Exam 2: Tonight 8:20-10:10pm
• Room Assignments:
• Breakdown of the 20 Problems
Last Name
Room
Material
# of Problems
A-G
CLB C130
Chapter 7
6
H-O
WEIM 1064
Chapter 8
5
P-Z
MCCC 100
Chapter 9
5
Chapter 10
4
• Crib Sheet:
• Calculator:
You may bring a single hand written formula sheet on 8½ x 11
inch paper (both sides).
You should bring a calculator (any type).
• Scratch Paper:
R. Field 4/1/2014
University of Florida
We will provide scratch paper.
PHY 2053
Page 1
Final Exam Spring 2011: Problem 12
• A uniform disk with mass M, radius R = 0.50 m, and moment of
inertia I = MR2/2 rolls without slipping along the floor at 2 revolutions
per second when it encounters a long ramp angled upwards at 45o
with respect to the horizontal. How high above its original level will
the center of the disk get (in meters)?
Answer: 3.0
% Right: 39%
v
2
H
θ = 45ο
R
Ei = Mv + Iω + MgR = E f = Mg ( H + R)
1
2
v
1
2
2
x-axis
v = Rω
ω = 2πf
1 ⎛ 2 I 2⎞ 1 ⎛ 2 2 I 2⎞
⎜v + ω ⎟ =
⎜R ω + ω ⎟
2g ⎝
M
M
⎠ 2g ⎝
⎠
R2 ⎛
I ⎞ 2 2π 2 R 2 ⎛
I ⎞ 2 2π 2 (0.5m) 2 ⎛ 3 ⎞
2
=
1
+
f
=
(
2
/
s
)
≈ 3.0m
=
ω
⎜1 +
⎟
⎜
⎟
⎜
⎟
2
2
2
2 g ⎝ MR ⎠
g ⎝ MR ⎠
9.8m / s ⎝ 2 ⎠
H=
R. Field 4/1/2014
University of Florida
PHY 2053
Page 2
Exam 2 Fall 2011: Problem 17
• A cloth tape is wound around the outside of a nonuniform solid cylinder (mass M, radius R) and
fastened to the ceiling as shown in the figure. The
cylinder is held with the tape vertical and then release
from rest. If the acceleration of the center-of-mass of
the cylinder is 3g/5, what is its moment of inertia
about its symmetry axis?
R
2
Answer: 23 MR
% Right: 48%
Mg − FT = Ma y
FT = M ( g − a y )
τ = RFT = Iα =
Ia y
R
R
FT
y-axis
⎛ g − 53 g ⎞
R 2 FT ⎛⎜ g − a y ⎞⎟
2
⎟ MR 2 = 23 MR 2
=
I=
MR = ⎜⎜ 3
⎟
⎜ a ⎟
ay
g
y
5
⎝
⎠
⎝
⎠
R. Field 4/1/2014
University of Florida
PHY 2053
Mg
Page 3
Final Exam Fall 2010: Problem 12
• A block of mass m is attached to a cord that is
wrapped around the rim of a flywheel of radius R
and hangs vertically, as shown. The rotational
inertia of the flywheel is I = MR2/2. If when the
block is released and the cord unwinds the
acceleration of the block is equal to g/2, what is
the mass m of the block?
Answer: M/2
% Right: 51%
s = Rθ
v = Rω
a = Rα
R. Field 4/1/2014
University of Florida
m
mg − FT = ma y
τ = Iα = RFT
I
I
FT = α = 2 a y
R
R
I
mg − 2 a y = ma y
R
I ⎛⎜ a y ⎞⎟ 12 MR 2 ⎛ 12 g ⎞ 1
⎜⎜
⎟⎟ = 2 M
m= 2
=
2
1
⎜
⎟
R ⎝ g − ay ⎠
R ⎝g−2g⎠
PHY 2053
R
M
FT
m
y-axis
mg
Page 4
Exam 2 Fall 2013: Problem 39
• Near the surface of the Earth, a cloth tape is wound around the
outside of the two uniform solid cylinders shown in the figure. Solid
cylinder 1 has mass M1, radius R1, and moment of inertia I1 = ½M1R12.
Solid cylinder 2 has mass M2, radius R2, and moment of inertia I2 =
½M2R22. Cylinder 2 is attached to the ceiling and rotates without
friction and cylinder 1 hangs vertically. If M1 = 2M2, when the
cylinders are released from rest and the tape unwinds off both
cylinders, what is the acceleration of the center-of-mass of cylinder 1
(in m/s2)?
y = Rθ + R θ
Answer: 8.4
% Right: 8%
1 1
2 2
v y = R1ω1 + R2ω2
a y = R1α1 + R2α 2
M 1 g − FT = M 1a y
τ 1 = R1 FT = I1α1
τ 2 = R2 FT = I 2α 2
M2
R2
FT
y
M1
R1
FT
y-axis
FT = M 1 ( g − a y )
α1 = R1 FT / I1
α 2 = R2 FT / I 2
R12 FT R22 FT ⎛ M 1 R12 M 1 R22 ⎞
⎟⎟( g − a y ) = X ( g − a y )
+
= ⎜⎜
+
a y = R1α1 + R2α 2 =
I1
I2
I2 ⎠
⎝ I1
⎛ M 1 R12 M 1 R22 ⎞ ⎛ M 1 ⎞
⎟⎟ = 2⎜⎜1 +
⎟⎟ ⎯M⎯=⎯
+
⎯→ 6
X = ⎜⎜
1 2M 2
I2 ⎠ ⎝ M 2 ⎠
⎝ I1
Xg
ay =
= 76 g = 76 (9.8m / s 2 ) = 8.4m / s 2
1+ X
R. Field 4/1/2014
University of Florida
PHY 2053
Page 5
Exam 2 Fall 2010: Problem 13
• A uniform plank is 6 m long and weighs 80 N. It is
balanced on a sawhorse at its center. An additional 160 N
weight is now placed on the left end of the plank. To keep
the plank balanced, the sawhorse must be moved what
distance to the left?
Answer: 2 m
% Right: 48%
Mg = 80 N
mg = 160 N
L = 6m
R. Field 4/1/2014
University of Florida
⎞
⎛L
mg ⎜ − d ⎟ − Mgd = 0
⎝2
⎠
L/2
d
mg
Mg
⎛ mg ⎞ L ⎛ 160 N
⎞L 1
⎟⎟ = ⎜
d = ⎜⎜
⎟ = L = 2m
mg
+
Mg
2
80
N
+
160
N
⎠2 3
⎝
⎝
⎠
PHY 2053
Page 6
Exam 2 Fall 2011: Problem 10
Before
• Two pendulum bobs have masses MA = 3 kg and
MB = 2 kg and equal lengths L as shown in the
figure. Bob A is initially held horizontally while bob
B hangs vertically at rest. Bob A is released and
collides with bob B. The two masses then stick
together and swing upward to the right to a
maximum angle θ. What is the maximum swing
angle θ?
v2 = 2 gL
E = M gL = E = 1 M v 2
50.2o
Answer:
% Right: 31%
1
A
2
2
Ei = 12 ( M A + M B )V 2 = E f = ( M A + M B ) gh
h=
2
⎛ MA ⎞
V
⎟⎟ L
= ⎜⎜
2g ⎝ M A + M B ⎠
2
R. Field 4/1/2014
University of Florida
h = L(1 − cos θ )
L
B
After
θ ≈ 50.2
1
L
A
90o
L
2
A
V
A+B
After
L
A+B
θ Lcosθ
h
o
PHY 2053
θ
L
A+B
2
⎛ MA ⎞
h
9 16
⎟⎟ = 1 −
cos θ = 1 − = 1 − ⎜⎜
=
L
25 25
⎝ MA + MB ⎠
A
90
A 2
pi = M Av2 = M A 2 gL = p f = ( M A + M B )V
L
o
Page 7
Exam 2 Spring 2012: Problem 39
• A modern sculpture has a large horizontal spring, that is
attached to a 60-kg piece of uniform metal at its end and
holds the metal at rest at an angle of θ = 60° above the
horizontal direction as shown in the figure. The other end
of the metal is wedged into a corner and is free to rotate.
If the spring is stretched 0.2 m what is the spring constant
k (in N/m)?
Answer: 849
% Right: 31%
sin( A ± B) = sin A cos B ± cos A sin B
kΔx
L
LkΔx sin θ − Mg cos θ = 0
2
θ
Mg
(60kg )(9.8m / s 2 )
k=
cot(60o ) ≈ 849 N / m
cot θ =
2(0.2m)
2Δx
Mg
R. Field 4/1/2014
University of Florida
PHY 2053
Page 8
Exam 2 Fall 2011: Problem 31
• A uniform solid disk with mass M and radius R is
mounted on a vertical shaft with negligible rotational
inertia and is initially rotating with angular speed ω. A
non-rotating uniform solid disk with mass M and
radius R/2 is suddenly dropped onto the same shaft as
shown in the figure. The two disks stick together and
rotate at the same angluar speed. What is the new
angular speed of the two disk system?
Answer: 4ω/5
% Right: 47%
M
R
M
Li = I iωi = 12 MR12ω = L f = I f ω f = ( 12 MR12 + 12 MR22 )ω f
R12
R2
ωf = 2
ω= 2
ω = 54 ω
2
2
R1 + R2
R + ( R / 2)
R. Field 4/1/2014
University of Florida
PHY 2053
Page 9
Exam 2 Fall 2010: Problem 16
• A mouse of mass M/6 lies on the rim of a solid uniform
disk of mass M that can rotate freely about its center like
a merry-go-round. Initially the mouse and disk rotate
together with an angular velocity of ω. If the mouse walks
to a new position that is at the center of the disk what is
the new angular velocity of the mouse-disk system?
Initial
m
Answer: 4ω/3
% Right: 58%
R
Final
L f = I f ω f = I disk ω f
Li = I iωi = ( I disk + mR 2 )ω
I disk ωnew = ( I disk + mR )ω
2
ωnew
Li = L f
PM
Note that energy is
not conserved in
this problem!
⎛
⎛ mR 2 ⎞
mR 2 ⎞
⎛ 2m ⎞
4
⎟
⎟⎟ω = ⎜⎜1 +
ω
ω
1
=
=
+
= ⎜⎜1 +
⎟
⎜
3ω
2
⎟
1
I
MR
M
⎠
⎝
disk ⎠
⎝
⎠
⎝ 2
R. Field 4/1/2014
University of Florida
PHY 2053
R
m
PM
ΔE = E f − Ei
L2f
L2i
=
−
= 19 MR 2ω 2
2I f 2Ii
Page 10
Exam 2 Fall 2010: Problem 20
• Suppose that you release a small ball from rest at a depth
of 39.2 m below the surface in a pool of water (with
density ρwater) near the surface of the Earth. If the density
of the ball is 1/3 the density of water, how long does it
take the ball to reach the surface?
Answer: 2 s
% Right: 53%
d = a yt
1
2
t=
2
2d
ay
d
y-axis
Fbuoyancy
ρ waterVball g − ρ ballVball g = ρ ballVball a y
⎛ ρ water
⎞
ρ water − ρ ball
⎜
− 1⎟⎟ g
ay =
g =⎜
ρ ball
⎝ ρ ball
⎠
Mg
t=
R. Field 4/1/2014
University of Florida
M water g − M ball g = M ball a y
water
2d
⎛ ρ water
⎞
⎜⎜
− 1⎟⎟ g
⎝ ρ ball
⎠
=
2(39.2m)
2
=
4
s
= 2s
2
(3 − 1)(9.8m / s )
PHY 2053
Page 11
Final Exam Fall 2010: Problem 13
• A block of wood has a mass of 4 kg and density of 600
kg/m3. It is loaded on top with lead (density = 11,400
kg/m3) so that the block of wood will float in water with
90% of its volume submerged. What is the mass of the
lead if the water density is 1,000 kg/m3?
Answer: 2 kg
% Right: 46%
Mlead
Fbuoyancy
M water g − ( M wood + M lead ) g = ( M wood + M lead )a y = 0
M lead = M water − M wood = Vsub ρ water − M wood =
⎛ M wood
Vsub ρ water ⎜⎜
⎝ Vwood ρ wood
⎞
⎛V ρ
⎞
⎟⎟ − M wood = ⎜⎜ sub water − 1⎟⎟ M wood
⎠
⎝ Vwood ρ wood
⎠
y-axis
Mwood
(Mwood+Mlead)g
water
⎛ 0.9(1,000kg / m 3 ) ⎞
1
⎟
= ⎜⎜
−
1
M
=
wood
2 M wood = 2kg
3
⎟
⎝ 600kg / m
⎠
R. Field 4/1/2014
University of Florida
PHY 2053
Page 12
Exam 2 Fall 2010: Problem 19
• An object hangs from a spring balance. When
submerged in water the object weighs one-half what it
weighs in air. If when submerged in an unknown liquid
with density ρX the object weighs three-fourths what it
weighs in air, what is the density of the unknown liquid?
Answer: 2 ρ water
% Right: 41%
1
Air
Fspring = W0
W0 = Mg
Mg
Liquid
Fspring = Wliquid
Fbuoyancy
Mg
Wliquid − Mg + M liquid g = 0
W0 − Wliquid = M liquid g = ρ liquidVg
W0 − Wwater = ρ waterVg
W0 − WX = ρ X Vg
W0 − 34 W0 1
W0 − WX
ρX
=
=
=
1
ρ water W0 − Wwater W0 − 2 W0 2
R. Field 4/1/2014
University of Florida
PHY 2053
Page 13
Exam 2 Spring 12: Problem 46
• What is the minimum radius (in m) that a
spherical helium balloon must have in order to
lift a total mass of m = 10-kg (including the mass
of the empty balloon) off the ground? The
density of helium and the air are, ρHE = 0.18 kg/m3
and ρair = 1.2 kg/m3, respectively.
Answer: 1.33
% Right: 37%
r
y-axis
m
Fbuoyancy − ( M HE + m) g = 0
(MHE +m)g
ρ airVballon g − ( ρ HEVballon + m) g = 0
Vballon = 43 πr 3 =
m
ρ air − ρ HE
1
3
1
3
⎛
⎞ ⎛
⎞
3m
3(10kg )
3
⎟⎟ = ⎜⎜
⎟
r = ⎜⎜
=
2
.
34
m
3 ⎟
⎝ 4π ( ρ air − ρ HE ) ⎠ ⎝ 4π (1.02kg / m ) ⎠
R. Field 4/1/2014
University of Florida
Fbuoyancy
PHY 2053
(
)
1
3
≈ 1.33m
Page 14
Exam 3 Fall 2012: Problem 43
• An ideal spring-and-mass system is undergoing simple
harmonic motion (SHM). If the speed of the block is 1.0
m/s when the displacement from equilibrium is 2.0 m, and
the speed of the block is 3.0 m/s when the displacement
from equilibrium is 1.0 m, what is the angular frequency of
the oscillations (in rad/s)?
Answer: 1.63
% Right: 33%
E = 12 kA = 12 mv + 12 kx
2
ω 2 A2 = v x2 (t 2 ) + ω 2 x 2 (t 2 )
ω 2 A2 = vx2 (t1 ) + ω 2 x 2 (t1 )
ω=
2
x
2
k 2
k 2
2
A = v x (t ) + x (t )
m
m
0 = v x2 (t 2 ) − v x2 (t1 ) + ω 2 ( x 2 (t 2 ) − x 2 (t1 ))
2
2
v
(
t
)
−
v
2
x 2
x (t1 )
ω = 2
x (t1 ) − x 2 (t 2 )
v x2 (t 2 ) − v x2 (t1 )
(3m / s ) 2 − (1m / s ) 2
=
≈ 1.63 / s
2
2
2
2
x (t1 ) − x (t 2 )
(2m) − (1m)
R. Field 4/1/2014
University of Florida
PHY 2053
Page 15