billing (cab4763) – hw 10 – opyrchal – (11109)
1
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printout
should
have
11
questions.
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before answering.
001 (part 1 of 3) 10.0 points
A 5
.
2 kg mass is connected by a light cord to
a 1
.
7 kg mass on a smooth surface as shown.
The pulley rotates about a frictionless axle
and has a moment of inertia of 0
.
67 kg
·
m
2
and a radius of 0
.
37 m.
5
.
2kg
1
.
7kg
F
1
F
2
R
What is the acceleration of the two masses?
The acceleration of gravity is 9
.
81 m
/
s
2
.
As
sume that the cord does not slip on the pulley.
Correct answer: 4
.
32522 m
/
s
2
.
Explanation:
F
2
a
F
1
m
1
g
a
Let :
I
= 0
.
67 kg
·
m
2
,
R
= 0
.
37 m
,
m
1
= 5
.
2 kg
,
m
2
= 1
.
7 kg
,
and
g
= 9
.
81 m
/
s
2
.
Applying Newton’s second law to
m
1
and
m
2
,
m
1
a
=
m
1
g

F
1
m
2
a
=
F
2
F
net
=
F
1

F
2
= (
m
1
g

m
1
a
)

m
2
a
=
m
1
g

(
m
1
+
m
2
)
a
a
=
αR
, so applying Newton’s second law to
the pulley,
F
net
R
=
I α
=
I
parenleftBig
a
R
parenrightBig
I a
=
F
net
R
2
I a
= [
m
1
g

(
m
1
+
m
2
)
a
]
R
2
a
=
m
1
g R
2
I
+ (
m
1
+
m
2
)
R
2
=
(5
.
2kg)(9
.
81m
/
s
2
)(0
.
37m)
2
0
.
67kg
·
m
2
+(5
.
2kg+1
.
7kg)(0
.
37m)
2
=
4
.
32522 m
/
s
2
.
002 (part 2 of 3) 10.0 points
What is the magnitude of the force
F
1
?
Correct answer: 28
.
5209 N.
Explanation:
F
1
=
m
1
g

m
1
a
= (5
.
2kg)(9
.
81m
/
s
2

4
.
32522 m
/
s
2
)
=
28
.
5209 N
.
003 (part 3 of 3) 10.0 points
What is the magnitude of the force
F
2
?
Correct answer: 7
.
35287 N.
Explanation:
F
2
=
m
2
a
= (1
.
7kg)(4
.
32522 m
/
s
2
)
=
7
.
35287 N
.
004
10.0 points
A uniform 9 kg rod with length 42 m has
a frictionless pivot at one end.
The rod is
released from rest at an angle of 30
◦
beneath
the horizontal.
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billing (cab4763) – hw 10 – opyrchal – (11109)
2
21 m
42 m
9 kg
30
◦
What is the angular acceleration of the rod
immediately after it is released? The moment
of inertia of a rod about the center of mass
is
1
12
m L
2
,
where
m
is the mass of the rod
and
L
is the length of the rod. The moment
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