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Solutions HW 10
S07
P112
1)
See textbook Figure 10.41.
Let the
z
axis be along the axis of rotation of the
wheel with the
z
+
direction pointing out of the page.
N
50
.
7
1
=
F
.
N
30
.
5
2
=
F
.
m
330
.
0
=
R
.
m
N
726
.
0
)
m
330
.
0
)(
N
50
.
7
(
)
m
330
.
0
)(
N
30
.
5
(
1
2
−
=
−
=
−
=
R
F
R
F
z
τ
.
2)
a)
a
rad
=
0,
a
tan
=
α
r
=
2
( )
0.300 m
( )
=
2
and so
a
=
2
.
b)
θ
=
π
3
rad, so
a
rad
=
ω
2
r
=
2 0.600 rad s
2
( )
3 rad
( )
0.300 m
( )
=
2
.
The tangential acceleration is still
0.180 m s
2
,
and so
a
=
2
( )
2
+
2
( )
2
=
0.418 m s
2
.
c)
For an angle of 120
°
a
rad
=
2
, and
a
=
2
, since
a
tan
is
still
0.180 m/s
2
.
3)
The moment of inertia can be calculated using the Parallel Axis Theorem:
I
=
I
COM
+
Md
2
, where I
COM
is the moment of inertia of the plate rotated about its center
of mass, and d is the distance from the COM to the new turning point. d
2
is given by
2(R/2)
2
, and so
I
=
1
6
MR
2
+
1
2
MR
2
=
2
3
MR
2
.
4) The magnitude of vector C is C=AB sin(140)=2.57.
The direction is in the
positive z direction.
6)
The angular momentum of a rotating body is given by L=I
ω
, where I is the moment of
inertia of a car tire (approximately a solid cylinder of mass 7kg and radius r= 0.3m, so
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View Full DocumentI = 0.5 (7 kg)(0.32 m)
2
= 0.3 kg m
2
).
ω
is the angular velocity on the highway, and in the
case of rolling without slipping:
ω
r =v.
Let’s estimate the linear velocity to about 80
km/h = 22m/s. The angular momentum is then approximately
L = (0.3 kg m
2
)(22 m/s)/(0.3 m) =
22 kg m
2
/s.
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 Spring '07
 LECLAIR,A
 mechanics

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