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 12-11 gives
!
f
=
!
0
2
+
2
"
(
#
f
$
#
0
)
=
0
+
2
(
18 rev
!
60
=
min
2
)(
2 rev
)
=
65
.
7 rpm
,
and (b) Equation 12-10 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 12-1 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 11-5 (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 11-2 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 s
,
!
"
=
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 12-12 with
!
=
90
°
.)

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