Environmental Engineering Reference
In-Depth Information
8.4 Work in a frame that rotates with the hoop. First obtain the equation of
motion for the bead, i.e. consider the torque about the centre of the hoop.
Coriolis force is cancelled by reaction from hoop (fixed
ω
) so net torque is
just
θ
where
θ
mω
2
R
2
sin
θ
cos
θ
mR
2
0 corresponds to
the bead at the bottom of the hoop. The bottom of the hoop corresponds
to
−
mgR
sin
θ
+
=
=
θ
¨
θ<
0forsmall(
θ >
0)
=
0 but to be a stable equilibrium we require
=
√
g/R
.If
ω>
the bead will
displacements. Thus
2
R
2
=
gR
hence
θ
ω
2
cos
θ
andsocos
θ
2
/ω
2
.
rise until
=
0, i.e.
g/R
=
=
PROBLEMS 9
M
9.1 Consider an element of the shell of mass d
M
4
π
sin
θ
d
θ
d
φ
. After inte-
grating over azimuth, the potential at a point a distance
x
from the centre of
the shell is
=
π
GM
2
sin
θ
y
=−
d
θ,
0
where
y
2
R
2
x
2
2
xR
cos
θ
. Now change variables to obtain an inte-
gral over
y
subject to
R
=
+
−
−
x<y<R
+
x
. After integration
=−
GM/R
,
which is constant.
9.2 When the body is at a distance
x
from the centre of the Earth, it feels a
force due only to the mass at smaller radii (the previous question proves that
there is no force from the mass at larger radii). Hence
G(x/R)
3
M/x
2
which corresponds to simple harmonic motion with angular frequency equal
to
(GM/R
3
)
1
/
2
.
9.3 Consider building up the sphere by adding successive shells brought in from
infinity. The total work done in adding a shell of thickness d
r
to a pre-existing
sphere of radius
r
and mass
Mr
3
/R
3
x
¨
=−
GM(r
3
/R
3
)
d
m/r
where d
m
=
M(
4
πr
2
d
r)/(
4
πR
3
/
3
)
is the mass of the shell. Integrating over 0
<r<R
gives the result. Equating to
is
−
1
2
Iω
2
with
I
2
5
MR
2
gives
ω
(
3
GM/R
3
)
1
/
2
.
=
=
10
15
m
2
s
−
1
since velocity is tangential at perihe-
lion. (ii) Kinetic energy divided by the mass is
v
0
/
2
9.4 (i)
L/m
=
v
0
R
=
4
.
46
×
10
9
Jkg
−
1
=
1
.
78
×
and the gravitational potential energy divided by the mass is
−
GM
/R
=
10
9
Jkg
−
1
. These are equal within errors and so we cannot tell if
the orbit is bound or unbound. (iii) Total energy is zero, i.e.
−
1
.
78
×
L
m
2
1
2
v
r
+
1
2
r
2
−
GM
r
0
=
,
where
v
r
is the radial component of the velocity when the comet is a dis-
tance
r
from the Sun. Putting the numbers in gives
v
r
=
29
.
8kms
−
1
.Toget
the speed we need also the tangential component which is, by the conserva-
tion of angular momentum,
(
7
.
48
/
15
)v
0
=
29
.
7kms
−
1
. Speed is therefore
42.1 km s
−
1
.
