kg-m2 and the moment of inertia of the bar is 1., = 2.98 x 10‘3 kg-mz. If the thread does not stretch significantly as the mass descends, then vmax = 1' com“, where r = 0.015 m is the radius of the smallest step pulley. Table 2. Energy Calculations (30 pts) 93 Lab 9 Circular Motion Part 2 ,./! GPEi calculation KEf calculation I v 1 “:4”: K5: = W93 ‘ (<- : ('0"’*5“i:31°‘~Wfiixééd : '/:z («was #1934 6573‘“??? ~ . A Z . .. v' . r x \ "wt , " Q i Vii/2 if; u t {is} v’ ivK GPEf calculation RKEf calculation GPEg= Wain {fiat giggly“): = (. anaemia/Woe Kflmml = 7.2 (7.5x :09 kgm‘)(b.5‘“:§ *’ : .Hoj‘ ‘ My vmax calculation ,, Emecgfjsalculation ,, me = W" = .mem ( M‘MG) 2 tools "3’s 5mm = mam thaws? Mamas: Mum Maommtmsz 2. Check how well the conservation principle of mechanlcal energy holds for the _L M __ platter/hanging mass system by calculating the percent difference A% between the initial "3 '5” ‘ My total mechanical energy of the main platter and hanging mass I Emech,i = m g and the total mechanical energy Emhyf=gmv§m +y21w§m +mghf at the instant the mass reaches its lowest position. Record the percent difference in Table 2. Show work for the 50 gram mass below. (2 pts) ix315m2€cgl 86+.15a ’2... 3. How well is the total mechanical energy of the main platter and hanging mass conserved as the hanging mass falls from rest to its lowest position? (A percent difference of 10% or less may be considered good agreement.) (2 pts) ’v' i n, MOO -‘ lab/l .— i l v: 2» ~ >.' Tr" Jr‘f *5 We”? Viki (swseiwfwi w “MM 4., What specific factors might explain why the total mechanical energy of the main platter and hanging mass was not conserved? (4 pts) .% , Ni“) “RNA C’rob ' / r - ' '2' ' .t 2-‘1,\%‘~’ , 5‘ l *c VK ‘lv‘ ‘ (Oi/“(MW sword ~ mu M La“ 4‘ i w 5 Li, vb“ I: J ;,.»ev;\; i. X t!» v ("A 1‘ {Gila-V‘s fwgufi u» h\}\ 94 Lab 9: Circular Motion Part 2 PART II: Ltot,i I—Itot,f (kg-mZ/s) (kg‘mz/s) Main platter and Aux11ia - latter Exercises and Questions for Part II 1. Calculate the total angular momentum Ltom of the two platters before the collision and the total angular momentum Lot; of the two platters after the collision. Likewise, calculate the total angular momentum me of the platter/bar system before the collision and the total angular momentum Ltot; after the collision. Record these values in Table 3. Calculate the angular momentum (equations, work and units) and record in Table 3 above. The moment of inertia for the main platter is Imp = 7.5x 10‘3 kg°m2, for the rectangular steel bar, It,ar = 2.98x 10'3 kg-mz, and for the auxiliary platter, Iap = 7.22x 10'3 kg°m2. You must show work for the main platter /auxiliary platter below. (6 pts) Main platter/auxiliary platter system Ltot,i (ant-‘3; 7» rm? CA).- = (,oms wgmfitaaafwf‘gi : '2‘ _ : “714‘ Kgfys Check how well the conservation principle of angular momentum holds by calculating the percent difference 13% between the total angular momentum Lmi before the collision and the total angular momentum Ltot; after the collision in each case. Record the percent differences in Table 3.Show work for one below. (2 pts) ‘ - g K V my a??? '2 How well is the total angular momentum of the main platter and auxiliary platter conserved during the collision between the two? The total angular momentum of the main platter and the rectangular bar? (A percent difference of 10% or less may be considered good agreement.) (2 pts) “5,; ‘ r -. 15,; W: ‘x , .E,, _4 i" x r i , 5 a 1+ :3 was, avaseiW-y is i t, m /; ci-tfifawc e 95 Lab 9 Circular Motion Part 2 4. What specific factors might explain why the total angular momentum of the two systems was not conserved during the collisions? (4 pts) ( ‘l‘l luck i/l UWM QIJQF 5. Calculate the initial and final rotational kinetic energies for each system. Show all work, including equations and units. (12 pts) Main platter/auxiliary platter system RKE,‘ RKEf WE; =- sze: no? Rm 6 Z A}: j , : V1 (007553173: +- tamz kw“) “9 {5% W: = ‘7’; (cc/75 42%;} fieaWVs = oz? 3‘ a 082’? Main platter/rectangular bar system RKEi RKEf QKE; : as)?” KWEG "1 We : ‘25200'2’?’ if??? ’42 “if = Veliwwb‘fiwz‘i Images”; glgm’?‘ : Ill/H T t: 6. Was the Rotational Kinetic Energy of the system conserved? (2 pts) W5 Qt €734; qw‘iifily Maia! 3’50 {926' ‘43» W o If not account for the decrease in RKE. (3 pts) ‘ I : .wl' ‘ ,= ; :4 “t c yea l W“ 1' 1' human e CoWAvaiw t“ f 7. What kind of collision takes place between the main platter and the auxiliary platter and the main platter and the bar? Explain. (4 pts) 9" ,_ , i $.79 F ,i; i- r :l- "'3‘ r: it”? 33 (jam? 313 l~l( (GM/“U m'gkgq 5"} “féls’i’iilfi'i zm’m‘. Va “CM!” W’QI‘SY * lb? law {'3 5m 5 i 6 ss 04? V0 larva“ «4i luv» at“ K- 9' :v/ 96 ...
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This note was uploaded on 01/10/2012 for the course PHYS 2014 taught by Professor Nandi during the Spring '08 term at Oklahoma State.

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Lab 9 Circular - Lab 9 Circular Motion Part 2 Lab Report(100 pts A t PHYS 1114 201 Day and Time Station Check Part I Table 1 Position and Maximum

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