Version 133 – exam3 – Chiu – (60180)
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001
10.0 points
A uniform 9 kg rod with length 12 m has
a frictionless pivot at one end.
The rod is
released from rest at an angle of 27
◦
beneath
the horizontal.
6 m
12 m
9 kg
27
◦
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
of inertia of a rod about either end is
1
3
m L
2
,
and the acceleration of gravity is 9
.
8 m
/
s
2
.
1. 1.29793
2. 0.324483
3. 0.803557
4. 0.457454
5. 1.09148
6. 2.23819
7. 0.471867
8. 1.62242
9. 0.486772
10. 0.353627
Correct answer: 1
.
09148 rad
/
s
2
.
Explanation:
Let :
m
= 9 kg
,
L
= 12 m
,
and
θ
= 27
◦
.
The rod’s moment of inertia about its end
point is
I
=
1
3
m L
2
,
so the angular accelera
tion of the rod is
α
=
τ
I
=
1
2
m g L
cos
θ
1
3
m L
2
=
3
2
g
L
cos
θ
=
3
2
9
.
8 m
/
s
2
12 m
cos 27
◦
=
1
.
09148 rad
/
s
2
.
002
10.0 points
A(n) 65 kg boat that is 7 m in length is
initially 8
.
9 m from the pier.
A 47 kg child
stands at the end of the boat closest to the
pier. The child then notices a turtle on a rock
at the far end of the boat and proceeds to
walk to the far end of the boat to observe the
turtle.
Assume:
There is no friction between boat
and water.
8
.
9 m
7 m
How far is the child from the pier when she
reaches the far end of the boat?
1. 12.9625
2. 15.4528
3. 12.1874
4. 11.6375
5. 13.3273
6. 15.9
7. 14.8316
8. 12.6674
9. 13.98
10. 12.0465
Correct answer: 12
.
9625 m.
Explanation:
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Version 133 – exam3 – Chiu – (60180)
2
Let
D
= distance of the boat from the pier
,
= 8
.
9 m
,
L
= length of the boat
,
= 7 m
,
M
= mass of the boat
,
= 65 kg
,
and
m
= mass of the child
,
and
= 47 kg
,
X
= change in position of the boat
.
ℓ
= final distance of child from pier
.
d
= final distance of boat from pier
.
D
L
X
d
ℓ
X
We will use the pier as the origin of the
x
coordinate.
Initially, the center of mass of the boatchild
system is
x
cm
=
parenleftbigg
D
+
L
2
parenrightbigg
M
+
D m
M
+
m
Finally, the center of mass of the boatchild
system is
x
cm
=
parenleftbigg
D
+
L
2
 X
parenrightbigg
M
+ (
D
+
L
 X
)
m
M
+
m
,
where
X
is the change in position of the center
of mass of the boat. Since the center of mass
of the system does not move, we can equate
the above two expressions for
x
cm
parenleftbigg
D
+
L
2
parenrightbigg
M
+
D m
M
+
m
=
parenleftbigg
D
+
L
2
 X
parenrightbigg
M
+ (
D
+
L
 X
)
m
M
+
m
and, solving for
X
, we have
parenleftbigg
D
+
L
2
parenrightbigg
M
+
D m
=
parenleftbigg
D
+
L
2
 X
parenrightbigg
M
+ (
D
+
L
 X
)
m
0 =
X
M
+ (
L
 X
)
m
X
(
M
+
m
) =
L m
X
=
m
m
+
M
L
=
(47 kg)
(47 kg) + (65 kg)
×
(7 m)
= 2
.
9375 m
.
The final distance
ℓ
of the child from the
pier is
ℓ
=
D
+
L
 X
= (8
.
9 m) + (7 m)

(2
.
9375 m)
=
12
.
9625 m
.
The final distance
d
of the near end of the
boat to the pier is
d
=
D
 X
= (8
.
9 m)

(2
.
9375 m)
= 5
.
9625 m
.
003
10.0 points
A uniform rod of mass 2
.
3 kg is 20 m long. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 20 m from the center of
mass of the rod.
Initially the rod makes an
angle of 45
◦
with the horizontal. The rod is
released from rest at an angle of 45
◦
with the
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 Spring '08
 Turner
 Friction, Momentum, τ, Wnc, xcm, 45◦

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