homework 24 – BAUTISTA, ALDO – Due: Mar 29 2006, 4:00 am
1
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Question 1
Part 1 of 3.
10 points.
Three particles of mass 8 kg, 5 kg, and 8 kg
are connected by rigid rods of negligible mass
lying along the
y
axis and are placed at 5 m,

3 m, and

7 m
,
respectively as in the figure.
The system rotates about the
x
axis with an
angular speed of 2
.
5 rad
/
s
.
Contrary to what is observed in the figure,
consider the masses to be point particles.
5 m

3 m

7 m
2
.
5 rad
/
s
x
8 kg
5 kg
8 kg
Find the moment of inertia about the
x
axis.
Correct answer: 637 kg m
2
(tolerance
±
1 %).
Explanation:
Let :
m
1
= 8 kg
,
m
2
= 5 kg
,
m
3
= 8 kg
,
y
1
= 5 m
,
y
2
=

3 m
,
y
3
=

7 m
,
and
ω
= 2
.
5 rad
/
s
.
The total rotational inertia of the system
about the
x
axis is
I
=
X
m
i
r
2
i
=
637 kg m
2
,
where,
r
i
=

y
i

.
Question 2
Part 2 of 3.
10 points.
Find the total rotational energy of the sys
tem.
Correct answer: 1990
.
62 J (tolerance
±
1 %).
Explanation:
Since
ω
= 2
.
5 rad
/
s
,
the total rotational
energy is
E
=
1
2
I ω
2
=
1
2
(637 kg m
2
) (2
.
5 rad
/
s)
2
=
1990
.
62 J
.
Question 3
Part 3 of 3.
10 points.
Find the linear speed of the top particle of
mass 8 kg in the figure.
Correct answer: 12
.
5 m
/
s (tolerance
±
1 %).
Explanation:
The linear speed of each particles are
v
i
=
r
i
ω
v
1
=
r
1
ω
= (5 m) (2
.
5 rad
/
s) =
12
.
5 m
/
s
.
Question 4
Part 1 of 1.
10 points.
A 1900 kg car is being lowered by a winch.
At the monent shown in the figure, the gear
box of the winch breaks and the car falls from
rest. During the car’s fall, there is no slipping
between the (massless) rope, the pulley, and
the winch drum.
The moment of inertia of
the winch drum is 257 kg
·
m
2
and that of the
pulley is 14 kg
·
m
2
. The radius of the drum
is 86 cm and the radius of the pulley is 9 cm.
The
acceleration
due
to
gravity
is
9
.
81 m
/
s
2
.
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homework 24 – BAUTISTA, ALDO – Due: Mar 29 2006, 4:00 am
2
1900 kg
86 cm
9 m
Find the speed of the car as it hits the
ground.
Correct answer: 9
.
18609
kN (tolerance
±
1
%).
Explanation:
Let :
I
d
= 257 kg
·
m
2
,
I
p
= 14 kg
·
m
2
,
r
d
= 86 cm = 0
.
86 m
,
r
p
= 9 cm = 0
.
09 m
,
m
b
= 1900 kg
,
and
Δ
h
= 9 m
.
Applying conservation of mechanical en
ergy (
K
i
=
U
f
= 0) and
v
=
r ω
,
1
2
m v
2
+
1
2
I
d
ω
d
+
1
2
I
p
ω
p

m g
Δ
h
= 0
1
2
m v
2
+
1
2
I
d
v
2
r
2
d
+
1
2
I
p
v
2
r
2
p

m g
Δ
h
= 0
v
=
v
u
u
u
t
2
m g
Δ
h
m
+
I
d
r
2
d
+
I
p
r
2
p
=
v
u
u
u
t
2 (1900 kg) (9
.
81 m
/
s
2
) (9 m)
1900 kg +
257 kg
·
m
2
(0
.
86 m)
2
+
14 kg
·
m
2
(0
.
09 m)
2
=
9
.
18609 m
/
s
.
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 Spring '08
 Turner
 Angular Momentum, Kinetic Energy, Mass, Work, Moment Of Inertia, Rigid Body

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