yindeemark (rry82) – Homework 11 – chelikowsky – (59005)
1
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001
10.0 points
A rod can pivot at one end and is free to
rotate without friction about a vertical axis,
as shown.
A force
vector
F
is applied at the other
end, at an angle
θ
to the rod.
L
m
F
θ
If
vector
F
were to be applied perpendicular to
the rod, at what distance
d
from the axis
of rotation should it be applied in order to
produce the same torque
vector
τ
?
1.
d
=
L
cos
θ
2.
d
=
L
sin
θ
correct
3.
d
=
L
4.
d
=
√
2
L
5.
d
=
L
tan
θ
Explanation:
The torque the force generates is
τ
=
F L
sin
θ .
Thus the distance is
L
sin
θ
.
002
10.0 points
Consider two masses of 3
.
2 kg and 7
.
8 kg
connected by a string passing over a pulley
having a moment of inertia 8
.
6 g
·
m
2
about
its axis of rotation, as in the figure below.
The string does not slip on the pulley, and the
system is released from rest.
The radius of
the pulley is 0
.
24 m.
39 cm
0
.
24 m
8
.
6 g
·
m
2
ω
7
.
8 kg
3
.
2 kg
Find the linear speed of the masses after
the 7
.
8 kg mass descends through a distance
39 cm.
Assume mechanical energy is con
served during the motion. The acceleration of
gravity is 9
.
8 m
/
s
2
.
Correct answer: 1
.
77589 m
/
s.
Explanation:
Let :
I
= 8
.
6 g
·
m
2
= 0
.
0086 kg
·
m
2
,
R
= 0
.
24 m
,
m
1
= 3
.
2 kg
,
m
2
= 7
.
8 kg
,
and
h
= 39 cm = 0
.
39 m
.
Consider the free body diagrams
7
.
8 kg
3
.
2 kg
T
2
T
1
m
2
g
m
1
g
a
a
If we neglect friction in the system, the me
chanical energy is constant and the increase
in kinetic energy equals the decrease in po
tential energy.
Since
K
i
= 0 (the system is
initially at rest), the masses have a common
speed and
v
=
R ω ,
Δ
K
=
K
f

K
i
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yindeemark (rry82) – Homework 11 – chelikowsky – (59005)
2
=
1
2
m
1
v
2
+
1
2
m
2
v
2
+
1
2
I ω
2
,
=
1
2
m
1
v
2
+
1
2
m
2
v
2
+
1
2
I
parenleftBig
v
R
parenrightBig
2
=
1
2
parenleftbigg
m
1
+
m
2
+
I
R
2
parenrightbigg
v
2
.
From the figure, we see that
m
2
loses potential
energy while
m
1
gains potential energy;
i.e.
,
Δ
U
2
=

m
2
g h
and Δ
U
1
=
m
1
g h
. Applying
the principle of conservation of energy in the
form
Δ
K
+ Δ
U
1
+ Δ
U
2
= 0
1
2
parenleftbigg
m
1
+
m
2
+
I
R
2
parenrightbigg
v
2
=
m
2
g h

m
1
g h .
v
=
radicaltp
radicalvertex
radicalvertex
radicalbt
2 (
m
2

m
1
)
g h
m
1
+
m
2
+
I
R
2
.
=
radicaltp
radicalvertex
radicalvertex
radicalvertex
radicalbt
2 (7
.
8 kg

3
.
2 kg) (9
.
8 m
/
s
2
)
3
.
2 kg + 7
.
8 kg +
0
.
0086 kg
·
m
2
(0
.
24 m)
2
×
√
0
.
39 m
=
1
.
77589 m
/
s
.
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 Fall '08
 Turner
 Energy, Work, Correct Answer, kg, τ

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