Environmental Engineering Reference
In-Depth Information
the centripetal acceleration of the centre of mass. Therefore, the torque has
magnitude
mv
2
r
R
mgr
sin
φ
−
cos
φ
and it causes the angular momentum to precess about the vertical. Thus
the component of
L
in the horizontal plane,
L
h
, rotates with angular speed
=
v/R
.Using
1
2
mr
2
v
cos
φ
L
h
=
,
r
gives
=
mrv
2
cos
φ
2
R
d
L
d
t
L
h
=
.
Equating this to the torque gives the required result.
10.3 Similar to the example with the pencil but with different moments of inertia.
Remember that the sphere is pivoted on the finger. For the solid sphere
I
3
=
2
2
7
5
mr
2
(by the Parallel Axis Theorem).
This will give a minimum spin of 48 rad s
−
1
. For the spherical shell we have
I
3
=
5
mr
2
5
mr
2
mr
2
and
I
=
+
=
2
5
giving a minimum spin of 31 rad s
−
1
.
3
mr
2
3
mr
2
and
I
=
10.4 With
x
3
=
0 for the whole plate we calculate:
2
a
2
a
0
2
a
M
1
3
Ma
2
,
x
2
d
x
1
d
x
2
=
I
11
=
0
2
a
2
a
2
a
M
4
3
Ma
2
,
x
1
I
22
=
d
x
1
d
x
2
=
0
0
2
a
2
a
2
a
M
5
3
Ma
2
,
(x
1
+
x
2
)
d
x
1
d
x
2
=
I
33
=
0
0
2
a
2
a
2
a
M
1
2
Ma
2
.
I
12
=
(
−
x
1
x
2
)
d
x
1
d
x
2
=−
0
0
So
2
−
30
Ma
2
6
.
I
=
−
38 0
00 0
Ma
2
ω
6
√
2
The angular momentum,
L
=
I
ω
=
(
−
1
,
5
,
0
)
. We have constant
ω
Ma
2
ω
2
2
so we can compute the torque:
τ
=
ω
×
L
=
(
0
,
0
,
1
).
The kinetic
energy is
1
2
ω
·
1
6
Ma
2
ω
2
.
T
=
L
=
