Pickering, Tracey – Homework 25 – Due: Apr 3 2006, noon – Inst: Drummond
1
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16
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before answering.
The due time is Central
time.
001
(part 1 of 3) 10 points
A string is wrapped around the stem of a Yo
Yo which has a mass
M
and a moment of
inertia
I
about its center axis. The radius of
the stem is
r
1
and the radius of the disk is
r
2
. Denote the tension of the string
T
and the
descending acceleration
a
.
h
r
2
r
1
M
ω
The correct torque equation is given by
1.
T r
1
=
a I
r
1
.
correct
2.
T r
2
=
2
a I
r
1
+
r
2
.
3.
T r
1
=
2
a I
r
1
+
r
2
.
4.
T
r
1
+
r
2
2
=
2
a I
r
1
+
r
2
.
5.
T
r
1
+
r
2
2
=
2
a I
r
2
.
6.
T r
2
=
a I
r
2
.
7.
T
r
2

r
1
2
=
2
a I
r
1
+
r
2
.
8.
T
r
1
+
r
2
2
=
2
a I
r
2

r
1
.
9.
T
r
1
+
r
2
2
=
2
a I
r
1
.
Explanation:
The torque created by the tension is
τ
=
T r
1
=
I α
=
I
a
r
1
=
a I
r
1
.
002
(part 2 of 3) 10 points
The correct vertical force equation is given by
1.
M g
+
T
=
M a .
2.
T

M g
=
M a .
3.
M g

T
=
M a .
correct
Explanation:
The net force on the YoYo is
F
net
=
X
F
:
M g

T
=
M a .
003
(part 3 of 3) 10 points
Release the YoYo from rest.
It reaches a
center of mass speed
v
when it falls for a
height Δ
h
. Using the same notation as in the
previous questions, the correct workenergy
equation is given by
1.
1
2
M v
2
=
1
2
Iv
2
.
2.
1
2
M v
2
=
1
2
I
v
r
1
¶
2
.
3.
M g
Δ
h
=
1
2
I
v
r
1
¶
2
.
4.
M g
Δ
h
=
1
2
M v
2
+
1
2
I
v
r
1
¶
2
.
correct
5.
1
2
M v
2
=
1
2
I
v
r
2
¶
2
.
6.
M g
Δ
h
=
1
2
I
v
r
2
¶
2
.
7.
M g
Δ
h
=
1
2
M v
2
.
8.
M g
Δ
h
=
1
2
M v
2
+
1
2
I
v
r
2
¶
2
.
Explanation:
The work done by the gravity is
W
=
M gh
.
The combined (translational and rotational)
kinetic energy is
K
=
1
2
M v
2
+
1
2
I ω
2
=
1
2
M v
2
+
1
2
I
v
r
1
¶
2
.
Thus the work energy theorem gives
M g
Δ
h
=
1
2
M v
2
+
1
2
I
v
r
1
¶
2
.
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Pickering, Tracey – Homework 25 – Due: Apr 3 2006, noon – Inst: Drummond
2
004
(part 1 of 2) 10 points
A hollow cylinder has length a 2
.
4 m, mass
0
.
6 kg, and radius 0
.
1 m.
The cylinder is
free to rotate about a vertical axis that passes
through its center and is perpendicular to the
cylinder’s axis.
Inside the cylinder are two
masses of 0
.
2 kg each, attached to springs
of spring constant
k
and unstretched lengths
0
.
4 m.
The inside walls of the cylinder are
frictionless.
2
.
4 m
0
.
8 m
0
.
2 kg
0
.
2 kg
0
.
6 kg
Determine the value of the spring constant
k
if the masses are located 0
.
6 m from the cen
ter of the cylinder when the cylinder rotates
at 20 rad
/
s.
Correct answer: 120 N
/
m.
Explanation:
Let :
m
= 0
.
6 kg
,
Δ
x
= 0
.
6 m
,
and
ω
= 20 rad
/
s
.
Applying Newton’s second law, the net in
ward force acting on each of the 0
.
2 kg masses
is
X
F
radial
=
k
Δ
x
=
m
(
x
+ Δ
x
)
ω
2
.
Thus
k
=
m
(
x
+ Δ
x
)
ω
2
Δ
x
=
(0
.
6 kg) (0
.
6 m) (20 rad
/
s)
2
0
.
4 m
=
120 N
/
m
.
005
(part 2 of 2) 10 points
How much work was needed to bring the sys
tem from rest to 20 rad
/
s?
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