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CHAPTER 12 SOLUTIONS
Problem
5. A wheel turns through 2.0 revolutions while being accelerated from rest at
18 rpm/s.
(a) What is the final angular speed?
(b) How long does it take to turn the 2.0 revolutions?
Solution
For constant angular acceleration, (a) Equation 1211 gives
!
f
=
0
2
+
2
"
(
#
f
$
0
)
=
+
2
(
18 rev
!
60
=
min
2
)(
2 rev
)
=
65
.
7 rpm
,
and (b) Equation 1210 gives
f
"
0
=
0
+
1
2
t
2
,
or
t
=
2
(
2 rev
)
=
(
18 rev
=
60 s
2
)
=
3
.
65 s
.
Problem
6. You switch a food blender from its high to its low setting; the blade speed drops from
3600 rpm
to
1800 rpm
in
1.4 s.
How many revolutions does it make during this time?
Solution
At constant angular acceleration, Equations 121 and 8 give
!
=
1
2
(
0
+
f
)
!
t
=
1
2
(3600
+
1800)(rev
=
60 s)(1.4 s)
=
63.0 rev.
Problem
15. The angular acceleration of a wheel in rad/s
2
is given by
24
t
2
!
16
t
3
,
where
t
is the time in seconds. The wheel starts
from rest at
t
=
0.
(a) When is it again at rest? (b) How many revolutions has it turned between
t
=
0
and when it is
again at rest?
Solution
(a) Integrating Equation 115 (with
0
=
0
),
we find
(
t
)
=
"
0
t
(
$
t
)
d
$
t
=
"
0
t
(
24
$
t
2
%
16
$
t
3
)
d
$
t
=
8
t
3
%
4
t
4
.
The wheel
is at rest when
(
t
)
=
(
8
"
4
t
)
t
3
=
0
,
or at
t
=
0
and
t
=
2 s
.
(b) Integrating Equation 112 for the angular displacement
(in radians), we find
(
t
)
"
0
=
#
0
t
$
(
%
t
)
d
%
t
=
#
0
t
(
8
%
t
3
"
4
%
t
4
)
d
%
t
=
2
t
4
"
4
t
5
=
5
.
For
t
=
,
!
=
2
5
(
1
#
4
=
5
)
=
6.4 radians
=
1.02 rev.
Problem
17. A torque of
110 N
!
m
is required to start a revolving door rotating. If a child can push with a maximum force of
90 N,
how far from the door’s rotation axis must she apply this force?
Solution
If the force is applied perpendicular to the door, the radial distance should be
r
=
=
F
=
110 N
"
m
=
90 N
=
1.22 m from the
axis. (See Equation 1212 with
=
90
°
.)
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View Full DocumentCHAPTER 12
279
Problem
23. A 1.5mdiameter wheel is mounted on an axle through its center. (a) Find the net torque about the axle due to the forces
shown in Fig. 1242. (b) Are there any forces that don’t contribute to the torque?
Solution
The displacements from the center of the wheel to the points of application of the forces, and the angles between these
displacements and the forces, are shown on Fig. 1242. (a) If we take positive torques counterclockwise, the net torque (sum
of Equation 1212) is
!
net
=
"
r
i
F
i
sin
#
i
=
r
1
F
1
sin0
° +
(0.25 m)(3.0 N)sin45
°
$
(0.50 m)(2.0 N)sin90
° +
(0.75 m)
%
(1.8 N)sin60
° +
0
!
F
5
=
0.699 N
"
m.
(b) Evidently,
F
1
and
F
5
, whose lines of action pass through the center of the wheel,
produce no torque about the center.
FIGURE 1242 Problem 23 Solution.
Problem
28. The full diameter of a wheel is 92 cm, and its rotational inertia is
7.8 kg
!
m
2
.
(a) What is the minimum mass it could
have? (b) How could it have more mass?
Solution
(a) Any part of the wheel has a distance from the center less than or equal to the maximum radius, so
I
=
!
m
i
r
i
2
"
(
!
m
i
)
r
max
2
.
Thus
M
=
!
m
i
"
I
=
r
max
2
=
M
min
=
(
7
.
8 kg
#
m
2
)
=
(
1
2
$
0
.
92 m
)
2
=
36
.
9 kg
.
(b) If not all the mass of the wheel
is concentrated at the rim, the total mass is greater than this minimum.
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 Winter '07
 Hicks
 Physics

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