Phys 2101
Homework 8 Solution Spring `11
These solutions use the parameter values from the problems printed in the book, not those that
appear in your personal homework assignment. The logic used to get to the answer is the same,
however.
1. We note that the component of
v
perpendicular to
r
has magnitude
v
sin
where
= 30°. A
similar observation applies to
F
.
(a) Eq. 1120 leads to
2
3.0 m
2.0 kg
4.0 m/s sin30
12 kg m
s.
rmv
(b) Using the righthand rule for vector products, we find
r
p
points out of the page, or along the
+
z
axis, perpendicular to the plane of the figure.
(c) Similarly, Eq. 1038 leads to
2
sin
3.0 m
2.0 N sin 30
3.0N m.
rF
(d) Using the righthand rule for vector products, we find
r
F
is also out of the page, or along the
+
z
axis, perpendicular to the plane of the figure.
2. (a) The acceleration vector is obtained by dividing the force vector by the (scalar) mass:
a
=
F
/
m
= (3.00 m/s
2
)i
^
–
(4.00 m/s
2
)j
^
+ (2.00 m/s
2
)k
^
.
(b) Use of Eq. 1118 leads directly to
L
=
(42.0 kg
.
m
2
/s)i
^
+ (24.0 kg
.
m
2
/s)j
^
+ (60.0 kg
.
m
2
/s)k
^
.
(c) Similarly, the torque is
r
F
= (
–
8.00
N m
)i
^
–
(26.0
N m
)j
^
–
(40.0
N m
)k
^
.
(d) We note (using the Pythagorean theorem) that the magnitude of the velocity vector is 7.35 m/s
and that of the force is 10.8 N.
The dot product of these two vectors is
v
.
F
=
–
48
(in SI units)
.
Thus, Eq. 320 yields
= cos
1
[
48.0/(7.35
10.8)] = 127
.
3. For the 3.1 kg particle, Eq. 1121 yields
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2
1
1
1
2.8 m
3.1kg
3.6 m/s
31.2 kg m
s.
r mv
Using the righthand rule for vector products, we find this
)
(
1
1
p
r
is out of the page, or along the
+
z
axis, perpendicular to the plane of Fig. 1141. And for the 6.5 kg particle, we find
2
2
2
2
1.5 m
6.5 kg
2.2 m/s
21.4 kg m
s.
r mv
And we use the righthand rule again, finding that this
)
(
2
2
p
r
is into the page, or in the
–
z
direction.
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 Spring '07
 GROUPTEST
 Angular Momentum, Kinetic Energy, Momentum, Work, Angular velocity, li, kg

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