hernandez (ejh742) – homework 23 – Turner – (58120)
1
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001
10.0 points
Two pulley wheels, of respective radii
R
1
=
0
.
37 m and
R
2
= 1
.
5 m are mounted rigidly
on a common axle and clamped together. The
combined moment of inertia of the two wheels
is
I
+ 1
.
5 kg m
2
.
Mass
m
1
= 6
.
3 kg is attached to a cord
wrapped around the first wheel, and another
mass
m
2
= 2
.
5 kg is attached to another cord
wrapped around the second wheel:
1
2
m
R
1
m
R
2
Find the angular acceleration of the system.
Take clockwise direction as positive.
The
acceleration of gravity is 9
.
8 m
/
s
2
.
Correct answer: 1
.
741 rad
/
s
2
.
Explanation:
1
2
m
g
R
1
m
g
R
2
T
1
2
T
2
T
T
1
Let’s assume
m
2
moves downward and take
motion downward as positive for
m
2
and mo
tion upward as positive for
m
1
, take clockwise
as positive for the pulley wheels.
Since the
two masses are connected to the pulleys, their
accelerations are
a
1
=
α R
1
and
a
2
=
α R
2
.
Applying Newton’s second law to
m
1
,
m
2
,
and the pulleys separately, we obtain
T
1

m
1
g
=
m
1
a
1
=
m
1
α R
1
(1)
m
2
g

T
2
=
m
2
a
2
=
m
2
α R
2
(2)
τ
net
=
T
2
R
2

T
1
R
1
=
I α .
(3)
Solving these equations, we find
α
=
m
2
g R
2

m
1
g R
1
I
+
m
1
R
2
1
+
m
2
R
2
2
=
1
.
741 rad
/
s
2
.
The fact that
α >
0 indicates that our pre
vious assumption that
m
2
moves downward
is right.
If
α <
0, our assumption was not
correct and
m
2
would move upward, but the
equations obtained to calculate
T
1
and
T
2
would be still correct and we need not reas
sume the direction of the motion and recalcu
late.
002
10.0 points
A bicycle wheel has a diameter of 62
.
9 cm
and a mass of 1
.
68 kg. The bicycle is placed
on a stationary stand and a resistive force of
115 N is applied tangent to the rim of the tire.
Assume that the wheel is a hoop with all of
the mass concentrated on the outside radius.
In order to give the wheel an acceleration
of 3
.
15 rad
/
s
2
, what force must be applied
by a chain passing over a 3
.
5 cm diameter
sprocket?
Correct answer: 2096
.
62 N.
Explanation:
Let :
r
w
= 31
.
45 cm = 0
.
3145 m
,
M
= 1
.
68 kg
,
F
f
= 115 N
,
r
s
= 1
.
75 m = 0
.
0175 m
,
and
α
= 3
.
15 rad
/
s
2
.
For a wheel with mass concentrated on the
rim,
I
=
M R
2
. Applying rotational equilib
rium
summationdisplay
τ
=
I α
F r

F
f
R
=
I α
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hernandez (ejh742) – homework 23 – Turner – (58120)
2
F
=
F
f
R
+
I α
r
=
F
f
R
+
M R
2
α
r
=
(115 N)(0
.
3145 m)
0
.
0175 m
+
(1
.
68 kg)(0
.
3145 m)
2
(3
.
15 rad
/
s
2
)
0
.
0175 m
=
2096
.
62 N
.
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 Spring '08
 Turner
 Physics, Acceleration, Angular Momentum, Work, Moment Of Inertia, Rotation

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