PHY4222: CLASSICAL MECHANICS
TEST II
April 7, 2008
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. On the other hand,
an answer obtained just by the dimensional analysis may give you up to 25% of the credit.
2
1. A rocket ship of mass M0 loaded with fuel of mass m0 takes off vertically in a uniform gravitational field with
zero initial velocity. The engine ejects fuel with speed u0 with respect to the ship. It is not known how the mass
of the fuel depends on time but it is known that the fuel is completely ejected in time t0 after the start.
• a) Find the equation of motion of the ship.
Suppose that −dM > 0 is the mass of fuel ejected over time dt. Momentum conservation gives
(M + dM )(v + dv) + (v − u0 )(−dM ) − M v = −M gdt
or
M
dM
dv
= −u0
− Mg
dt
dt
b) What is the speed of the ship at time t0 , when all the fuel has been ejected?
Re-write the equation of motion as
dv = −u0
dM
− gdt
M
and integrate
v = −u0 ln M − gt + C,
where C is a constant. C is determined from the initial condition v = 0 at t = 0, or
C = u0 ln (M0 + m0 )
v = u0 ln
M0 + m 0
− gt.
M
At t = t0 , M = M0 . The speed of the ship at this time is
v = u0 ln
M0 + m 0
− gt0 .
M0
c) Take-off happens if v > 0 anytime in between 0 and t0 . The condition is
gt0
m > M0 exp
−1 .
u0
2. a) Find the tensor of inertia for a hollow cone of mass M, radius R, and height h about its apex.
Suppose that the material the cone is made has area mass density σ. The mass of the cone is then
Z
M = σ d2 r,
where the integral goes over the surface of the cone. Consider a thin strip of width dz/ cos α at height z from
the apex. The area of the strip is 2πz(tan α/ cos α)dz,
√
where tan α = h/R and cos α = h/ R2 + h2 . Then,
Z h
p
tan α
tan α 2
σ
dzz = π
σh = πR R2 + h2 σ.
cos α
cos α
0
R
R
0
0
Components I10 = I11
= σ d2 r y 2 + z 2 and I20 = I22
= σ d2 r x2 + z 2 are the same by symmetry. Adding
them up and dividing by two,
Z
σ
0
0
I1 = I 2 =
d2 r r2 + 2z 2 .
2
M = 2π
3
For any point of the surface of the cone, r = z tan α. Then,
Z
Z h
h
1 tan α
π tan ασ
tan α
tan2 α + 2 h4
2π
σ
σ tan2 α + 2
dzz 3 =
dzz z 2 tan2 α + 2z 2 = π
2 cos α
cos α
4 cos α
0
0
2
1
1
R
M
= M h2 tan2 α + 2 = M h2
+2 =
R2 + 2h2 .
4
4
h2
4
I10 =
The remaining component
0
I30 = I33
=σ
Z
d2 rr2 = σ2π tan α (tan α)
2
Z
h
dzz 3 =
0
1
1
1
2
2
σπ tan α (tan α) h4 = M h2 (tan α) = M R2 .
2
2
2
b) Suppose that the cone is suspended at its apex as shown in the Figure. Find the frequency of small oscillations
about the vertical.
The Lagrangian
L=
I10 2
θ̇ + M gzc.m. cos θ,
2
where zc.m. is the position of the center of mass with respect to the apex
zc.m.
R
R 2
α h
2
2π tan
d rz
2
cos α 0 dzz
R
= h.
=
=
tan α 2
3
d2 r
π cos α h
Expanding cos θ in L, dropping a constant, and using the result for zc.m. , we obtain
L=
I10 2 1 2 2
θ̇ − θ M gh.
2
2 3
Equation of motion
2
I10 θ̈ = − M ghθ
3
from which we find
ω2 =
(2/3) M gh
=
I10
(2/3) M gh
gh
8
.
=
2
2
2
3 R + 2h2
(R + 2h )
M
4
4
axis of rotation
h
R
FIG. 1: