PHY4222/Spring 2008: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #7: SOLUTIONS
due by 12:50 p.m. Wed 03/05
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without
such a check, no more than 75\% of the credit will be given even for an otherwise
correct solution.
10.6
z
z = f(r)
m
r
Consider a small mass m on the surface of the water. From Eq. (10.25)
&& − mω& × r − mω × ( ω × r ) − 2mω × v
Feff = F − mR
f
r
In the rotating frame, the mass is at rest; thus, Feff = 0 . The force F will consist of gravity
and the force due to the pressure gradient, which is normal to the surface in equilibrium.
&& = ω& = v = 0
R
r
Since f
, we now have
0 = mg + Fp − mω × ( ω × r )
where
Fp
is due to the pressure gradient.
Fp
θ
mω2r
θ′
mg
Since Feff = 0 , the sum of the gravitational and centrifugal forces must also be normal to
the surface.
Thus θ ′ = θ.
tan θ ′ = tan θ =
ω 2r
g
but
tan θ =
dz
dr
Thus
z=
ω2
2g
r 2 + constant
The shape is a circular paraboloid.
10.7
For a spherical Earth, the difference in the gravitational field strength between the poles
and the equator is only the centrifugal term:
g poles − g equator = ω 2 R
For ω = 7.3 × 10 −5 rad ⋅ s −1 and R = 6370 km, this difference is only 34 mm ⋅ s −2 . The
disagreement with the true result can be explained by the fact that the Earth is really an
oblate spheroid, another consequence of rotation. To qualitatively describe this effect,
approximate the real Earth as a somewhat smaller sphere with a massive belt about the
equator. It can be shown with more detailed analysis that the belt pulls inward at the
poles more than it does at the equator.
10.9
Choosing the same coordinate system as in Example 10.3 of Thornton and Marion (see
Fig. 10-9), we see that the lateral deflection of the projectile is in the x direction and that
the acceleration is
ax = &&
x = 2 ω z vy = 2 (ω sin λ )(V0 cos α ) .
(1)
Integrating this expression twice and using the initial conditions, x& ( 0 ) = 0 and x ( 0 ) = 0 ,
we obtain
x ( t ) = ω V0 t 2 cos α sin λ
(2)
Now, we treat the z motion of the projectile as if it were undisturbed by the Coriolis
force. In this approximation, we have
z ( t ) = V0 t sin α −
1 2
gt ,
2
from which the time T of impact is obtained by setting z = 0:
(3)
T=
2V0 sin α
.
g
(4)
Substituting this value for T into (2), we find the lateral deflection at impact to be
x (T ) =
4ωV03
sin λ cos α sin 2 α
2
g
Bonus 10.17
Due to the centrifugal force, the water surface of the lake is not exactly perpendicular to
the Earth’s radius (see figure).
B
ge
m
g
β
Ta
n
nt
Wa
ter
sur
β
to
fac
rth
A
α
C
e
Ea
su
rfa
ce
Forces acting an a small element of water of mass m (gravity, centrifugal, and the
pressure gradient) form the triangle ABC. The length BC is (using cosine theorem)
BC =
AC 2 + (mg ) 2 − 2 ACmg cos α
where AC is the centrifugal force AC = mω 2 R cos α with α = 47° and Earth’s radius
R ≅ 6400 km ,
Using the sine theorem, the angle β that the water surface is deviated from the direction
tangential to the Earth’s surface is found from
BC
AC
=
sin α sin β
⇒ sin β =
AC sin α
= 4.3 × 10−5
BC
So the distance the lake falls at its center is h = r sin β where r = 162 km is the lake’s
radius.
So finally we find h = 7 m.
(5)