Quiz 5

Name________________________________________
Physics 110 Quiz #5, November 1, 2013
Please show all work, thoughts and/or reasoning in order to receive partial credit. The
quiz is worth 10 points total.
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1. Two objects of equal mass and radius are shown on the
right. A disk (on the right with moment of inertia
I = 12 MR 2 ) and a hoop (on the left with moment of inertial
I = MR 2 ) have a string wrapped around their
circumferences. Hanging from the string, halfway between
the disk and the hoop, is a block of mass m . The disk and
the hoop are free to rotate about their centers. When the block is released from rest
and allowed to fall, the block
a.
b.
c.
d.
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moves to the right of the centerline.
moves to the left of the centerline.
moves along the centerline.
moves both left and right of the centerline or it oscillates about the centerline.
2. Suppose that we have the set-up shown below consisting of a shaft of radius rshaft =
0.5cm around which a massless string is wound and a disk of moment of inertial I is
placed. The string is passed over a massless pulley where a mass mh = 200g is hung
and allowed to fall from rest through a height h = 1m acquiring a speed of v =
1.84m/s.
a. Using energy ideas derive an expression for the moment of inertia of the disk and
then evaluate your expression using the information given in the problem above.
ΔE = 0 = ΔKE T + ΔKE R + ΔU g = ( 12 mh v 2f − 0) + ( 12 Iω 2f − 0) + (0 − mh gy i )
 v 

0 = ( 12 mh v 2f − 0) +  12 I f  − 0 + (0 − mh gh )
r
  sh 



m
 2gh 
2 × 9.8 s2 ×1m 
2
I =  2 −1 mh rsh2 = 
−1 (0.2kg)(0.005m) = 2.4 ×10−5 kg ⋅ m 2
2
 1.84 m

 vf

s


Disk
2
(
)
rshaft
mh
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b.
Using ideas of torque and forces derive an expression for the moment of inertia
of the disk and then evaluate your expression using the information given in the
start of the problem.
Disk
rshaft
mh
∑τ :
rsh FT = rsh ( mh g − mh a) = Iα = I
a
rsh




 2gh 
g 
g


→ I =  −1mh rsh2 =  2 −1 mh rsh2 =  2 −1 mh rsh2 = 2.4 ×10−5 kg ⋅ m 2
v 
a 
 vf

 f  
 2h

  
∑F
y
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Useful formulas:
: FT − mh g = −mh a → FT = mh g − mh a
Motion in the r = x, y or z-directions
Uniform Circular Motion
v2
ar =
r
v2
Fr = mar = m
r
2πr
v=
T
mm
FG = G 1 2 2
r
rf = r0 + v 0r t + 12 ar t 2
v fr = v 0r + ar t
v fr 2 = v 0r 2 + 2arΔr
Vectors
Circles
Triangles
Spheres
C = 2πr
A = 12 bh
A = 4 πr 2
A = πr 2
V = 43 πr 3
Quadratic equation : ax 2 + bx + c = 0,
whose solutions are given by : x =
−b ± b 2 − 4ac
2a
Useful Constants
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2
magnitude of avector = v x + v 2 y
g = 9.8 m s2
Linear Momentum/Forces
→
→
N A = 6.02×10
→
p f = p i + F Δt
→
→
F = ma
→
1
2
K R = Iω
Heat
2
TC = 59 [TF − 32]
2
TF = 95 TC + 32
Lnew = Lold (1+ αΔT )
U g = mgh
→
Fs = −k x
U S = 12 kx 2
F f = µFN
WT = FdCosθ = ΔE T
Anew = Aold (1+ 2αΔT )
Vnew = Vold (1+ βΔT ) : β = 3α
PV = Nk B T
W R = τθ = ΔE R
W net = W R + WT = ΔE R + ΔE T
3
2
ΔQ kA
= ΔT
Δt L
ΔQ
PR =
= εσAΔT 4
ΔT
ΔU = ΔQ − ΔW
PC =
ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss
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Fluids
kB T = 12 mv 2
ΔQ = mcΔT
ΔE R + ΔE T + ΔU g + ΔU S = 0
Rotational Motion
kg 2
vsound = 343 m s
Work/Energy
KT = 2 mv
→
2
kB = 1.38×10−23 J K
mole
σ = 5.67×10−8 W m 2 K 4
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p = mv
→
G = 6.67×10−11 Nm
23 atoms
v 
direction of avector → φ = tan−1 y 
 vx 
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Geometry /Algebra
Simple Harmonic Motion/Waves
ω=
€
k
2π
= 2πf =
m
T
TS = 2π
m
k
TP = 2π
l
g
1
v =±
k  x 2 2
A1− 
m  A2 
x ( t ) = x max sin(ωt ) or x max cos(ωt )
v ( t ) = v max cos(ωt ) or − v max sin(ωt )
a( t ) = −amax sin(ωt ) or − amax cos(ωt )
v max = ωx max ; amax = ω 2 x max
Sound
v = fλ =
v = fλ = (331+ 0.6T) ms
v
2L
I = 2π 2 f 2 ρvA 2
I
; Io = 1×10−12 mW2
I0
v
v
f n = nf1 = n ; f n = nf1 = n
2L
4L
f n = nf1 = n
β = 10log
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FT
µ