Classical Mechanics I
A particle is launched from afar toward a repulsive Coulomb scattering center (potential
energy U = α/r, α > 0), with initial energy E and impact parameter b. Find
a) (10 points) the nearest approach distance rmin ,
b) (10 points) the magnitude of particle’s speed in the nearest approach point.
1
Classical Mechanics II
A uniform solid semisphere of radius R and mass m is placed on a horizontal surface with
finite (non-zero) friction coefficient.
a) (3 points) Find the center of mass of the hemisphere.
b) (7 points) Find the moment of inertia of the hemisphere with respect to an axis PX
lying on the horizontal surface (see Figure).
c) (10 points) Find the period of small oscillations of the semisphere.
Y
P
2
X
Classical Mechanics III
A ball is bouncing vertically and perfectly elastically in a standing elevator. The maximal
height of bouncing motion is h0 . The upward acceleration of the elevator is changing very
slowly from 0 to g/2. Find the new maximal height of bouncing motion.
3
Solution I: Kepler problem (Kostya Likharev)
a) The point of the nearest approach may be found from the equation Ueff (rmin ) = E, where
Ueff (r) = U (r) +
L2z
α Eb2
+ 2 ,
=
2mr2
r
r
since L2z = (mvi b)2 = 2mEb2 . Solving the resulting quadratic equation, we get
rmin
α
+
=
2E
s
α
2E
2
+ b2 > b.
(The negative sign before the square root would give a non-physical, negative value for rmin .)
b) The simplest way to find the velocity vx in the nearest approach point is to realize that its vector
√ is perpendicular to the radius-vector and hence Lz = mvx rmin . Since
Lz = const = b 2mE, we get
s
vx =
2E b
b
= vi
< vi .
m rmin
rmin
4
Solution II: Semisphere (Sasha Abanov)
a) The density of the semisphere is ρ = m/( 23 πR3 ). The center of mass of the semisphere is
obviously on the axis of symmetry of the semisphere. Introducing the vertical coordinate z
directed from the center of the semisphere downwards we have:
O
R
z cm
z
r
P
Z
zcm =
1 ZR
3 ZR
3
2
2
zρπ(R2 − z 2 ) dz =
zπ(R
−
z
)
dz
=
R.
m 0
2R3 0
8
3
zcm = R.
8
b) The moment of inertia of the solid ball with respect to its axis of symmetry is 52 mR2 .
Therefore, the moment of inertia of the semisphere with respect to the axis going through
2
. We have
the point O and orthogonal to the axis of symmetry is IO = 52 mR2 = Icm + mzcm
2
2
2
Icm = 5 mR − mzcm . On the other hand the moment of inertial relative to the instantaneous
2
axis of rotation P Y is IP Y = Icm + m(R − zcm )2 = 25 mR2 − mzcm
+ m(R − zcm )2 = 13
mR2 .
20
IP Y =
13
mR2 .
20
c) If φ is the angular deviation of the semisphere from its equilibrium
position
we have
q
q
I
26 R
I φ̈ = −mgzcm cos φ and the period of small oscillations T = 2π mgzcm = 2π 15 g .
s
T = 2π
5
26 R
.
15 g
Solution III: Bouncing ball (Sasha Abanov)
2
p
The trajectory of the q
bouncing ball in the phase space is given by parabola 2m
+ mgx =
E = mgh or by p = m 2g(h − x) and by the straight line x = 0. The adiabatic invariant is
the area of the phase space bounded by that trajectory
√
Z
Z h q
4 2 1/2 3/2
I = p dx = 2
m 2g(h − x) dx =
mg h .
3
0
p
h
x
In the moving elevator g → gef f = a + g and we have
I = (a + g)1/2 h3/2 = const.
3/2
Comparing adiabatic invariant I with its initial value we obtain (a + g)1/2 h3/2 = g 1/2 h0
and
!1
3
g
.
h = h0
g+a
We substitute a = g/2 and obtain the final result
1
h = h0
6
2
3
3
.