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Gozick, Brandon – Homework 15 – Due: Nov 3 2006, 5:00 pm – Inst: D Weathers
1
This printout should have 9 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
A 27 kg mass and a 17 kg mass are suspended
by a pulley that has a radius oF 6
.
3 cm and
a mass oF 8 kg.
The cord has a negligible
mass and causes the pulley to rotate without
slipping. The pulley rotates without Friction.
The masses start From rest 1
.
3 m apart. Treat
the pulley as a uniForm disk.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
1
.
3 m
6
.
3 cm
8 kg
ω
27 kg
17 kg
Determine the speeds oF the two masses as
they pass each other.
Correct answer: 1
.
62916 m
/
s.
Explanation:
Let :
M
= 8 kg
,
R
= 6
.
3 cm
,
m
1
= 27 kg
,
m
2
= 17 kg
,
h
= 1
.
3 m
,
v
=
ω R ,
I
=
1
2
M R
2
,
and
K
disk
=
1
2
I ω
2
=
1
4
M v
2
.
±rom conservation oF energy
K
1
+
K
2
+
K
disk
=

Δ
U
or
m
1
v
2
2
+
m
2
v
2
2
+
M v
2
4
= (
m
1

m
2
)
g
h
2
µ
m
1
+
m
2
+
M
2
¶
v
2
= (
m
1

m
2
)
g h ,
where
h
2
is the height. Taking no slipping into
account, we can solve For
v
v
=
s
2(
m
1

m
2
)
g h
2(
m
1
+
m
2
) +
M
(1)
=
s
127
.
4 J
48 kg
=
1
.
62916 m
/
s
.
Alternative Solution:
The Forces in the
vertical directions For
m
1
, m
2
,
and the torque
on the pulley give us
m
1
g

T
1
= +
m
1
a
m
2
g

T
2
=

m
2
a ,
so
T
1
=
m
1
(
g

a
)
T
2
=
m
2
(
g
+
a
)
,
and
I α
= [
T
1

T
2
]
R
1
2
M R
2
‡
a
R
·
=
h
m
1
(
g

a
)]
R

[
m
2
(
g
+
a
)
i
R
1
2
M a
=
m
1
g

m
1
a

m
2
a

m
2
g
a
=
2(
m
1

m
2
)
g
M
+ 2(
m
1
+
m
2
)
.
The velocity is
v
2
f
=
v
2
i
+
a
Δ
y
v
f
=
r
2
a
h
2
=
s
2(
m
1

m
2
)
g h
2(
m
1
+
m
2
) +
M
,
which is the same as Eq. 1.
keywords:
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View Full DocumentGozick, Brandon – Homework 15 – Due: Nov 3 2006, 5:00 pm – Inst: D Weathers
2
002
(part 1 of 1) 10 points
A uniform disk of radius 1
.
3 m and mass
1
.
3 kg is suspended from a pivot 0
.
13 m above
its center of mass.
The acceleration of gravity is 9
.
8 m
/
s
2
.
axis
Find the angular frequency
ω
for small os
cillations.
Correct answer: 1
.
21578 rad
/
s.
Explanation:
Basic Concepts
The physical pendulum:
τ
=
I α
=

m g d
sin
θ
α
=
d
2
θ
dt
2
so that the angular frequency for small oscil
lations (sin
θ
≈
θ
) is
ω
=
r
m g d
I
.
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 Weathers
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