56.
REASONING AND SOLUTION
a. If the wheel does not slip, a point on the rim rotates about the axle with a speed
v
T
=
v
= 15.0 m/s
For a point on the rim
ω
=
v
T
/
r
= (15.0 m/s)/(0.330 m) =
45.5 rad/s
b.
v
T
=
r
ω
= (0.175 m)(45.5 rad/s) =
7.96 m/s
57.
REASONING
a.
The constant angular acceleration
α
of the wheel is defined by Equation 8.4 as the change
in the angular velocity,
ω

ω
0
, divided by the elapsed time
t
, or
(
)
0
/
t
α
ω
ω
=

. The time is
known. Since the wheel rotates in the positive direction, its angular velocity is the same as
its angular speed. However, the angular speed is related to the linear speed
v
of a wheel and
its radius
r
by
v
=
r
ω
(Equation 8.12). Thus,
/
v r
ω
=
, and we can write for the angular
acceleration that
0
0
0
v
v
v
v
r
r
t
t
r t
ω
ω
α



=
=
=
b.
The angular displacement
θ
of each wheel can be obtained from Equation 8.7 of the
equations of kinematics:
2
1
0
2
t
t
θ
ω
α
=
+
, where
ω
0
=
v
0
/
r
and
α
can be obtained as
discussed in part (a).
SOLUTION
a.
The angular acceleration of each wheel is
(
)(
)
2
0
2.1 m/s
6.6 m/s
1.4 rad/s
0.65 m
5.0 s
v
v
r t
α


=
=
= 
Chapter 8
Problems
423
b.
The angular displacement of each wheel is
(
)
(
)
(
)
2
2
0
1
1
0
2
2
2
2
1
2
6.6 m/s
5.0 s
1.4 rad/s
5.0 s
33 rad
0.65 m
v
t
t
t
t
r
θ
ω
α
α
=
+
=
+
=
+

= +
______________________________________________________________________________
58.
REASONING
For a wheel that rolls without slipping, the relationship between its linear
speed
v
and its angular speed
ω
is given by Equation 8.12 as
v
=
r
ω
, where
r
is the radius of
a wheel.
For a wheel that rolls without slipping, the relationship between the magnitude
a
of its linear
acceleration and the magnitude
α
of the angular acceleration is given by Equation 8.13 as
a
=
r
α
,
where
r
is the radius of a wheel.
The linear acceleration can be obtained using the
equations of kinematics for linear motion, in particular, Equation 2.9.
SOLUTION
a.
From Equation 8.12 we have that
(
)(
)
0.320 m
288 rad /s
92.2 m/s
v
r
ω
=
=
=
b.
The magnitude of the angular acceleration is given by Equation 8.13 as
α
=
a/r
.
The
linear acceleration
a
is related to the initial and final linear speeds and the displacement
x
by
Equation 2.9 from the equations of kinematics for linear motion;
2
2
0
2
v
v
a
x

=
.
Thus, the
magnitude of the angular acceleration is
(
)
(
)
(
)
(
)
(
)(
)
2
2
0
2
2
2
2
2
0
/ 2
92.2 m/s
0 m/s
34.6 rad /s
2
2 384 m
0.320 m
v
v
x
a
r
r
v
v
x r
α

=
=


=
=
=
59.
REASONING AND SOLUTION
a.
If the rope is not slipping on the cylinder, then the tangential speed of the teeth on the
larger gear
(gear 1) is 2.50 m/s. The angular speed of gear 1 is then
ω
1
=
v
/
r
1
= (2.50 m/s)/(0.300 m) =
8.33 rad/s
424
ROTATIONAL KINEMATICS
The direction of the larger gear is
counterclockwise
.
b. The gears are in contact and do not slip.
This requires that the teeth on both gears
move
with the same tangential speed.
v
T1
=
v
T2
or
ω
1
r
1
=
ω
2
r
2
So
(
)
1
2
1
2
0.300 m
8.33 rad/s
14.7 rad/s
0.170 m
r
r
ω
ω
=
=
=
The direction of the smaller gear is
clockwise
.
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 Spring '16
 Ahmed
 Physics, Homework Key