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Unformatted text preview: Physics 2101, Exam #3, Fall 2009  October 27 , 2009 Name: SQUID 001 t S ID#: Section: [Circle one]
1 {Chastaim MWF 8:40 AM) 4 [Plummen TTh 9:10]
2 {Chastaim MWF 10:40 AM) 5 (Adams, TTh 12:10] 3 (Rupnik, MWF 12:40 PM} ' Please be sure to write [print] your name and circle your section above. ° Please turn OFF your cell phone and MP3 player!  Feel free to detach, use, and keep the formuia sheet. No other reference material is ;
allowed during the exam. ' You may use either a scientiﬁc or a graphing calculator... ' GOOD LUCK! l‘) l. (10 pts) A cart with mass 0.34 kg moving on a frictionless air track at an initial speed
of 1.2 m/s undergoes an elastic collision with a stationary cart on unknown mass. After
the collision the first cart continues moving in the original direction with a speed of 0.66
m/s. What is the rnass of the second cart? t 9 _ W7 _ A
(a) 1.34 kg [3% KS; 22> l :2. H
, mamma 2Q U éQIé m
WWW VA ,4 Q
2 \Cﬁgwipﬁa ‘gLi— M? \liii
ml magnum. “ _. ##4##“ . . WW_ Warm; ,ge em? ‘3 Qii 31:} ﬁgléé '5 (m u
in: 2. (10 pts) Show your work on this problem. The solid disk shown in the ﬁgure has a radius R and a mass M. One end of the block of
mass m is connected to a spring of force constant k, and the other end is fastened to a
chord wrapped around the disk. The disk is initially wound counterclockwise so that the
spring stretches a distance d from its unstretched position. The disk is then released.
Calculate the acceleration of the block immediately after the disk is released assuming
that the table surface is frictionless. Your answer should be in terms of M m, k, and d. d165,. E k
ratej : Qt; _ disk 1‘ wt,» 25: if M at n m T lllclil’J/lll’ 3. (Spts) Just before a jet shuts down its engines the rotor of one of the engines has an
initial angular velocity of 2000 rad/s. The rotor’s rotation slows at a rate of 80 rad/$2. How long does it take for the rotor to ceme to rest? (a) 2000 s. 3 L0 w 6;;
(b) 1000 5. Ni: {’3 7t
(0)200 s. to «to q, 91.53% *9 .6 i(e)25:) “$5: 4. The uniform rod in the ﬁgure below is released from the horizontal position and then
rotates in the plane of the ﬁgure about an axis through one end. it has a rotational inertia
I. As the rod swings through its lowest position, it collides with a putty wad of mass m
that sticks to the end of the rod. What is the correct answer to the following two questions? (neglect friction) A. (5 pts) From the instant the rod is released until the instant immediately before the rod
hits the putty, which of the following quantities are conserved? ' (a) Mechanical“ Energy mmms{g “mum ‘ i (b) Kinetic energy “W Rnlzniuu axis Rod (0) Angular Momentum
(d) Both (a) and (c) z (e) (a), (b), and (c) (1) None of the above are conserved B. (5 pts) From the instant immediately before the rod hits the putty to the instant just
after the putty and rod have stuck together, which of the following quantities are
conserved for the system? (Assume the coliision occurs over a very short time.) (a) Kinetic Energy L
(b) Torque ' EMT T T (I? 60 W 3er M (d) Both (b) and (c) (it
Thigh) if”; Q  in
(e) (a), (b), and (c) 50 m \MC 6 Moswen ‘ (f) None of the above are conserved 5. Show your work. In the ﬁgure below, a solid cylinder of radius R and mass M starts
from rest and rolls without slioping a distance L along the incline. A) (Spts) What force(s) acting on the cylinder result(s) in a torque about its axis of
rotation? (21) Fn b F and mg u 11 of them (e) none of them B) (15pts) Find translational speed of the cylinder when is reaches the bottorn of the
incline. Your answer should be in terms of g, L, and 6. (g = acceleration of gravity) , .7:
 5%,;
E’WGN x5 mosses): god 6. Show your work. A solid pkg is only supported by a centrally placed pivot as shown
below. The plank has a length of l m and a mass of 5 kg and the pivot is located 0.5 m
from the end. A 10 N force is suddenly applied to one end as shown below. A) (5 pts) What is the net torque on the plank about the pivot point? if g. m; we, e669
«£st93 JO cos of 2:. 1.5 NW} B) (5 pts) What is the angular acceleration of the plank?
.m s “Em
VT *2: he? : (gmt, cc
QQ e: 2.15”; _. éVad 3%: (LCNK wing: 7. (5 pts) In the ﬁgure below 1111 I 4 kg, m; = 1 kg, and m; = 2 kg. The Xpositions are x
= 0 (nil), X = 1.0 m (m3), and X = 1.5 m (1112). Find the magnitude of the net force on 1113
due to the gravitation attraction of the other two masses.  $5236 a
. 10‘10 N (0)106 x 10'10N i (d) 6.67 x 10'11 N (e) 2.5 x 10'9 N m1 m3 m2 7‘ wastq ._.:a go 8. (Spts) Which of the following statements is necessarily true for an object in static
equilibrium? l““‘3§3’“"*m .3met await/W Wit a; was (a) The net torque on the object is zero. 5
(b) The net force on the object is zero. (d) The frictional forces cancel each other out. (e) The object is moving in a straight line. 9. A rod of length 2m and mass 5 kg is supported at one end by a hinge and the other end
by a cable as shown below. Fx and Fy represent the horizontal and vertical components of the force of the hinge on the rod. A) (5 pts) Which of the foilowing forces produce a torque about the hinge? 3) mg
b) T B (lOpts) What is the magnitude of the F" tension in the cable? a) 24.5 N 6) none of the above {:33 Aggy a m
T = 1 C es Lies to. (10 pts) A thin solid rod of length 0.4 In and mass 6 kg is initially at rest, but is free to
rotate in a vertical plane without friction about a horizontal axis through its center. A 2 kg
wad of wet putty drops onto one englgf the rod and sticks to it. The system then begins to rotate with an angular velocity 0 5 Iradi‘sﬁyvhat was the speed of the putty just before it
hit the TOd? Gama; {at} {5 TwJ‘WEW (um; Wm me an}: in a)4m/s “in: mm”: QExZEixVG
(b) 2 S ‘ «1 pivot C S “ﬂiﬁw ' {fl ' g V\ (d) zero r26, (31% {a if, AC2 J i Q rim”. “Name! my, 7 I. (e) none of the above “Egg; "3:51:a y era :2? a» r; Formula Sheet for LSU Physics 2101 Exams, Fall ’09 Units:
1m : 39.4in : 3.28ft lmi = 5280ft 1min : 60$, 1day = 24h 1rev : 360° = 27. rad 0]? latm = 1.013X105Pa lcal = 4.187J T = TO + 273.15K TF = (:00) TC + 32°F
Constants: g : gem/S2 REM. = 6.37 x 106m M Emh : 5.98x1024kg G = 6.67x10’11 ms/(kgs2) RMM = 1.74 x 106m MMW = 7.36x1022kg Earth—Sun distance : 1.50x1011 m MSW I 1.99x1030kg Earth—Moon distance : 3.82x108m k = 1.38 X 10"23 J/K R = 8.31 J/(mOlK) Avogadro’s 7% = 6.02 X1023 particles/mol
Properties of H20: Density: ,Owater : 1000 kg/mS Speciﬁc heat: cwater : 4187 J/(kg K) cice I 2220 J / (kg K) Heats of transformation: Lvapoﬁzanion : 2.256 x 106 J/kg qusion = 3.33 X 105 J/kg —b :l: Vb2 — 4ac Quadratic formula: for 0.262 + bx + c = 0, $13 = 2
a Magnitude of a vector: [5 = mg + a; + 0: Bot Product: El?  3 = ambﬂ3 + ayby + azbz 2 {6 cos(q5) (gt is smaller angle between EL’ and Cross Product: a x 5’ = (aybz — azbyﬁ + (azbm — awbz)? + (Gaby m aybmy‘c, [a x 8 = gay [5'] sin(c'>) Equations of Constant Acceleration: linear equation along X missing missing rotational equation
vwsz+amt 33—30 6—60 w=wo+amt
a: — 320 = vowt + 5am? 11m to 9 — 90 = wot + East",
of} m vim + 2aw(:1: — $0) t t (J2 = w?) + 2o:(6 — 60)
a: — are = —:—('uam + ow)t am a: 6 +190 = $660 + w)t
a: — 3:0 2 vmt — gawt2 vow too 6 — 90 = wt — gatZ Vector Equations of Motion for Constant Acceleration: '1" = 7"}, + fiat —§— 561:2, 13" = 179 + it Projectile Motion: (with + direction pointing up from Earth) m — mo 2 (v0 cos 60)t y — yo : (ii0 sin 00)t — Egg
11m = no cos 6,, '0,9 = (no sin 60) — gt
2 2 o
gm 1) 3111(290)
2  2 o
vzvsmﬂ —2 — =tan6 m——— 2m
y ( o o) 90; yo) y ( a) 2% ms 90? g
Newton’s Second Law: 2 E = 1115
_ . _ inn2 2 11' 1"
Uniform Circular motlon: F.: = m mac T =
r 1; Force of Friction: Static: f5 5 fs’mm = ,LLSFN, Kinetic: fk = mgFN Elastic (Spring) Force: Hooke’s Law F = —k:1: (k :_ spring (force) constant)
1
Kinetic Energy (nonrelativistic): Translational K = Emu2 Work:
_. _. wr 7‘: ..
W = F  d (constant force), W = f F($) dw (variable 1—D force), W = f F07)  d“? (variable 3D force)
rs, mi Work  Kinetic Energy Theorem: W : AK = Kf — Ki Where W is the network Work done by weight (gravity close to the Earth surface): W = m g?  d 2 2
. “if 31)..» 1;.
Work done by a spring force (F = —k:;c): W m —k 3 da: 2 wk 3"— — 3‘
$22
Power:
W a q dW a ﬂ )
Average: Pm,g = Kt, P 2 F  navy (const. force) Instantaneous: P = W, P = F  1; (const. force)
dU
Potential En. Change: AU 2 —W (only conservative force) PotentialForce Relation: = — den)
:1: Gravitational (near Earth) Potential Energy: U(y) = mgy (at the height y) 1
Elastic (Spring) Potential Energy: U = Ekm2 (relative to the relaxed spring) Mechanical Energy: Emec = K + U
Change of Mech. Energy due to nonconservative forces: WM 2 (K f + Uf) — + Ui) Conservation of energy: Whetaext = AK+AU+AEth+AEint, where Whenext is the net, ertemal work
done on the system, and AEth = —Wfk = (ﬁred for constant friction), or Wnet,ext+Ki+Ué—AEth = K f+Uf N 1 N 1 N 1 N
Center of mass: M = Z mi, mcom 2 — Z “mtsci, ycom = — Z mzyé, zcom : — Z mizz
._ M ._ M ._ M .g
2—1 zwl 2—1 1—1
1 N 1 N 1 N 1 N a
Fcom = ‘ﬁ ﬁcom = M miﬁi acorn : M mtg?) : “M Ft
2—1 2—.1 zal 2—1
_, N
Deﬁnition of Linear Momentum: one particle: 15’ 2 m6, system of particles: P = Z 13?; = M 6mm
i=1
:1 . « _, d13
Newton’s '2n Law for a System of Partlcles: Fnet = M acom = E
Conservation of Linear Momentum of an Isolated System: 213;; = 215}
._, t2 _, _.
Impulse  Linear Momentum Theorem: A351 2 J12 2 F12 (t)dt = Favg,12At
_ h
m — m 2m ' 2m m — m
Elastic Collision (1 Dim):° elf = #vli + —2—v2i vgf = “em—l—vh + #1121
m1+m2 m1+m2 m1+m2 m1+m2
Linear and Angular Variables Related:
2
v
s = r6 '0 = car at = oar (1,: = 7 = (.027: (magnitude of the radial or centripetal acceieration)
'r
Rotation: Rotational Inertia (1mm) for Simple Shapes: see next page
N
Rotational Inertia: Descrete particles: I = Z I: Continuous object: I 2 / rzdm
i=1
Parallei Axis Theorem: I = 1mm + M h2
Torque: ’F='F)<E TZTE=TLF=TFSi11¢
Angular Momentum: rigid body, ﬁxed axis: E = I <3 point—like particle: I: = r’ X 35’
at
Newton’s 2Ild Law: f’net = 15; .Fnet = E
Conservation Law (isolated system, 2 1 m 0): Z = E if
9} 1
Rotational Work: W = Tdﬁ = TangQ Kinetic Energy: K = 51002
37:
. dW W
Rotational Power: Instantaneous: P = “gt— = To; Average: Paw = T = Tﬂvgwmg
Rolling: vcom = wR 0mm 2 (11% 1 1
Kinetic Energy of Rolling: 7K 2 Emvfom + Elm,an 2 =55 Some Rotational Inertias through COM <: A203 Hoop about Annular ij’iizlder : Solid cylinder
central axis (oiring) about 7 ' (or disk) about centre} axis ' _ central axis Soiid cylinder Thin rod about ‘W Solid sphere
{or disk) about _  ~ zoos through center about any
central diameter _7  _ perpendicular to diameter
' ' length Thin ' Hoop about any ' Slab about spherical shell diameter
about any
'j: 2 R diameter , i l perpendicular
. axis sh rough Static equilibrium: Fﬂet m 0 f'net =
Gravity:
Newton’s law: = G—2— Grav1tat10na1 acceleration (planet of mass M): ﬂy 2 2
r r
' 471—2 m in
Law of periods: T2 = 7'3 Potential Energy: U = —G——~lw~—2
GM 7*
m m m m m m
Potential Energy of a System (more than 2 masses): U = — (G%lm—2 + G'—1—:3 + Gﬁ + 7‘12 7'13 T23
Static Fluids:
D '1: Amp AF Hcltt‘P + h
ens: : = — ressure: = _ ms a 10 ressure: =
Y P AV P AA y P P0 99
Pressure Variation with Height or Depth: p2 2 191 + pg(y1 — yg) Archimedes’ Principle: Fb = prdispIacedQ = mfg weightapparentz mg — F1, ...
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This note was uploaded on 05/20/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.
 Spring '07
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