Overton, Mays – Homework 26 – Due: Apr 4 2005, 4:00 am – Inst: Turner
1
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printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
A projectile of mass
m
= 1
.
11 kg moves to the
right with speed
v
0
= 15 m
/
s. The projectile
strikes and sticks to the end of a stationary
rod of mass
M
= 6
.
08 kg and length
d
=
2
.
89 m that is pivoted about a frictionless
axle through its center.
d
O
O
v
0
ω
m
M
Find the angular speed of the system right
after the collision.
Correct answer: 3
.
67348 rad
/
s.
Explanation:
The initial angular momentum of the pro
jectile about the pivot O is
L
i
=
d
2
m v
0
. With
the projectile stuck to the end of the rod, the
rotational inertia of the projectile and the rod
combined about O is,
I
=
m
d
2
¶
2
+
1
12
M d
2
=
•
m
4
+
M
12
‚
d
2
(1)
=
•
(1
.
11 kg)
4
+
(6
.
08 kg)
12
‚
(2
.
89 m)
2
= 6
.
54944 kg m
2
.
Using conservation of angular momentum, we
have
L
i
=
I
f
ω
(2)
=
~r
×
~p
=
d
2
m v
0
(3)
Therefore, combining Eqs. (1), (2), and (3),
we have
ω
=
d
2
m v
0
m d
2
4
+
M d
2
12
·
12
12
=
6
m v
0
(3
m
+
M
)
d
=
6 (1
.
11 kg) (15 m
/
s)
[3 (1
.
11 kg) + (6
.
08 kg)] (2
.
89 m)
= 3
.
67348 rad
/
s
.
002
(part 2 of 2) 10 points
How much kinetic energy is lost in the col
lision relative to the initial kinetic energy?
(Determine the ratio of the kinetic energy lost
to the initial kinetic energy)
Correct answer: 0
.
646121 .
Explanation:
The initial energy of the system is
E
0
=
1
2
m v
2
0
=
1
2
(1
.
11 kg) (15 m
/
s)
2
= 124
.
875 J
.
The final energy of the system is
E
=
1
2
I ω
2
=
1
2
•
3
m
+
M
12
¶
d
2
‚ ‰
6
m v
0
[3
m
+
M
]
d
2
=
1
2
m v
2
0
•
3
m
3
m
+
M
‚
=
1
2
(1
.
11 kg) (15 m
/
s)
2
×
•
3 (1
.
11 kg)
3 (1
.
11 kg) + (6
.
08 kg)
‚
= 44
.
1906 J
.
Therefore, the fractional loss of the energy is
f
=
E
0

E
E
0
=
1
2
m v
2
0
•
1

3
m
3
m
+
M
‚
1
2
m v
2
0
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Overton, Mays – Homework 26 – Due: Apr 4 2005, 4:00 am – Inst: Turner
2
=
M
3
m
+
M
=
(6
.
08 kg)
3 (1
.
11 kg) + (6
.
08 kg)
= 0
.
646121
.
003
(part 1 of 2) 10 points
A small blob of putty of mass
m
falls from
the ceiling and lands on the outer rim of a
turntable of radius
R
and moment of inertia
I
0
that is rotating freely with angular speed
ω
i
about its vertical fixed symmetry axis.
What is the postcollision angular speed of
the turntable plus putty?
1.
ω
f
=
ω
i
1 +
m R
I
0
2.
ω
f
=
ω
i
2 +
m R
2
I
0
3.
ω
f
=
ω
i
1 +
m R
3
I
0
4.
ω
f
=
ω
i
1 +
m R
2
I
0
correct
Explanation:
The final rotational inertia of the turntable
plusblob is
I
f
=
I
0
+
I
blob
=
I
0
+
m R
2
Using conservation of angular momentum,
I
0
ω
i
=
I
f
ω
f
ω
f
=
I
0
ω
i
I
f
ω
f
=
I
0
ω
i
I
0
+
m R
2
=
ω
i
1 +
m R
2
I
0
004
(part 2 of 2) 10 points
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 Spring '08
 Turner
 Physics, Angular Momentum, Work, Correct Answer, kg, Overton, Mays

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