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001
10.0 points
A uniform rod of mass 3
.
5 kg is 15 m long. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 15 m from the center of
mass of the rod.
Initially the rod makes an
angle of 49
◦
with the horizontal. The rod is
released from rest at an angle of 49
◦
with the
horizontal, as shown in the figure below
15 m
15 m
3
.
5 kg
O
49
◦
What is the angular acceleration of the rod
at the instant the rod is in a horizontal posi
tion? The acceleration of gravity is 9
.
8 m
/
s
2
and the moment of inertia of the rod about
its center of mass is
1
12
m ℓ
2
.
Correct answer: 0
.
603077 rad
/
s
2
.
Explanation:
Let :
ℓ
= 15 m
,
d
=
ℓ
= 15 m
,
θ
= 49
◦
,
and
m
= 3
.
5 kg
.
I
=
I
CM
+
m d
2
=
1
12
m ℓ
2
+
m ℓ
2
=
13
12
m ℓ
2
.
Since the rod is uniform, its center of mass
is located at
ℓ .
Recalling that the weight
m g
acts at the center of mass, the magnitude of
the torque at the horizontal position is
τ
=
I α
r F
=
parenleftbigg
13
12
m ℓ
2
parenrightbigg
α
ℓ
(
m g
) =
13
12
m ℓ
2
α
α
=
12
g
13
ℓ
=
12 (9
.
8 m
/
s
2
)
13 (15 m)
=
0
.
603077 rad
/
s
2
.
002
(part 1 of 2) 10.0 points
A uniform flat plate of metal is situated in the
reference frame shown in the figure below.
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
x
( m )
y
( m )
Calculate the
x
coordinate of the center of
mass of the metal plate.
Correct answer: 1
.
33333 m.
Explanation:
Let :
(
x
1
, y
1
) = (0 m
,
0 m)
(
x
2
, y
2
) = (4 m
,
0 m)
(
x
3
, y
3
) = (0 m
,
3 m)
.
The equation for the hypotenuse is
y

y
2
=
s
(
x

x
2
)
=
y
3
x
2
(
x

x
2
)
.
Since
dm
=
ρ y dx
, for the areal density, the
x
coordinate of the center of mass is
x
cm
=
integraldisplay
x dm
integraldisplay
dm
=
σ
integraldisplay
x
2
0
x y dx
σ
integraldisplay
x
2
0
y dx
oldhomework 25 – Turner – (58185)
1
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hinojosa (jlh3938) – oldhomework 25 – Turner – (58185)
2
=
integraldisplay
x
2
0
x s
(
x

x
2
)
dx
integraldisplay
x
2
0
s
(
x

x
2
)
dx
=
integraldisplay
x
2
0
x
(
x

x
2
)
dx
integraldisplay
x
2
0
(
x

x
2
)
dx
=
1
3
x
3

1
2
(
x
2
)
x
2
vextendsingle
vextendsingle
vextendsingle
vextendsingle
x
2
0
1
2
x
2

(
x
2
)
x
vextendsingle
vextendsingle
vextendsingle
vextendsingle
x
2
0
=
1
3
x
3
2

1
2
(
x
2
)
x
2
2
1
2
x
2
2

(
x
2
)
x
2
=
x
3
2
3
x
2
2
=
1
3
x
2
=
1
3
(4 m)
=
1
.
33333 m
.
003
(part 2 of 2) 10.0 points
If the mass of the plate is 2 kg, find the
moment of inertia of the triangle with the
y

axis as the axis of rotation.
Correct answer: 5
.
33333 kg
·
m
2
.
Explanation:
The moment of inertia of the triangle about
the
x
coordinate is
I
x
=0
y
≡
integraldisplay
x
2
0
x
2
dm
=
σ
integraldisplay
x
2
0
x
2
y dx
=
σ s
integraldisplay
x
2
0
x
2
(
x

x
2
)
dx
=
σ s
parenleftbigg
1
4
x
4

1
3
x
2
x
3
parenrightbiggvextendsingle
vextendsingle
vextendsingle
vextendsingle
x
2
0
=
σ s
parenleftbigg
1
4
x
4
2

1
3
x
4
2
parenrightbigg
=
parenleftbigg
2
m
x
2
y
3
parenrightbigg parenleftbigg

y
3
x
2
parenrightbigg parenleftbigg

1
12
x
4
2
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 Spring '08
 Turner
 Kinetic Energy, Mass, Work, Moment Of Inertia, Rigid Body, kg, Hinojosa

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