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pasha (sep635) – HW 91 – Antoniewicz – (56445)
1
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beFore answering.
001 (part 1 of 2) 10.0 points
Amanwho
s
ema
s
si
s80kgandaw
oman
whose mass is 50 kg sit at opposite ends oF a
canoe 5 m long, whose mass is 35 kg.
Assume the man is seated at
x
=0and
the boat extends along the positive
x
axis
with the woman at the other end. Where is
the center oF mass oF the system consisting oF
man, woman, and canoe?
Correct answer: 2
.
04545 m.
Explanation:
With the man (
m
1
)attheorigin,wehave
x
CM
=
m
1
x
1
+
m
2
x
2
+
m
3
x
3
m
1
+
m
2
+
m
3
=
0+(50kg)(5m)+(35kg)(2
.
5m)
80 kg + 50 kg + 35 kg
=2
.
04545 m
.
002 (part 2 of 2) 10.0 points
Suppose that the man moves quickly to the
center oF the canoe and sits down there. How
Far does the canoe move in the water?
Correct answer: 1
.
21212 m.
Explanation:
Assume that the net Force on the canoe
is zero. Then its center oF mass momentum
(and velocity) remains constant, which is zero
in this case.
So the center oF mass oF the
system remains stationary.
Place the origin at the leFt end oF the canoe
(where the man was sitting in part 1) and the
new center oF mass is at
x
CM
=
m
1
x
1
+
m
2
x
2
+
m
3
x
3
m
1
+
m
2
+
m
3
=
(80 kg)(2
.
5m)+(50kg)(5m)
80 kg + 50 kg + 35 kg
+
(35 kg)(2
.
5m)
80 kg + 50 kg + 35 kg
=3
.
25758 m
.
At frst, the center oF mass was 2
.
04545 m
From the leFt end. Now the center oF mass is
3
.
25758 m From the leFt end. But the center
oF mass never moved! This means the boat
moved
3
.
25758 m

2
.
04545 m =
1
.
21212 m
to the leFt as the man walked 1
.
21212 m to
the right.
003 (part 1 of 3) 7.0 points
A string is wound around a uniForm disc
whose mass is 3
.
5kgandrad
iu
si
s0
.
58 m
,
see fgure below. The disc is released From
rest with the string vertical and its top end
tied to a fxed support.
The acceleration oF gravity is 9
.
8m
/
s
2
.
h
0
.
58 m
3
.
5kg
ω
As the disc descends, calculate the tension
in the string.
Correct answer: 11
.
4333 N.
Explanation:
Let :
R
=0
.
58 m
,
M
=3
.
5kg
,
and
g
=9
.
8m
/
s
2
.
Basic Concepts
X
~
F
=
m
~
a
X
~τ
=
I
~α
Δ
U
+
Δ
K
rot
+
Δ
K
trans
=0
Solution
X
F
=
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View Full Documentpasha (sep635) – HW 91 – Antoniewicz – (56445)
2
X
τ
=
TR
=
I
α
=
1
2
MR
2
±
a
R
²
.
(2)
Solving for
a
in (2),
a
=
2
T
M
.
(3)
Using
a
from Eq. (3) and solving for
T
in (1),
T
=
M
(
g

a
)
=
M
³
g

2
T
M
´
=
Mg

2
T
3
T
=
Mg
T
=
Mg
3
(4)
=
(3
.
5kg)(9
.
8m
/
s
2
)
3
=
11
.
4333 N
.
004 (part 2 of 3) 7.0 points
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 Fall '08
 Turner
 Mass

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