hernandez (ejh742) – oldhomework 20 – Turner – (58120)
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001
(part 1 of 2) 10.0 points
Given:
A uniform flexible chain whose mass
is 5
.
4 kg and length is 7 m. A table whose top
is frictionless.
Initially you are holding the chain at rest
and onehalf of the length of the chain is hung
over the edge of the table. When you let loose
of the chain it falls downward.
The acceleration of gravity is 9
.
8 m
/
s
2
.
a
7 m
4
.
27 m
Radius of table
is negligible
compared to the
length of chain
Mass of chain
is 5
.
4 kg
.
Find the acceleration
a
of the chain when
the length of the chain hanging vertically is
4
.
27 m
.
Correct answer: 5
.
978 m
/
s
2
.
Explanation:
Note:
The initial condition does not enter
into the consideration for the acceleration.
Let :
g
= 9
.
8 m
/
s
2
,
L
= 7 m
,
ℓ
= 4
.
27 m
,
and
m
= 5
.
4 kg
.
The linear density of the chain is
λ
=
m
L
=
5
.
4 kg
7 m
= 0
.
771429 kg
/
m
.
F
=
ℓ λ g
cm
The free body diagram in the vertical di
rection gives
summationdisplay
F
y
=
ℓ λ g
=
L λ a .
Therefore
a
=
ℓ
L
g
(1)
=
4
.
27 m
7 m
(9
.
8 m
/
s
2
)
= 5
.
978 m
/
s
2
.
002
(part 2 of 2) 10.0 points
Find the magnitude of the velocity of the of
the chain when 4
.
27 m of the chain is hanging
vertically.
Correct answer: 2
.
89414 m
/
s.
Explanation:
The change in kinetic energy is
Δ
K
=
1
2
m v
2
=
1
2
λ L v
2
.
(2)
Let
ℓ
i
=
L
2
and
ℓ
f
=
ℓ
.
Using the table top as the origin of the
y
coordinate and down as the positive
y
di
rection
y
cm
=
m
on table
parenleftBig
0
parenrightBig
+
m
hanging
parenleftbigg
ℓ
2
parenrightbigg
m
on table
+
m
hanging
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hernandez (ejh742) – oldhomework 20 – Turner – (58120)
2
y
cm
i
=
parenleftbigg
L

L
2
parenrightbigg
λ
0 +
L
2
λ
parenleftbigg
L
4
parenrightbigg
λ L
y
cm
f
=
(
L

ℓ
)
λ
0 +
ℓ λ
parenleftbigg
ℓ
2
parenrightbigg
λ L
The vertical center of mass difference Δ
y
cm
is
Δ
y
cm
=
y
cm
f

y
cm
i
=
λ ℓ
ℓ
2

λ
L
2
L
4
λ L
=
1
8
L
[4
ℓ
2

L
2
]
.
(3)
The change in potential energy is
Δ
U
=
λ L g
Δ
y
cm
=
1
8
λ g
[4
ℓ
2

L
2
]
.
(4)
From conservation of energy Δ
K
= Δ
U
, Eq.
2 and Eq. 4, we have
1
2
λ L v
2
=
1
8
λ g
(4
ℓ
2

L
2
)
v
2
=
g
4
L
[4
ℓ
2

L
2
]
.
(5)
Therefore
v
=
radicalbigg
g
4
L
[4
ℓ
2

L
2
]
(6)
=
radicalBigg
(9
.
8 m
/
s
2
)
4 (7 m)
[4 (4
.
27 m)
2

(7 m)
2
]
= 2
.
89414 m
/
s
.
Alternative:
You can use the kinematic
expression and remember that the center of
mass accelerates at
g
[not Eq. 1], since the
acceleration of gravity is not a function of
mass.
v
2
= 2
a
Δ
y
cm
= 2
g
1
8
L
[4
ℓ
2

L
2
]
=
g
4
L
[4
ℓ
2

L
2
]
,
(7)
where Δ
y
cm
is obtained from Eq. 3.
Note:
Equation 7 is identical to Eq. 5.
003
10.0 points
Two particles of masses
m
and 3
m
are
moving toward each other along the
x
axis
with the same speed
v
. They undergo a head
on elastic collision and rebound along the
x

axis.
m
v
3
m
v
Determine the final speed of the heavier
object.
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 Spring '08
 Turner
 Physics, Energy, Kinetic Energy, Momentum, Work, Correct Answer

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