homework 25 – BAUTISTA, ALDO – Due: Mar 31 2006, 4:00 am
1
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V1:1, V2:4, V3:2, V4:4, V5:3.
Question 1
Part 1 of 3.
10 points.
A cylindrical 9
.
67 kg pulley with a radius of
0
.
643 m is used to lower a 6
.
36 kg bucket into
a well. The bucket starts from rest and falls
for 4
.
04 s.
The acceleration of gravity is 9
.
81 m
/
s
2
.
a) What is the linear acceleration of the
falling bucket?
Correct answer: 5
.
57317 m
/
s
2
(tolerance
±
1
%).
Explanation:
Basic Concepts:
Newton’s second law for the bucket:
F
net
=
ma
=
mg

F
T
τ
=
Iα
=
1
2
MR
2
α
τ
=
F
T
R
since
F
T
is perpendicular to
R
.
a
=
αR
Given:
R
= 0
.
643 m
M
= 9
.
67 kg
m
= 6
.
36 kg
g
= 9
.
81 m
/
s
2
Solution:
F
T
=
τ
R
=
(
1
2
MR
2
) (
a
R
)
R
=
1
2
Ma
Thus from Newton’s second law,
ma
=
mg

1
2
Ma
2
ma
+
Ma
= 2
mg
a
=
2
mg
(2
m
+
M
)
=
2(6
.
36 kg)(9
.
81 m
/
s
2
)
2(6
.
36 kg) + 9
.
67 kg
= 5
.
57317 m
/
s
2
Question 2
Part 2 of 3.
10 points.
b) How far does it drop?
Correct answer:
45
.
4815
m (tolerance
±
1
%).
Explanation:
Basic Concept:
Δ
y
=
v
i
Δ
t
+
1
2
a
(Δ
t
)
2
=
1
2
a
(Δ
t
)
2
since
v
i
= 0 m
/
s.
Given:
Δ
t
= 4
.
04 s
Solution:
Δ
y
=
1
2
(5
.
57317 m
/
s
2
)(4
.
04 s)
2
= 45
.
4815 m
Question 3
Part 3 of 3.
10 points.
c) What is the angular acceleration of the
cylindrical pulley?
Correct answer: 8
.
66744 rad
/
s
2
(tolerance
±
1 %).
Explanation:
Basic Concept:
a
=
αR
Solution:
α
=
a
R
=
5
.
57317 m
/
s
2
0
.
643 m
= 8
.
66744 rad
/
s
2
Question 4
Part 1 of 2.
10 points.
A 1
.
7 kg particle moves in a circle of ra
dius 1
.
7 m.
The magnitude of its angular
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homework 25 – BAUTISTA, ALDO – Due: Mar 31 2006, 4:00 am
2
momentum relative to the center of the circle
depends on time according to
L
= (6 N
·
m)
t
.
Find the magnitude of the torque acting on
the particle.
Correct answer: 6 N
·
m (tolerance
±
1 %).
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 Spring '08
 Turner
 Energy, Kinetic Energy, Work, Moment Of Inertia, Correct Answer, Physical quantities

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