Chapter 11 – Student Solutions Manual
5. By Eq. 1052, the work required to stop the hoop is the negative of the initial kinetic
energy of the hoop. The initial kinetic energy is
K
I
mv
=
+
1
2
2
1
2
2
ω
(Eq. 115), where
I =
mR
2
is its rotational inertia about the center of mass,
m
= 140 kg, and
v
= 0.150 m/s is the
speed of its center of mass. Eq. 112 relates the angular speed to the speed of the center of
mass:
ω
=
v
/
R
. Thus,
(
)(
2
2
2
2
2
2
1
1
140 kg
0.150 m/s
2
2
v
K
mR
mv
mv
R
⎛
⎞
=
+
=
=
⎜
⎟
⎝
⎠
)
,
which implies that the work required is – 3.15 J.
23. If we write
then (using Eq. 330) we find
G
r
x
y
z
=
+
+
±
±
±
i
j
k
G
G
r
F
×
is equal to
yF
zF
zF
xF
xF
yF
z
y
x
z
y
x
−
+
−
+
−
d
i
b
g
d
i
±
±
i
j
±
k.
(a) Plugging in, we find
(
)(
)
(
)(
)
ˆ
ˆ
3.0m
6.0N
4.0m
8.0N
k
(50N m) k.
τ
=
−
−
=
⎡
⎤
⎣
⎦
G
⋅
(b) We use Eq. 327,
where
φ
is the angle between


sin
,
G
G
r
F
rF
×
=
φ
G
r
and
. Now
G
F
r
x
y
=
+
=
2
2
50
.
m
and
F
F
F
x
y
=
+
=
2
2
10 N. Thus,
rF
=
=
50
10
50
.
m
N
N m,
b
gb
g
⋅
the same as the magnitude of the vector product calculated in part (a). This implies sin
φ
= 1 and
φ
= 90°.
29. (a) We use
where
G
A
G
G
=
×
mr
v
,
G
r
is the position vector of the object,
G
v
is its velocity
vector, and
m
is its mass. Only the
x
and
z
components of the position and velocity
vectors are nonzero, so Eq. 330 leads to
G
G
r
v
xv
zv
z
z
×
= −
+
b
±
j.
g
Therefore,
(
)
(
)
(
)
(
)
(
)
(
)
(
)
ˆ
ˆ
j
0.25 kg
2.0 m
5.0 m s
2.0 m
5.0 m s
j
0.
z
x
m
xv
zv
=
−
+
=
−
+ −
−
=
G
A
(b) If we write
then (using Eq. 330) we find
G
r
x
y
z
=
+
+
±
±
±
i
j
k,
G
G
r
F
×
is equal to
yF
zF
zF
xF
xF
yF
z
y
x
z
y
x
−
+
−
+
−
d
i
b
g
d
i
±
±
.
i
j
±
k
With
x
= 2.0,
z
= –2.0,
F
y
= 4.0 and all other components zero (and SI units understood)
the expression above yields
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document