CM: Classical Mechanics
problem. Indeed, consider two particles that interact via a conservative central force F21 = –F12 = n rF( r), where r and n r are, respectively, the magnitude and the direction of the distance vector r r1 – r2
connecting the two particles (Fig. 3).
m 1
F
12
r
r r - r
1
2
1
F21
r
2
Fig. 3.3. Vectors in the planetary problem.
m 2
0
Generally, two particles moving without constraints in 3D space, have 3 + 3 = 6 degrees of freedom, which may be described, e.g., by their Cartesian coordinates { x 1, y 1, z 1, x 2, y 2, z 2} However, for this particular form of interaction, the following series of tricks allows the number of essential degrees of freedom to be reduced to just one.
First, the conservative force of particle interaction may be described by a time-independent potential energy U( r), such that F( r) = – U( r)/ r.11 Hence the Lagrangian function of the system is m
m
L T U r
1
2
2
2
r
r U ( r).
(3.30)
2 1
2 2
Let us perform the transfer from the initial six scalar coordinates of the particles to the following six generalized coordinates: three Cartesian components of the distance vector
r r1 – r2,
(3.31)
and three scalar components of the following vector:
m r m
1 1
2 2
R
r , with M m m ,
(3.32) Center of
1
2
M
mass
which defines the position of the center of mass of the system, with the total mass M. Solving the system of two linear equations (31) and (32) for r1 and r2, we get
m
m
2
r R
r,
1
r R
.
r
(3.33)
1
2
M
M
Plugging these relations into Eq. (30), we see that it is reduced it to
M
m
2
2
L
R
r U ( r),
(3.34)
2
2
where m is the so-called reduced mass:
m m
1
1
1
1
2
m
,
that
so
.
(3.35) Reduced
M
m
m
m
mass
1
2
10 This name is very conditional, because this group of problems includes, for example, charged particle scattering (see Sec. 3.7 below).
11 See, e.g., MA Eq. (10.8) with / = / = 0.
Chapter 3
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Note that according to Eq. (35), the reduced mass is lower than that of the lightest component of the two-body system. If one of m 1,2 is much less than its counterpart (like it is in most star-planet or planet-satellite systems), then with a good precision m min [ m 1, m 2].
Since the Lagrangian function (34) depends only on R rather than R itself, according to our discussion in Sec. 2.4, all Cartesian components of R are cyclic coordinates, and the corresponding generalized momenta are conserved:
L
P
R
M const,
j
3.
2,
1,
(3.36)
j
R
j
j
Physically, this is just the conservation law for the full momentum P MR of our system, due to the absence of external forces. Actually, in the axiomatics used in Sec. 1.3 this law is postulated – see Eq.
(1.10) – but now we may attribute the momentum P to a certain geometric point, with the center-of-mass radius vector R. In particular, since according to Eq. (36) the center moves with a constant velocity in the inertial reference frame used to write Eq. (30), we may consider a new inertial frame with the origin at point R. In this new frame, R 0, so that the vector r (and hence the scalar r) remain the same as in the old frame (because the frame transfer vector adds equally to r1 and r2, and cancels in r = r1 – r2), and the Lagrangian (34) is now reduced to
m
2
L
r U ( r).
(3.37)
2
Thus our initial problem has been reduced to just three degrees of freedom – three scalar components of the vector r. In other words, Eq. (37) shows that the dynamics of the vector r of our initial, two-particle system is identical to that of the radius vector of a single particle with the effective mass m, moving in the central potential field U ( r).
Two more degrees of freedom may be excluded from the planetary problem by noticing that according to Eq. (1.35), the angular momentum L = rp of our effective single particle of mass m is also conserved, both in magnitude and direction. Since the direction of L is, by its definition, perpendicular to both r and v = p/ m, this means that the particle’s motion is confined to the plane whose orientation is determined by the initial directions of the vectors r and v. Hence we can completely describe the particle’s position by just two coordinates in that plane, for example by the distance r to the origin, and the polar angle In these coordinates, Eq. (37) takes the form identical to Eq. (2.49): m
L
2 2 2
r r U ( r) .
(3.38)
2
Moreover, the latter coordinate, polar angle , may be also eliminated by using the conservation of angular momentum’s magnitude, in the form of Eq. (2.50): 12
2
L mr
(3.39)
z
.
const
A direct corollary of this conservation is the so-called 2nd Kepler’s law:13 the radius vector r sweeps equal areas A in equal time periods. Indeed, in the linear approximation in dA << A, the area 12 Here index z stands for the coordinate perpendicular to the motion plane. Since other components of the angular momentum equal zero, this index is not really necessary, but I will still use it – just to make a clear distinction between the angular momentum Lz and the Lagrangian function L.
Chapter 3
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differential dA is equal to the area of a narrow right triangle with the base being the arc differential rd, and the height equal to r – see Fig. 4. As a result, according to Eq. (39), the time derivative of the area, dA
r( rd) / 2
1
L
2
r
z
,
(3.40)
dt
dt
2
2 m
remains constant. Since the factor Lz/2 m is constant, integration of this equation over an arbitrary (not necessarily small!) time interval t proves the 2nd Kepler’s law: A t.
rd
r
dA
Fig. 3.4. The area differential dA in
the polar coordinates.
0
Now note that since L/ t = 0, the Hamiltonian function H is also conserved, and since, according to Eq. (38), the kinetic energy of the system is a quadratic-homogeneous function of the generalized velocities r and , we have H = E, so that the system’s energy E, m 2 m 2 2
E
r r U ( r) ,
(3.41)
2
2
is also the first integral of motion.14 But according to Eq. (39), the second term on the right-hand side of Eq. (41) may be represented as
2
m
L
2
2
r
z
,
(3.42)
2
2
2
mr
so that the energy (41) may be expressed as that of a 1D particle moving along axis r, m
2
E
r U ( r),
(3.43)
2
ef
in the following effective potential:
2
L
Effective
U ( r) U ( r)
z
.
(3.44) potential
ef
2
2 mr
energy
So the planetary motion problem has been reduced to the study of an effectively-1D system.15
13 This is one of the three laws deduced, from the extremely detailed astronomical data collected by Tycho Brahe (1546-1601), by Johannes Kepler in the early 17th century. In turn, the three Kepler’s laws have become the main basis for Newton’s discovery, a few decades later, of the gravity law (1.15). That relentless march of physics…
14 One may argue that this fact should have been evident earlier because the effective particle of mass m moves in a potential field U( r), which conserves energy.
15 Note that this reduction has been done in a way different from that used for our testbed problem (Fig.
2.1) in Sec. 2 above. (The reader is encouraged to analyze this difference.) To emphasize this fact, I will keep writing E instead of H here, though for the planetary problem we are discussing now, these two notions coincide.
Chapter 3
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Now we may proceed just like we did in Sec. 3, with due respect to the very specific effective potential (44) which, in particular, diverges at r 0 – besides the very special case of an exactly radial motion, Lz = 0. In particular, we may solve Eq. (43) for dr/ dt to get 1/ 2
m
dr
dt
.
(3.45)
2 [ E U ( r)]1/2
ef
This equation allows us not only to get a direct relation between time t and distance r, similarly to Eq.
(26),
m 1/2
dr
m 1/2
dr
t
,
(3.46)
1/ 2
2
[ E U ( r)]
2
[ E U ( r) L / 2 mr ]
ef
2
2 1/ 2
z
but also do a similar calculation of the angle of the effective particle. Indeed, integrating Eq. (39), L
dt
dt z
,
(3.47)
2
m r
and plugging dt from Eq. (45), we get an explicit expression for the particle’s trajectory ( r): L
dr
L
dr
z
z
(3.48)
2 m1/ 2
2
r [ E U ( r)]1/ 2
m
r E U r L
mr
ef
2
.
1/ 2
2 [
( )
2 / 2
2 ]1/2
z
Note that according to Eq. (39), the derivative d/ dt does not change sign at the reflection from any classical turning point r 0, so that, in contrast to Eq. (46), the sign on the right-hand side of Eq. (48) is uniquely determined by the initial conditions and cannot change during the motion.
Let us use these results, valid for any interaction law U( r), for the planetary motion’s classification. (Following a good tradition, in what follows I will select the arbitrary constant in the potential energy in the way to provide U 0 and hence U ef 0, at r .) The following cases should be distinguished.
If U( r) < 0, i.e. the particle interaction is attractive (as it always is in the case of gravity), and the divergence of the attractive potential at r 0 is faster than 1/ r 2, then U ef( r) – at r 0, so that at appropriate initial conditions the particle may drop on the center even if Lz 0 – the event called the capture.16 On the other hand, with U( r) either converging or diverging slower than 1/ r 2, at r 0, the effective energy profile U ef( r) has the shape shown schematically in Fig. 5. This is true, in particular, for the very important case
Attractive
U ( r) , with ,
0
(3.49)
Coulomb
r
potential
which describes, in particular, the Coulomb (electrostatic) interaction of two particles with electric charges of opposite signs, and Newton’s gravity law (1.15). This particular case will be analyzed in detail below, but for now let us return to the analysis of an arbitrary attractive potential U( r) < 0 leading to the effective potential shown in Fig. 5 when the angular-momentum term in Eq. (44) dominates at small distances r.
16 In order to analyze what exactly happens at the capture, i.e. at r = 0, we would need a model more specific than Eq. (30).
Chapter 3
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U ( r)
ef
E 0
0
E 0
r
U( r)
Fig. 3.5. Effective potential profile of an attractive
r
r
central field, and two types of motion in it.
max
min r
0
According to the analysis in Sec. 3, such potential profile, with a minimum at some distance r 0, may sustain two types of motion, depending on the energy E (determined by initial conditions): (i) If E > 0, there is only one classical turning point where E = U ef, so that the distance r either grows with time from the very beginning or (if the initial value of r was negative) first decreases and then, after the reflection from the increasing potential U ef, starts to grow indefinitely. The latter case, of course, describes the scattering of the effective particle by the attractive center.17
(ii) On the opposite, if the energy is within the range
U ( r ) E ,
0
(3.50)
ef
0
the system moves periodically between two classical turning points r min and r max – see Fig. 5. These oscillations of the distance r correspond to the bound orbital motion of our effective particle about the attracting center.
Let us start with the discussion of the bound motion, with the energy within the range (50). If the energy has its minimal possible value,
E U ( r ) min[ U ( r)],
(3.51)
ef
0
ef
the distance cannot change, r = r 0 = const, so that the particle’s orbit is circular, with the radius r 0
satisfying the condition dU ef/ dr = 0. Using Eq. (44), we see that the condition for r 0 may be written as 2
L
dU
z
.
(3.52)
3
r r 0
mr
dr
0
Since at circular motion, the velocity v is perpendicular to the radius vector r, Lz is just mr 0 v, the left-hand side of Eq. (52) equals mv 2/ r 0, while its right-hand side is just the magnitude of the attractive force, so that this equality expresses the well-known 2nd Newton’s law for the circular motion. Plugging this result into Eq. (47), we get a linear law of angle change, t const, with the angular velocity L
v
z
,
(3.53)
2
mr
r
0
0
and hence the rotation period T 2/ obeys the elementary relation
17 In the opposite case when the interaction is repulsive, U( r) > 0, the addition of the positive angular energy term only increases the trend, and the scattering scenario is the only one possible.
Chapter 3
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2 r
0
T
.
(3.54)
v
Now let the energy be above its minimum value (but still negative). Using Eq. (46) just as in Sec. 3, we see that the distance r oscillates with the period
r
m 1/2 max
dr
T 2
(3.55)
r
.
2
[ E U ( r) 2
2 1/ 2
L / 2
]
r
mr
min
z
This period is not necessarily equal to another period, T, that corresponds to the 2-change of the angle.
Indeed, according to Eq. (48), the change of the angle between two sequential points of the nearest approach,
r max
L
dr
2
z
(3.56)
2 m1/ 2
2
r
r E U ( r)
2
L /
2
mr
z
,
2
1/ 2
min
is generally different from 2. Hence, the general trajectory of the bound motion has a spiral shape –
see, e.g., an illustration in Fig. 6.
r
0
Fig. 3.6. A typical open orbit of a particle
moving in a non-Coulomb central field.
The situation is special, however, for a very important particular case, namely that of the Coulomb potential described by Eq. (49).18 Indeed, plugging this potential into Eq. (48), we get L
dr
z
.
(3.57)
2 m1/ 2 2
r E / r 2
2
L / 2 mr
z
1/2
This is a table integral,19 giving
2
L
m r
1
-
/
1
cos z
(3.58)
1 2
2
2
EL
m
z
const.
/
1/ 2
18 For the power-law interaction, U r, the orbits are closed curves only if = –1 (the Coulomb potential) or if
= +2 (the 3D harmonic oscillator) – the so-called Bertrand theorem, proved by J. Bertrand only in 1873.
19 See, e.g., MA Eq. (6.3a).
Chapter 3
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Hence the reciprocal function, r(), is 2-periodic:
p
r
,
(3.59)
1 e cos( const)
Elliptic
orbit
so that at E < 0, the orbit is a closed line characterized by the following parameters:20
and its
parameters
1/ 2
2
2
L
2 EL
p
z ,
e 1
z
.
(3.60)
2
m
m
The physical meaning of these parameters is very simple. Indeed, the general Eq. (52), in the Coulomb potential for which dU/ dr = / r 2, shows that p is just the circular orbit radius21 for the given L
2
z: r 0 = Lz / m p, so that
2
m
min U ( r) U ( r )
.
(3.61)
ef
ef
0
2
2 Lz
Using this equality together with the second of Eqs. (60), we see that the parameter e (called the eccentricity) may be represented just as
1/ 2
E
e 1
.
(3.62)
min[ U ( r)]
ef
Analytical geometry tells us that Eq. (59), with e < 1, is one of the canonical representations of an ellipse, with one of its two focuses located at the origin. The fact that planets have such trajectories is known as the 1st Kepler’s law. Figure 7 shows the relations between the dimensions of the ellipse and the parameters p and e.22
y r sin
center
focus (one of the two)
p
aphelion
perihelion
b
a
p
0
p
x r cos
1 e
ae
1 e
Fig. 3.7. Ellipse, and its special
p
points and dimensions.
In particular, the major semi-axis a and the minor semi-axis b are simply related to p and e and hence, via Eqs. (60), to the motion integrals E and Lz:
p
p
L
a
,
b
z
(3.63)
1
2
e
2 E
1 2
e 1/2
m E .
2
1/ 2
20 Let me hope that the difference between the parameter p and the particle momentum’s magnitude is absolutely clear from the context, so that using the same (traditional) notation for both notions cannot lead to confusion.
21 Mathematicians prefer a more solemn terminology: the parameter 2 p is called the latus rectum of the ellipse.
22 In this figure, the constant participating in Eqs. (58)-(59) is assumed to be zero. A different choice of the constant corresponds just to a different origin of , i.e. a constant turn of the ellipse about the origin.
Chapter 3
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As was mentioned above, at E min [ U ef( r)] the orbit is almost circular, with r() r 0 p. On the contrary, as E is increased to approach zero (its maximum value for the closed orbit), then e 1, so that the aphelion point r max = p/(1 – e) tends to infinity, i.e. the orbit becomes extremely extended – see the magenta lines in Fig. 8.
(a)
(b)
e 0
.
1
25
.
1
5
.
1
01
.
1
e 00
.
1
20 p
75
.
0
99
.
0
5
.
0
0
.
0
p
Fig.3.8.(a) Zoom-in and (b) zoom-out on the Coulomb-
field trajectories corresponding to the same parameter p
(i.e., the same Lz), but different values of the eccentricity
parameter e, i.e. of the energy E – see Eq. (60): ellipses
( e < 1, red lines), a parabola ( e = 1, magenta line), and
hyperbolas ( e > 1, blue lines). Note that the transition
from closed to open trajectories at e = 1 is dramatic only
at very large distances, r >> p.
The above relations enable, in particular, a ready calculation of the rotation period T Tr = T.
(In the case of a closed trajectory, Tr and T coincide.) Indeed, it is well known that the ellipse’s area A =
ab. But according to the 2nd Kepler’s law (40), dA/ dt = Lz/2 m = const. Hence A
ab
T
.
(3.64a)
dA / dt
L / 2 m
z
Using Eqs. (60) and (63), this important result may be represented in several other forms: 1/ 2
1/ 2
2
p
m
m
3 / 2
T
2 a .
(3.64b)
2 3 / 2
3
1
( e )
( L / 2 m)
z
2
E
Since for the Newtonian gravity (1.15), = Gm 1 m 2 = GmM, at m 1 << m 2 (i.e. m << M) this constant is proportional to m, and the last form of Eq. (64b) yields the 3rd Kepler’s law: periods of motion of different planets in the same central field, say that of our Sun, scale as T a 3/2. Note that in contrast to the 2nd Kepler’s law (which is valid for any central field), the 1st and the 3rd Kepler’s laws are potential-specific.
Now reviewing the above derivation of Eqs. (59)-(60), we see that they are also valid in the case of E 0 – see the top horizontal line in Fig. 5 and its discussion above, if we limit the results to the Chapter 3
Page 15 of 22
CM: Classical Mechanics
physically meaningful range r 0. This means that if the energy is exactly zero, Eq. (59) (with e 1) is still valid for all values of (except for one special point = where r becomes infinite) and describes a parabolic (i.e. open) trajectory – see the magenta lines in Fig. 8.
Moreover, if E > 0, Eq. (59) is still valid within a certain sector of angles , 1
/ 2
EL
1
1
1
2
2
2cos
2cos 1
z
,
for E 0
,
(3.65)
2
e
m
and describes an open, hyperbolic trajectory (see the blue lines in Fig. 8). As was mentioned earlier, such trajectories are typical, in particular, for particle scattering.
3.5. Elastic scattering
If E > 0, the motion is unbound for any realistic interaction potential. In this case, the two most important parameters of the particle trajectory are the impact parameter b and the scattering angle
(Fig. 9), and the main task of the theory is to find the relation between them in the given potential U( r).
r min
0
("
b impact parameter") 0
2 ("
scattering angle"
)
0
0
Fig. 3.9. Main geometric parameters of the scattering problem.
For that, it is convenient to note that b is related to the two conserved quantities, the particle’s energy23 E and its angular momentum Lz, in a simple way. Indeed, at r >> b, the definition L = r( mv) yields Lz = bmv, where v = (2 E/ m)1/2 is the initial (and hence the final) speed of the particle, so that L b
(3.66)
z
2 mE1/2.
Hence the angular contribution to the effective potential (44) may be represented as
2
2
L
b
z
E
.
(3.67)
2
2
2
mr
r
Next, according to Eq. (48), the trajectory sections going from infinity to the nearest approach point ( r = r min) and from that point to infinity, have to be similar, and hence correspond to equal angle changes 0 – see Fig. 9. Hence we may apply the general Eq. (48) to just one of the sections, say [ r min,
], to find the scattering angle:
23 The energy conservation law is frequently emphasized by calling such process elastic scattering.
Chapter 3
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CM: Classical Mechanics
L
dr
bdr
2 2
z
(3.68)
0
2 m1/ 2
2
r
r E U ( r)
2
L /
2
mr
z
2
2
1/ 2
2
r
r 1 U ( r) /
2
2
E b
r .
/
1/ 2
min
min
In particular, for the Coulomb potential (49), now with an arbitrary sign of , we can use the same table integral as in the previous section to get24
Eb
1
/ 2
2cos
(3.69a)
1 ( / 2
2
Eb .
) 1/ 2
This result may be more conveniently rewritten as
tan
.
(3.69b)
2
2 Eb
Very clearly, the scattering angle’s magnitude increases with the potential strength , and decreases as either the particle energy or the impact parameter (or both) are increased.
The general result (68) and the Coulomb-specific relations (69) represent a formally complete solution of the scattering problem. However, in a typical experiment on elementary particle scattering, the impact parameter b of a single particle is unknown. In this case, our results may be used to obtain the statistics of the scattering angle , in particular, the so-called differential cross-section 25
Differential
d
1 dN
cross-
,
(3.70)
section
d
n
d
where n is the average number of the incident particles per unit area, and dN is the average number of the particles scattered into a small solid angle interval d. For a uniform beam of initial particles, d/ d may be calculated by counting the average number of incident particles that have the impact parameters within a small range db:
dN n 2
.
bdb
(3.71)
Scattered by a spherically-symmetric center, which provides an axially-symmetric scattering pattern, these particles are scattered into the corresponding small solid angle interval d = 2sin d .
Plugging these two equalities into Eq. (70), we get the following general geometric relation: d
db
b
.
(3.72)
d
sin d
In particular, for the Coulomb potential (49), a straightforward differentiation of Eq. (69) yields the so-called Rutherford scattering formula (reportedly derived by R. H. Fowler): 2
Rutherford
d
1
scattering
.
(3.73)
formula
d
4 E sin 4 ( / 2)
24 Alternatively, this result may be recovered directly from the first form of Eq. (65), with the eccentricity e expressed via the same dimensionless parameter (2 Eb/): e = [1 + (2 Eb/)2]1/2 > 1.
25 This terminology stems from the fact that an integral (74) of d/ d over the full solid angle, called the total cross-section , has the dimension of the area: = N/ n, where N is the total number of scattered particles.
Chapter 3
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CM: Classical Mechanics
This result, which shows very strong scattering to small angles (so strong that the integral that expresses the total cross-section
d
d
(3.74) Total cross-
d
section
4
is diverging at 0),26 and very weak backscattering (to angles ) was historically extremely significant: in the early 1910s, its good agreement with -particle scattering experiments carried out by Ernest Rutherford’s group gave a strong justification for his introduction of the planetary model of atoms, with electrons moving around very small nuclei – just as planets move around stars.
Note that elementary particle scattering is frequently accompanied by electromagnetic radiation and/or other processes leading to the loss of the initial mechanical energy of the system. Such inelastic scattering may give significantly different results. (In particular, the capture of an incoming particle becomes possible even for a Coulomb attracting center.) Also, quantum-mechanical effects may be important at the scattering of light particles with relatively low energies,27 so that the above results should be used with caution.
3.6. Exercise problems
2 d
3.1. For the system considered in Problem 2.6 (a bead sliding T
T
along a string with fixed tension T, see the figure on the right),
analyze small oscillations of the bead near the equilibrium.
g
m
2 d
3.2. For the system considered in Problem 2.7 (a bead sliding
along a string of a fixed length 2 l, see the figure on the right), analyze small oscillations near the equilibrium.
l
2
g
m
3.3. A bead is allowed to slide, without friction, along an
inverted cycloid in a vertical plane – see the figure on the right.
Calculate the frequency of its free oscillations as a function of
R
g
their amplitude.
m
3.4. For a 1D particle of mass m, placed into potential U( q) = q 2 n (where > 0, and n is a positive integer), calculate the functional dependence of the particle oscillation period T on its energy E.
Explore the limit n .
3.5. Two small masses m 1 and m 2 m 1 may slide, without friction, over a horizontal surface.
They are connected with a spring with equilibrium length l and elastic constant , and at t < 0 are at rest.
26 This divergence, which persists at the quantum-mechanical treatment of the problem (see, e.g., QM Chapter 3), is due to particles with very large values of b, and disappears at an account, for example, of any non-zero concentration of the scattering centers.
27 Their discussion may be found in QM Secs. 3.3 and 3.8.
Chapter 3
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At t = 0, the first of these masses receives a very short kick with impulse P F( t) dt in the direction normal to the spring line. Calculate the largest and smallest magnitude of its velocity at t > 0.
3.6. Explain why the term
2
2
mr / 2, recast in accordance with Eq. (42), cannot be merged with
U( r) in Eq. (38), to form an effective 1D potential energy U( r) – L 2
z /2 mr 2, with the second term’s sign
opposite to that given by Eq. (44). We have done an apparently similar thing for our testbed bead-on-rotating-ring problem at the very end of Sec. 1 – see Eq. (6); why cannot the same trick work for the planetary problem? Besides a formal explanation, discuss the physics behind this difference.
R
3.7. A system consisting of two equal masses m on a light rod of length l (frequently called a dumbbell) can slide without friction along a vertical ring of radius R, rotated about its vertical diameter with a constant angular velocity – see the figure on the right. Derive the condition of stability of the lower horizontal m
l
position of the dumbbell.
m
g
3.8. Analyze the dynamics of the so-called spherical pendulum – a point mass hung, in a uniform gravity field g, on a light cord of length l, with no motion’s confinement to a vertical plane. In particular:
(i) find the integrals of motion and reduce the problem to a 1D one,
(ii) calculate the time period of the possible circular motion around the vertical axis, and (iii) explore small deviations from the circular motion. (Are the pendulum orbits closed?)28
3.9. If our planet Earth were suddenly stopped on its orbit around Sun, how long would it take it to fall on our star? Solve this problem using two different approaches, while neglecting the Earth’s orbit eccentricity and the Sun’s size.
3.10. The orbits of Mars and Earth around the Sun may be well approximated as circles,29 with a radii ratio of 3/2. Use this fact, and the Earth’s year duration (which you should know :-), to calculate the time of travel to Mars spending the least energy on the spacecraft’s launch. Neglect the planets' size and the effects of their gravitational fields.
3.11. Derive first-order and second-order differential equations for the reciprocal distance u 1/ r as a function of , describing the trajectory of the particle’s motion in a central potential U( r). Spell out the latter equation for the particular case of the Coulomb potential (49) and discuss the result.
3.12. For the motion of a particle in the Coulomb attractive field ( U( r) = –/ r, with > 0), calculate and sketch the so-called hodograph 30 – the trajectory followed by the head of the velocity vector v, provided that its tail is kept at the origin.
28 Solving this problem is very good preparation for the analysis of the symmetric top’s rotation in Sec. 4.5.
29 Indeed, their eccentricities are close to, respectively, 0.093 and 0.0167.
30 The use of this notion for the characterization of motion may be traced back at least to an 1846 treatise by W.
Hamilton. Nowadays, it is most often used in applied fluid mechanics, in particular meteorology.
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3.13. Prove that for arbitrary motion of a particle of mass m in the Coulomb field U = –/ r, vector A pL – mn r is conserved.31 After that: (i) spell out the scalar product rA and use the result for an alternative derivation of Eq. (59), and for a geometric interpretation of the vector A;
(ii) spell out (A – pL)2, and use the result for an alternative derivation of the hodograph diagram discussed in the previous problem.
3.14. For motion in the following central potential:
U ( r)
,
2
r
r
(i) find the orbit r(), for positive and , and all possible ranges of energy E; (ii) prove that in the limit 0, and for energy E < 0, the orbit may be represented as a slowly rotating ellipse;
(iii) express the angular velocity of this slow orbit rotation via the parameters and , the particle’s mass m, its energy E, and the angular momentum Lz.
3.15. A star system contains a much smaller planet and an even much smaller amount of dust.
Assuming that the attractive gravitational potential of the dust is spherically symmetric and proportional to the square of the distance from the star,32 calculate the slow precession it gives to a basically circular orbit of the planet.
3.16. A particle is moving in the field of an attractive central force, with potential
U r
,
where n 0 .
r n
For what values of n, the circular orbits are stable?
3.17. Determine the condition for a particle of mass m, moving under the effect of a central attractive force
r
r
F
exp
,
3
r
R
where and R are positive constants, to have a stable circular orbit.
3.18. A particle of mass m, with angular momentum Lz, moves in the field of an attractive central force with a distance-independent magnitude F. If the particle's energy E is slightly higher than the value E min corresponding to the circular orbit of the particle, what is the time period of its radial oscillations? Compare the period with that of the circular orbit at E = E min.
3.19. A particle may move without friction, in the uniform gravity field g = – gnz, over an axially-symmetric surface that is described, in cylindrical coordinates {, , z}, by a smooth function 31 This fact, first proved in 1710 by Jacob Hermann, was repeatedly rediscovered during the next two centuries.
As a result, the most common name of A is, rather unfairly, the Runge-Lenz vector.
32 As may be readily shown from the gravitation version of the Gauss law (see, e.g., the model solution of Problem 1.7), this approximation is exact if the dust density is constant between the star and the planet.
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z
Z() – see the figure on the right. Derive the condition of stability of circular orbits of the particle around the symmetry axis z with respect to small m
perturbations. For the cases when the condition is fulfilled, find out whether
the weakly perturbed orbits are open or closed. Spell out your results for the
following particular cases:
g
(i) a conical surface with Z = ,
z Z
(ii) a parabaloid with Z = 2/2, and
(iii) a spherical surface with Z 2 + 2 = R 2, for < R.
0
3.20. The gravitational potential (i.e. the gravitational energy of a unit probe mass) of our Milky Way galaxy, averaged over interstellar distances, is reasonably well approximated by the following axially-symmetric function:
2
V
r, z
ln 2
2
r z ,
2
where r is the distance from the galaxy’s symmetry axis and z is the distance from its central plane, while V and > 0 are constants.33 Prove that circular orbits of stars in this gravity field are stable, and calculate the frequencies of their small oscillations near such orbits, in the r- and z-directions.
3.21. For particle scattering by a repulsive Coulomb field, calculate the minimum approach distance r min and the velocity v min at that point, and analyze their dependence on the impact parameter b (see Fig. 9) and the initial velocity v of the particle.
3.22. A particle is launched from afar, with impact parameter b, toward an attracting center with
U ( r)
, with n 2, 0.
r n
(i) Express the minimum distance between the particle and the center via b, if the initial kinetic energy E of the particle is barely sufficient for escaping its capture by the attracting center.
(ii) Calculate the capture’s total cross-section; explore the limit n 2.
3.23. A meteorite with initial velocity v approaches an atmosphere-free planet of mass M and radius R.
(i) Find the condition on the impact parameter b for the meteorite to hit the planet’s surface.
(ii) If the meteorite barely avoids the collision, what is its scattering angle?
3.24. Calculate the differential and total cross-sections of the classical elastic scattering of small particles by a hard sphere of radius R.
3.25. The most famous34 confirmation of Einstein’s general relativity theory has come from the observation, by A. Eddington and his associates, of star light’s deflection by the Sun, during the May 33 Just for the reader’s reference, these constants are close to, respectively, 2.2105 m/s and 6.
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1919 solar eclipse. Considering light photons as classical particles propagating with the speed of light, v 0 c 3.00108m/s, and the astronomic data for the Sun’s mass, M S 1.991030kg, and radius, R S
6.96108m, calculate the non-relativistic mechanics’ prediction for the angular deflection of the light rays grazing the Sun’s surface.
3.26. Generalize the expression for the small angle of scattering, obtained in the solution of the previous problem, to a spherically-symmetric but otherwise arbitrary potential U( r). Use the result to calculate the differential cross-section of small-angle scattering by potential U = C/ rn, with integer n > 0.
d
1/ 2
n/ 2 1/ 2
Hint: You may like to use the following table integral:
.
1/ 2
n 1
2
n n
1
1
/ 2
34 It was not the first confirmation, though. The first one came four years earlier from Albert Einstein himself, who showed that his theory may qualitatively explain the difference between the rate of Mercury orbit’s precession, known from earlier observations, and the non-relativistic theory of that effect.
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Chapter 4. Rigid Body Motion
This chapter discusses the motion of rigid bodies, with a heavy focus on its most nontrivial part: rotation. Some byproducts of this analysis enable a discussion, at the end of the chapter, of the motion of point particles as observed from non-inertial reference frames.
4.1. Translation and rotation
It is natural to start a discussion of many-particle systems from a (relatively :-) simple limit when the changes of distances rkk’ rk –rk’ between the particles are negligibly small. Such an abstraction is called the ( absolutely) rigid body; it is a reasonable approximation in many practical problems, including the motion of solid samples. In other words, this model neglects deformations – which will be the subject of the next chapters. The rigid-body approximation reduces the number of degrees of freedom of the system of N particles from 3 N to just six – for example, three Cartesian coordinates of one point (say, 0), and three angles of the system’s rotation about three mutually perpendicular axes passing through this point – see Fig. 1.1
ω
n3
A
“moving”
n2
frame
“lab”
0
frame
Fig. 4.1. Deriving Eq. (8).
n1
As it follows from the discussion in Secs. 1.1-1.3, any purely translational motion of a rigid body, at which the velocity vectors v of all points are equal, is not more complex than that of a point particle. Indeed, according to Eqs. (1.8) and (1.30), in an inertial reference frame, such a body moves exactly as a point particle upon the effect of the net external force F(ext). However, the rotation is a bit more tricky.
Let us start by showing that an arbitrary elementary displacement of a rigid body may be always considered as a sum of the translational motion and of what is called a pure rotation. For that, consider a
“moving” reference frame {n1, n2, n3}, firmly bound to the body, and an arbitrary vector A (Fig. 1). The vector may be represented by its Cartesian components Aj in that moving frame:
3
A A n .
(4.1)
j
j
j1
1 An alternative way to arrive at the same number six is to consider three points of the body, which uniquely define its position. If movable independently, the points would have nine degrees of freedom, but since three distances rkk’ between them are now fixed, the resulting three constraints reduce the number of degrees of freedom to six.
© K. Likharev
CM: Classical Mechanics
Let us calculate the time derivative of this vector as observed from a different (“lab”) frame, taking into account that if the body rotates relative to this frame, the directions of the unit vectors n j, as seen from the lab frame, change in time. Hence, in each product contributing to the sum (1), we have to differentiate both operands:
3
3
dA
dA
dn
j
j
n
A
.
(4.2)
lab
in
j
j
dt
j1 dt
j1
dt
On the right-hand side of this equality, the first sum obviously describes the change of vector A as observed from the moving frame. In the second sum, each of the infinitesimal vectors dn j may be represented by its Cartesian components:
3
dn
d n ,
(4.3)
j
jj' j'
j' 1
where d jj’ are some dimensionless scalar coefficients. To find out more about them, let us scalar-multiply each side of Eq. (3) by an arbitrary unit vector n j” , and take into account the obvious orthonormality condition:
n n ,
(4.4)
j'
j"
j'j"
where j’j” is the Kronecker delta symbol.2 As a result, we get
dn n d .
(4.5)
j
j"
jj"
Now let us use Eq. (5) to calculate the first differential of Eq. (4):
dn n n dn d
d
;
0
2
,
particular
in
dn n 2 d 0 . (4.6)
j'
j"
j'
j"
j'j"
j"j'
j
j
jj
These relations, valid for any choice of indices j, j’, and j” of the set {1, 2, 3}, show that the matrix with elements d jj’ is antisymmetric with respect to the swap of its indices; this means that there are not nine just three non-zero independent coefficients d jj’, all with j j’. Hence it is natural to renumber them in a simpler way: d jj’ = – d j’j d j” , where the indices j, j’, and j” follow in the
“correct” order – either {1,2,3}, or {2,3,1}, or {3,1,2}. It is straightforward to verify (either just by a component-by-component comparison or by using the Levi-Civita permutation symbol3) that in this new notation, Eq. (3) may be represented just as a vector product:
dn dφn ,
(4.7)
Elementary
j
j
rotation
where d is the infinitesimal vector defined by its Cartesian components d j in the rotating reference frame {n1, n2, n3}.
This relation is the basis of all rotation kinematics. Using it, Eq. (2) may be rewritten as 3
dA
dA
dφ
dA
dφ
Vector’s
A
n
ω A,
where ω
.
(4.8)
evolution
lab
in
mov
in
j
j
mov
in
dt
dt
in time
j 1
dt
dt
dt
To reveal the physical sense of the vector , let us apply Eq. (8) to the particular case when A is the radius vector r of a point of the body, and the lab frame is selected in a special way: its origin has the 2 See, e.g., MA Eq. (13.1).
3 See, e.g., MA Eq. (13.2). Using this symbol, we may write d jj’ = – d j’j jj’j”d j” for any choice of j, j’, and j” .
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same position and moves with the same velocity as that of the moving frame, in the particular instant under consideration. In this case, the first term on the right-hand side of Eq. (8) is zero, and we get r
d
ω r ,
(4.9)
frame
lab
special
in
dt
were vector r itself is the same in both frames. According to the vector product definition, the particle velocity described by this formula has a direction perpendicular to the vectors and r (Fig. 2), and magnitude r sin. As Fig. 2 shows, the last expression may be rewritten as , where = r sin is the distance from the line that is parallel to the vector and passes through point 0. This is of course just the pure rotation about that line (called the instantaneous axis of rotation), with the angular velocity .
According to Eqs. (3) and (8), the angular velocity vector is defined by the time evolution of the moving frame alone, so it is the same for all points r, i.e. for the rigid body as a whole. Note that nothing in our calculations forbids not only the magnitude but also the direction of the vector , and thus of the instantaneous axis of rotation, to change in time; hence the name.
ω r
ω
r
Fig. 4.2. The instantaneous axis and
the angular velocity of rotation.
0
Now let us generalize our result a step further, considering two reference frames that do not rotate versus each other: one (“lab”) frame arbitrary, and another one selected in the special way described above, so that for it Eq. (9) is valid in it. Since their relative motion of these two reference frames is purely translational, we can use the simple velocity addition rule given by Eq. (1.6) to write Body
point’s
v
v
v
v
ω r,
(4.10)
lab
in
0
lab
in
frame
lab
special
in
0
lab
in
velocity
where r is the radius vector of a point is measured in the body-bound (“moving”) frame 0.
4.2. Inertia tensor
Since the dynamics of each point of a rigid body is strongly constrained by the conditions rkk’ =
const, this is one of the most important fields of application of the Lagrangian formalism discussed in Chapter 2. For using this approach, the first thing we need to calculate is the kinetic energy of the body in an inertial reference frame. Since it is just the sum of the kinetic energies (1.19) of all its points, we can use Eq. (10) to write:4
m
m
m
m
2
T v v ωr v v
m
ω r ωr (4.11)
0
2
2
(
)
(
)2.
2
2
2 0
0
2
4 Actually, all symbols for particle masses, coordinates, and velocities should carry the particle’s index, over which the summation is carried out. However, in this section, for the notation simplicity, this index is just implied.
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Let us apply to the right-hand side of Eq. (11) two general vector analysis formulas listed in the Math Appendix: the so-called operand rotation rule MA Eq. (7.6) to the second term, and MA Eq. (7.7b) to the third term. The result is
m
m
T
2
v
r
m (v
ω)
r
(ω r) .
(4.12)
0
0
2 2 2
2
2
This expression may be further simplified by making a specific choice of the point 0 (from which the radius vectors r of all particles are measured), namely by using for this point the center of mass of the body. As was already mentioned in Sec. 3.4 for the two-point case, the radius vector R of this point is defined as
MR mr,
M
with
m,
(4.13)
so that M is the total mass of the body. In the reference frame centered at this point, we have R = 0, so that the second sum in Eq. (12) vanishes, and the kinetic energy is a sum of just two terms: M
m
T T
T , T
2
V , T
r
(ω r) ,
(4.14)
tran
rot
tran
rot
2 2 2
2
2
where V dR/ dt is the center-of-mass velocity in our inertial reference frame, and all particle positions r are measured in the center-of-mass frame. Since the angular velocity vector is common for all points of a rigid body, it is more convenient to rewrite the rotational part of the energy in a form in that the summation over the components of this vector is separated from the summation over the points of the body:
3
1
T
I ,
(4.15) Kinetic
rot
jj' j j'
2
energy of
j, j'1
rotation
where the 33 matrix with elements
I
m r
r r
(4.16) Inertia
jj'
2
jj'
j j'
tensor
represents, in the selected reference frame, the inertia tensor of the body.5
Actually, the term “tensor” for the construct described by this matrix has to be justified, because in physics it implies a certain reference-frame-independent notion, whose matrix elements have to obey certain rules at the transfer between reference frames. To show that the matrix (16) indeed described such notion, let us calculate another key quantity, the total angular momentum L of the same body.6
Summing up the angular momenta of each particle, defined by Eq. (1.31), and then using Eq. (10) again, in our inertial reference frame we get
L r p r
m v mr v ω r
.
(4.17)
0
mr v
0
r
m ω r
We see that the momentum may be represented as a sum of two terms. The first one,
5 While the ABCs of the rotational dynamics were developed by Leonhard Euler in 1765, an introduction of the inertia tensor’s formalism had to wait very long – until the invention of the tensor analysis by Tullio Levi-Civita and Gregorio Ricci-Curbastro in 1900 – soon popularized by its use in Einstein’s general relativity.
6 Hopefully, there is very little chance of confusing the angular momentum L (a vector) and its Cartesian components Lj (scalars with an index) on one hand, and the Lagrangian function L (a scalar without an index) on the other hand.
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L mr v MR v ,
(4.18)
0
0
0
describes the possible rotation of the center of mass about the inertial frame’s origin. This term vanishes if the moving reference frame’s origin 0 is positioned at the center of mass (where R = 0). In this case, we are left with only the second term, which describes a pure rotation of the body about its center of mass:
L L
.
(4.19)
rot
r
m ω r
Using one more vector algebra formula, the “bac minis cab” rule,7 we may rewrite this expression as L mω 2
r rr ω .
(4.20)
Let us spell out an arbitrary Cartesian component of this vector:
3
3
L
m r 2 r
r
m
r 2
r r .
(4.21)
j
j
j j'
j'
j'
jj'
j j'
j' 1
j' 1
By changing the summation order and comparing the result with Eq. (16), the angular momentum may be conveniently expressed via the same matrix elements Ijj’ as the rotational kinetic energy: 3
Angular
L
I .
(4.22)
j
momentum
jj' j'
j'1
Since L and are both legitimate vectors (meaning that they describe physical vectors independent of the reference frame choice), the matrix of elements Ijj’ that relates them is a legitimate tensor. This fact, and the symmetry of the tensor ( Ijj’ = Ij’j), evident from its definition (16), allow the tensor to be further simplified. In particular, mathematics tells us that by a certain choice of the coordinate axes’ orientations, any symmetric tensor may be reduced to a diagonal form
I I ,
(4.23)
jj '
j
jj'
where in our case
Principal
moments of
I
m r
r
m r
r
m ,
(4.24)
j
2 2 j 2 2
j'
j"
2
j
inertia j being the distance of the particle from the j th axis, i.e. the length of the perpendicular dropped from the point to that axis. The axes of such a special coordinate system are called the principal axes, while the diagonal elements Ij given by Eq. (24), the principal moments of inertia of the body. In such a special reference frame, Eqs. (15) and (22) are reduced to very simple forms:
3 I
T rot and L in
T
principal-axes
j 2
,
(4.25)
rot
j
j1 2
frame
L I .
(4.26)
j
j
j
Both these results remind the corresponding relations for the translational motion, T tran = MV 2/2 and P =
MV, with the angular velocity replacing the linear velocity V, and the tensor of inertia playing the role of scalar mass M. However, let me emphasize that even in the specially selected reference frame, with 7 See, e.g., MA Eq. (7.5).
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its axes pointing in principal directions, the analogy is incomplete, and rotation is generally more complex than translation, because the measures of inertia, Ij, are generally different for each principal axis.
Let me illustrate the last fact on a simple but instructive system of three similar massive particles fixed in the vertices of an equilateral triangle (Fig. 3).
r 2
m
a
0
a
r
Fig. 4.3. Principal moments of
h
1
inertia: a simple case study.
/ 6
m
m
a
Due to the symmetry of the configuration, one of the principal axes has to pass through the center of mass 0 and be normal to the plane of the triangle. For the corresponding principal moment of inertia, Eq.
(24) readily yields I 3 = 3 m2. If we want to express this result in terms of the triangle’s side a, we may notice that due to the system’s symmetry, the angle marked in Fig. 3 equals /6, and from the shaded right triangle, a/2 = cos(/6) 3/2, giving = a/3, so that, finally, I 3 = ma 2.
Let me use this simple case to illustrate the following general axis shift theorem, which may be rather useful – especially for more complex systems. For that, let us relate the inertia tensor elements Ijj’
and I’jj’, calculated in two reference frames – one with its origin at the center of mass 0, and another one (0 ’) translated by a certain vector d (Fig. 4a), so that for an arbitrary point, r ’ = r + d. Plugging this relation into Eq. (16), we get
I'
m r d
r
d
r
d
jj'
2
jj '
j j j' j'
(4.27)
m 2
r 2
2
r d d r r r d r d d d
jj '
j j' j j' j' j j j'.
Since in the center-of-mass frame, all sums mrj equal zero, we may use Eq. (16) to finally obtain I' I M (
2
d d d ) .
(4.28)
jj'
jj '
jj'
j
j'
In particular, this equation shows that if the shift vector d is perpendicular to one (say, j th) of the principal axes (Fig. 4b), i.e. dj = 0, then Eq. (28) is reduced to a very simple formula: Principal
2
I' I Md .
(4.29)
j
j
axis'
shift
(a)
(b)
m
r'
r
j
j
'
r
r
d
Fig. 4.4. (a) A general coordinate
frame’s shift from the center of
mass, and (b) a shift perpendicular
'
0
d
to one of the principal axes.
0
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Now returning to the particular system shown in Fig. 3, let us perform such a shift to the new (“primed”) axis passing through the location of one of the particles, still perpendicular to their common plane. Then the contribution of that particular mass to the primed moment of inertia vanishes, and I’ 3 =
2 ma 2. Now, returning to the center of mass and applying Eq. (29), we get I 3 = I’ 3 – M2 = 2 ma 2 –
(3 m)( a/3)2 = ma 2, i.e. the same result as above.
The symmetry situation inside the triangle’s plane is somewhat less obvious, so let us start by calculating the moments of inertia for the axes shown vertical and horizontal in Fig. 3. From Eq. (24), we readily get:
2
2
2
2
2
a a
ma
a
ma
I 2
2
2
mh m m2
,
I 2 m
,
(4.30)
1
2 3
3
2
2
2
2
where h is the distance from the center of mass and any side of the triangle: h = sin(/6) = /2 =
a/ 23. We see that I 1 = I 2, and mathematics tells us that in this case, any in-plane axis (passing through the center-of-mass 0) may be considered as principal, and has the same moment of inertia. A rigid body Symmetric
top: with this property, I 1 = I 2 I 3, is called the symmetric top. (The last direction is called the main principal definition axis of the system.)
Despite the symmetric top’s name, the situation may be even more symmetric in the so-called spherical tops, i.e. highly symmetric systems whose principal moments of inertia are all equal, Spherical
top:
I I I I ,
(4.31)
1
2
3
definition
Mathematics says that in this case, the moment of inertia for rotation about any axis (but still passing through the center of mass) is equal to the same I. Hence Eqs. (25) and (26) are further simplified for any direction of the vector :
Spherical
I
top:
T
2
,
L ω
I ,
(4.32)
rot
properties
2
thus making the analogy of rotation and translation complete. (As will be discussed in the next section, this analogy is also complete if the rotation axis is fixed by external constraints.)
Evident examples of a spherical top are a uniform sphere and a uniform spherical shell; its less obvious example is a uniform cube – with masses either concentrated in vertices, or uniformly spread over the faces, or uniformly distributed over the volume. Again, in this case any axis passing through the center of mass is a principal one and has the same principal moment of inertia. For a sphere, this is natural; for a cube, rather surprising – but may be confirmed by a direct calculation.
4.3. Fixed-axis rotation
Now we are well equipped for a discussion of the rigid body’s rotational dynamics. The general equation of this dynamics is given by Eq. (1.38), which is valid for dynamics of any system of particles
– either rigidly connected or not:
L τ ,
(4.33)
where is the net torque of external forces. Let us start exploring this equation from the simplest case when the axis of rotation, i.e. the direction of vector , is fixed by some external constraints. Directing Chapter 4
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the z-axis along this vector, we have x = y = 0. According to Eq. (22), in this case, the z-component of the angular momentum,
L I ,
(4.34)
z
zz
z
where Izz, though not necessarily one of the principal moments of inertia. still may be calculated using Eq. (24):
I
,
(4.35)
zz
2
m
z
m 2 x 2 y
with z being the distance of each particle from the rotation axis z. According to Eq. (15), in this case the rotational kinetic energy is just
I
zz
2
T
.
(4.36)
rot
2
z
Moreover, it is straightforward to show that if the rotation axis is fixed, Eqs. (34)-(36) are valid even if the axis does not pass through the center of mass – provided that the distances z are now measured from that axis. (The proof is left for the reader’s exercise.)
As a result, we may not care about other components of the vector L,8 and use just one component of Eq. (33),
L ,
(4.37)
z
z
because it, when combined with Eq. (34), completely determines the dynamics of rotation: I ,
i.e. I ,
(4.38)
zz
z
z
zz
z
z
where z is the angle of rotation about the axis, so that z = . The scalar relations (34), (36), and (38), describing rotation about a fixed axis, are completely similar to the corresponding formulas of 1D
motion of a single particle, with z corresponding to the usual (“linear”) velocity, the angular momentum component Lz – to the linear momentum, and Iz – to the particle’s mass.
The resulting motion about the axis is also frequently similar to that of a single particle. As a simple example, let us consider what is called the physical (or “compound”) pendulum (Fig. 5) – a rigid body free to rotate about a fixed horizontal axis that does not pass through the center of mass 0, in a uniform gravity field g.
'
0
r
'
0
in
l
r
Fig. 4.5. Physical pendulum: a rigid
i 0
n
body with a fixed (horizontal) rotation
0
axis 0’ that does not pass through the
g
M
center of mass 0. (The plane of
drawing is normal to that axis.)
8 Note that according to Eq. (22), other Cartesian components of the angular momentum, Lx and Ly, may be different from zero, and even evolve in time. The corresponding torques x and y, which obey Eq. (33), are automatically provided by the external forces that keep the rotation axis fixed.
Chapter 4
Page 8 of 32
CM: Classical Mechanics
Let us drop the perpendicular from point 0 to the rotation axis, and call the oppositely directed vector l – see the dashed arrow in Fig. 5. Then the torque (relative to the rotation axis 0 ’) of the forces keeping the axis fixed is zero, and the only contribution to the net torque is due to gravity alone: τ
0
in '
r
F
0
in '
l r g
m
0
in
mlg mr g l
M g
0
