in
.
(4.39)
(The last step used the facts that point 0 is the center of mass, so that the second term on the right-hand side equals zero, and that the vectors l and g are the same for all particles of the body.) This result shows that the torque is directed along the rotation axis, and its (only) component z is equal to – Mgl sin, where is the angle between the vectors l and g, i.e. the angular deviation of the pendulum from the position of equilibrium – see Fig. 5 again. As a result, Eq. (38) takes the form, I' Mgl sin ,
(4.40)
where I’ is the moment of inertia for rotation about the axis 0 ’ rather than about the center of mass. This equation is identical to Eq. (1.18) for the point-mass (sometimes called “mathematical”) pendulum, with small-oscillation frequency
1/ 2
1/ 2
Physical
Mgl
g
I'
pendulum:
Ω
, with l
.
(4.41)
ef
frequency
I'
l
Ml
ef
As a sanity check, in the simplest case when the linear size of the body is much smaller than the suspension length l, Eq. (35) yields I’ = Ml 2, i.e. l ef = l, and Eq. (41) reduces to the well-familiar formula
= ( g/ l)1/2 for the point-mass pendulum.
Now let us discuss the situations when a rigid body not only rotates but also moves as a whole.
As was mentioned in the introductory chapter, the total linear momentum of the body,
d
P v
m r
m
r
m ,
(4.42)
dt
satisfies the 2nd Newton’s law in the form (1.30). Using the definition (13) of the center of mass, the momentum may be represented as
P MR M ,
V
(4.43)
so Eq. (1.30) may be rewritten as
C.o.m.:
law of
V
M F ,
(4.44)
motion
where F is the vector sum of all external forces. This equation shows that the center of mass of the body moves exactly like a point particle of mass M, under the effect of the net force F. In many cases, this fact makes the translational dynamics of a rigid body absolutely similar to that of a point particle.
The situation becomes more complex if some of the forces contributing to the vector sum F
depend on the rotation of the same body, i.e. if its rotational and translational motions are coupled.
Analysis of such coupled motion is rather straightforward if the direction of the rotation axis does not change in time, and hence Eqs. (34)-(36) are still valid. Possibly the simplest example is a round cylinder (say, a wheel) rolling on a surface without slippage (Fig. 6). Here the no-slippage condition may be represented as the requirement to the net velocity of the particular wheel’s point A that touches the surface to equal zero – in the reference frame bound to the surface. For the simplest case of plane Chapter 4
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surface (Fig. 6a), this condition may be spelled out using Eq. (10), giving the following relation between the angular velocity of the wheel and the linear velocity V of its center:
V r 0.
(4.45)
(a)
(b)
R
V
0
V
0
Fig. 4.6. Round cylinder
'
0
r
r
rolling over (a) a plane
A
surface and (b) a concave
A
surface.
Such kinematic relations are essentially holonomic constraints, which reduce the number of degrees of freedom of the system. For example, without the no-slippage condition (45), the wheel on a plane surface has to be considered as a system with two degrees of freedom, making its total kinetic energy (14) a function of two independent generalized velocities, say V and : M
I
2
2
T T
T
V .
(4.46)
tran
rot
2
2
Using Eq. (45) we may eliminate, for example, the linear velocity and reduce Eq. (46) to M
I
I
T
r
2
2
ef
2
,
where
2
I I Mr .
(4.47)
2
2
2
ef
This result may be interpreted as the kinetic energy of pure rotation of the wheel about the instantaneous rotation axis A, with I ef being the moment of inertia about that axis, satisfying Eq. (29).
Kinematic relations are not always as simple as Eq. (45). For example, if a wheel is rolling on a concave surface (Fig. 6b), we need to relate the angular velocities of the wheel’s rotation about its axis 0 ’ (say, ) and that (say, ) of its axis’ rotation about the center 0 of curvature of the surface. A popular error here is to write = –( r/ R) [WRONG!]. A prudent way to derive the correct relation is to note that Eq. (45) holds for this situation as well, and on the other hand, the same linear velocity of the wheel’s center may be expressed as V = ( R – r). Combining these formulas, we get the correct relation r
.
(4.48)
R r
Another famous example of the relation between translational and rotational motion is given by the “sliding-ladder” problem (Fig. 7). Let us analyze it for the simplest case of negligible friction, and the ladder’s thickness being small in comparison with its length l.
l / 2
R X, Y
l / 2
l / 2
Fig. 4.7. The sliding-ladder problem.
0
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To use the Lagrangian formalism, we may write the kinetic energy of the ladder as the sum (14) of its translational and rotational parts:
M
I
T
2 2
X Y
2
,
(4.49)
2
2
where X and Y are the Cartesian coordinates of its center of mass in an inertial reference frame, and I is the moment of inertia for rotation about the z-axis passing through the center of mass. (For the uniformly distributed mass, an elementary integration of Eq. (35) yields I = Ml 2/12). In the reference frame with the center in the corner 0, both X and Y may be simply expressed via the angle : l
X cos
l
, Y sin.
(4.50)
2
2
(The easiest way to obtain these relations is to notice that the dashed line in Fig. 7 has length l/2, and the same slope as the ladder.) Plugging these expressions into Eq. (49), we get
2
I
l
1
ef
2
2
T
,
I I M Ml .
(4.51)
ef
2
2
3
Since the potential energy of the ladder in the gravity field may be also expressed via the same angle, l
U MgY Mg sin,
(4.52)
2
may be conveniently used as the (only) generalized coordinate of the system. Even without writing the Lagrange equation of motion for that coordinate, we may notice that since the Lagrangian function L
T – U does not depend on time explicitly, and the kinetic energy (51) is a quadratic-homogeneous function of the generalized velocity , the full mechanical energy,
I
l
Mgl l
ef
2
2
E T U
Mg sin
sin ,
(4.53)
2
2
2 3 g
is conserved, giving us the first integral of motion. Moreover, Eq. (53) shows that the system’s energy (and hence dynamics) is identical to that of a physical pendulum with an unstable fixed point 1 = /2, a stable fixed point at 2 = –/2, and frequency
1/ 2
3 g
(4.54)
2
l
of small oscillations near the latter point. (Of course, this fixed point cannot be reached in the simple geometry shown in Fig. 7, where the ladder’s fall on the floor would change its equations of motion.
Moreover, even before that, the left end of the ladder may detach from the wall. The analysis of this issue is left for the reader’s exercise.)
4.4. Free rotation
Now let us proceed to more complex situations when the rotation axis is not fixed. A good illustration of the complexity arising in this case comes from the case of a rigid body left alone, i.e. not subjected to external forces and hence with its potential energy U constant. Since in this case, according Chapter 4
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to Eq. (44), the center of mass (as observed from any inertial reference frame) moves with a constant velocity, we can always use a convenient inertial reference frame with the origin at that point. From the point of view of such a frame, the body’s motion is a pure rotation, and T tran = 0. Hence, the system’s Lagrangian function is just its rotational energy (15), which is, first, a quadratic-homogeneous function of the components j (which may be taken for generalized velocities), and, second, does not depend on time explicitly. As we know from Chapter 2, in this case the mechanical energy, here equal to T rot alone, is conserved. According to Eq. (15), for the principal-axes components of the vector , this means 3 I
Rotational
T
j
const
.
(4.55) energy’s
rot
2
j
conservation
j1 2
Next, as Eq. (33) shows, in the absence of external forces, the angular momentum L of the body is conserved as well. However, though we can certainly use Eq. (26) to represent this fact as 3
Angular
L I n const
,
(4.56)
j
j
j
momentum’s
j 1
conservation
where n j are the principal axes, this does not mean that all components j a re constant, because the principal axes are fixed relative to the rigid body, and hence may rotate with it.
Before exploring these complications, let us briefly mention two conceptually easy, but practically very important cases. The first is a spherical top ( I 1 = I 2 = I 3 = I). In this case, Eqs. (55) and (56) imply that all components of the vector = L/ I, i.e. both the magnitude and the direction of the angular velocity are conserved, for any initial spin. In other words, the body conserves its rotation speed and axis direction, as measured in an inertial frame. The most obvious example is a spherical planet. For example, our Mother Earth, rotating about its axis with angular velocity = 2/(1 day) 7.310-5 s-1, keeps its axis at a nearly constant angle of 2327’ to the ecliptic pole, i.e. to the axis normal to the plane of its motion around the Sun. (In Sec. 6 below, we will discuss some very slow motions of this axis, due to gravity effects.)
Spherical tops are also used in the most accurate gyroscopes, usually with gas-jet or magnetic suspension in vacuum. If done carefully, such systems may have spectacular stability. For example, the gyroscope system of the Gravity Probe B satellite experiment, flown in 2004-2005, was based on quartz spheres – round with a precision of about 10 nm and covered with superconducting thin films (which enabled their magnetic suspension and monitoring). The whole system was stable enough to measure the so-called geodetic effect in general relativity (essentially, the space curving by the Earth’s mass), resulting in the axis’ precession by only 6.6 arc seconds per year, i.e. with an angular velocity of just
~10-11s-1, with experimental results agreeing with theory with a record ~0.3% accuracy.9
The second simple case is that of the symmetric top ( I 1 = I 2 I 3) with the initial vector L aligned with the main principal axis. In this case, = L/ I 3 = const, so that the rotation axis is conserved.10 Such tops, typically in the shape of a flywheel (heavy, flat rotor), and supported by gimbal systems (also called the “Cardan suspensions”) that allow for virtually torque-free rotation about three mutually 9 Still, the main goal of this rather expensive (~$750M) project, an accurate measurement of a more subtle relativistic effect, the so-called frame-dragging drift (also called “the Schiff precession”), predicted to be about 0.04 arc seconds per year, has not been achieved.
10 This is also true for an asymmetric top, i.e. an arbitrary body (with, say, I 1 < I 2 < I 3), but in this case the alignment of the vector L with the axis n2 corresponding to the intermediate moment of inertia, is unstable.
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perpendicular axes,11 are broadly used in more common gyroscopes. Invented by Léon Foucault in the 1850s and made practical later by H. Anschütz-Kaempfe, such gyroscopes have become core parts of automatic guidance systems, for example, in ships, airplanes, missiles, etc. Even if its support wobbles and/or drifts, the suspended gyroscope sustains its direction relative to an inertial reference frame.12
However, in the general case with no such special initial alignment, the dynamics of symmetric tops is more complicated. In this case, the vector L is still conserved, including its direction, but the vector is not. Indeed, let us direct the n2-axis normally to the common plane of the vector L and the current instantaneous direction n3 of the main principal axis (in Fig. 8 below, the plane of the drawing); then, in that particular instant, L 2 = 0. Now let us recall that in a symmetric top, the axis n2 is a principal one. According to Eq. (26) with j = 2, the corresponding component 2 has to be equal to L 2/ I 2, so it is equal to zero. This means that in the particular instant we are considering, the vector lies in this plane (the common plane of vectors L and n3) as well – see Fig. 8a.
(a)
n L
(b)
ω n
ω
3
n
L
L
3
3
n
L
n
pre
1
3
1
rot
L
1
1
Fig. 4.8. Free rotation of a symmetric top:
1
0
0
(a) the general configuration of vectors,
and (b) calculating the free precession
frequency.
Now consider some point located on the main principal axis n3, and hence on the plane [n3, L].
Since is the instantaneous axis of rotation, according to Eq. (9), the instantaneous velocity v = r of the point is directed normally to that plane. This is true for each point of the main axis (besides only one, with r = 0, i.e. the center of mass, which does not move), so the axis as a whole has to move normally to the common plane of the vectors L, , and n3, while still passing through point 0. Since this conclusion is valid for any moment of time, it means that the vectors and n3 rotate about the space-fixed vector L
together, with some angular velocity pre, at each moment staying within one plane. This effect is called the free (or “torque-free”, or “regular”) precession, and has to be clearly distinguished it from the completely different effect of the torque-induced precession, which will be discussed in the next section.
To calculate pre, let us represent the instant vector as a sum of not its Cartesian components (as in Fig. 8a), but rather of two non-orthogonal vectors directed along n3 and L (Fig. 8b): L
ω n n ,
n
.
(4.57)
rot
3
pre
L
L
L
11 See, for example, a nice animation available online at http://en.wikipedia.org/wiki/Gimbal.
12 Currently, optical gyroscopes are becoming more popular for all but the most precise applications. Much more compact but also much less accurate gyroscopes used, for example, in smartphones and tablet computers, are based on the effect of rotation on 2D mechanical oscillators (whose analysis is left for the reader’s exercise), and are implemented as micro-electro-mechanical systems (MEMS) – see, e.g., Chapter 22 in V. Kaajakari, Practical MEMS, Small Gear Publishing, 2009.
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Fig. 8b shows that rot has the meaning of the angular velocity of rotation of the body about its main principal axis, while pre is the angular velocity of rotation of that axis about the constant direction of the vector L, i.e. is exactly the frequency of precession that we are trying to find. Now pre may be readily calculated from the comparison of two panels of Fig. 8, by noticing that the same angle
between the vectors L and n3 participates in two relations:
L
sin
1
1
.
(4.58)
L
pre
Since the n1-axis is a principal one, we may use Eq. (26) for j = 1, i.e. L 1 = I 11, to eliminate 1 from Eq. (58), and get a very simple formula
L
Free
.
(4.59) precession:
pre
I
lab frame
1
This result shows that the precession frequency is constant and independent of the alignment of the vector L with the main principal axis n3, while its amplitude (characterized by the angle ) does depend on the initial alignment, and vanishes if L is parallel to n3.13 Note also that if all principal moments of inertia are of the same order, pre is of the same order as the total angular speed of the rotation.
Now let us briefly discuss the free precession in the general case of an “asymmetric top”, i.e. a body with arbitrary I 1 I 2 I 3. In this case, the effect is more complex because here not only the direction but also the magnitude of the instantaneous angular velocity may evolve in time. If we are only interested in the relation between the instantaneous values of j and Lj, i.e. the “trajectories” of the vectors and L as observed from the reference frame {n1, n2, n3} of the principal axes of the body, rather than in the explicit law of their time evolution, they may be found directly from the conservation laws. (Let me emphasize again that the vector L, being constant in an inertial reference frame, generally evolves in the frame rotating with the body.) Indeed, Eq. (55) may be understood as the equation of an ellipsoid in the Cartesian coordinates {1, 2, 3 }, so that for a free body, the vector has to stay on the surface of that ellipsoid.14 On the other hand, since the reference frame’s rotation preserves the length of any vector, the magnitude (but not the direction!) of the vector L is also an integral of motion in the moving frame, and we can write
3
3
2
2
2
2
L L
I
.
(4.60)
j
const
j
j
j 1
j 1
Hence the trajectory of the vector follows the closed curve formed by the intersection of two ellipsoids, (55) and (60) – the so-called Poinsot construction. It is evident that this trajectory is generally
“taco-edge-shaped”, i.e. more complex than a planar circle, but never very complex either.15
The same argument may be repeated for the vector L, for whom the first form of Eq. (60) descries a sphere, and Eq. (55), another ellipsoid:
13 For our Earth, free precession’s amplitude is so small (corresponding to sub-10-m linear deviations of the symmetry axis from the vector L at the surface) that this effect is of the same order as other, more irregular motions of the axis, resulting from turbulent fluid flow effects in the planet’s interior and its atmosphere.
14 It is frequently called the Poinsot’s ellipsoid, named after Louis Poinsot (1777-1859) who has made several important contributions to rigid body mechanics.
15 Curiously, the “wobbling” motion along such trajectories was observed not only for macroscopic rigid bodies but also for heavy atomic nuclei – see, e.g., N. Sensharma et al. , Phys. Rev. Lett. 124, 052501 (2020).
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3
1
T
L
const
.
(4.61)
rot
2
j
j1 2 I j
On the other hand, if we are interested in the trajectory of the vector as observed from an inertial frame (in which the vector L stays still), we may note that the general relation (15) for the same rotational energy T rot may also be rewritten as
3
3
1
T
I .
(4.62)
rot
j jj' j'
2 j1
j'1
But according to the Eq. (22), the second sum on the right-hand side is nothing more than Lj, so that 1 3
1
T
L
.
(4.63)
rot
ω L
2
j
j
j1
2
This equation shows that for a free body ( T rot = const, L = const), even if the vector changes in time, its endpoint should stay on a plane normal to the angular momentum L. Earlier, we have seen that for the particular case of the symmetric top – see Fig. 8b, but for an asymmetric top, the trajectory of the endpoint may not be circular.
If we are interested not only in the trajectory of the vector but also in the law of its evolution in time, it may be calculated using the general Eq. (33) expressed in the principal components j. For that, we have to recall that Eq. (33) is only valid in an inertial reference frame, while the frame {n1, n2, n3} may rotate with the body and hence is generally not inertial. We may handle this problem by applying, to the vector L, the general kinematic relation (8):
L
d
L
d
ω .
L
(4.64)
lab
in
mov
in
dt
dt
Combining it with Eq. (33), in the moving frame we get
L
d
ω L τ ,
(4.65)
dt
where is the external torque. In particular, for the principal-axis components Lj, related to the components j by Eq. (26), the vector equation (65) is reduced to a set of three scalar Euler equations Euler
I ( I I
)
,
(4.66)
j
j
j"
j'
j'
j"
j
equations
where the set of indices { j, j’ , j” } has to follow the usual “right” order – e.g., {1, 2, 3}, etc.16
In order to get a feeling how do the Euler equations work, let us return to the particular case of a free symmetric top (1 = 2 = 3 = 0, I 1 = I 2 I 3). In this case, I 1 – I 2 = 0, so that Eq. (66) with j = 3 yields
3 = const, while the equations for j = 1 and j = 2 take the following simple form:
Ω ,
Ω ,
(4.67)
1
pre
2
2
pre
1
where pre is a constant determined by both the system parameters and the initial conditions: 16 These equations are of course valid in the simplest case of the fixed rotation axis as well. For example, if =
n z, i.e. x = y = 0, Eq. (66) is reduced to Eq. (38).
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I I
Free
3
1
Ω
.
(4.68)
pre
3
precession:
I 1
body frame
The system of two equations (67) has a sinusoidal solution with frequency pre, and describes a uniform rotation of the vector , with that frequency, about the main axis n3. This is just another representation of the free precession analyzed above, but this time as observed from the rotating body.
Evidently, pre is substantially different from the frequency pre (59) of the precession as observed from the lab frame; for example, pre vanishes for the spherical top (with I 1 = I 2 = I 3), while pre, in this case, is equal to the rotation frequency.17
Unfortunately, for the rotation of an asymmetric top (i.e., an arbitrary rigid body) the Euler equations (66) are substantially nonlinear even in the absence of external torque, and may be solved analytically only in just a few cases. One of them is a proof of the already mentioned fact: the free top’s rotation about one of its principal axes is stable if the corresponding principal moment of inertia is either the largest or the smallest one of the three. (This proof is easy, and is left for the reader’s exercise.) 4.5. Torque-induced precession
The dynamics of rotation becomes even more complex in the presence of external forces. Let us consider the most counter-intuitive effect of torque-induced precession, for the simplest case of an axially-symmetric body (which is a particular case of the symmetric top, I 1 = I 2 I 3), supported at some point A of its symmetry axis, that does not coincide with the center of mass 0 – see Fig. 9.
z
(a)
(b)
pre
n3
L
0
L
rot
xy
0
l
Fig. 4.9. Symmetric top in the gravity field:
(a) a side view at the system and (b) the top
view at the evolution of the horizontal
A
g
M
component of the angular momentum vector.
The uniform gravity field g creates bulk-distributed forces that, as we know from the analysis of the physical pendulum in Sec. 3, are equivalent to a single force Mg applied in the center of mass – in Fig. 9, point 0. The torque of this force relative to the support point A is
τ r
g
M
n g .
(4.69)
0
3
A
in
Ml
Hence the general equation (33) of the angular momentum evolution (valid in any inertial frame, for example the one with its origin at point A) becomes
17 For our Earth with its equatorial bulge (see Sec. 6 below), the ratio ( I 3 – I 1)/ I 1 is ~1/300, so that 2/pre is about 10 months. However, due to the fluid flow effects mentioned above, the observed precession is not very regular.
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CM: Classical Mechanics
L Mln g .
(4.70)
3
Despite the apparent simplicity of this (exact!) equation, its analysis is straightforward only in the limit when the top is spinning about its symmetry axis n3 with a very high angular velocity rot. In this case, we may neglect the contribution to L due to a relatively small precession velocity pre (still to be calculated), and use Eq. (26) to write
L I ω I n .
(4.71)
3
3
rot
3
Then Eq. (70) shows that the vector L is perpendicular to both n3 (and hence L) and g, i.e. lies within a horizontal plane and is perpendicular to the horizontal component L xy of the vector L – see Fig. 9b.
Since, according to Eq. (70), the magnitude of this vector is constant, L = Mgl sin, the vector L (and hence the body’s main axis) rotates about the vertical axis with the following angular velocity: Torque-induced
L
Mgl sin
Mgl
Mgl
precession:
.
(4.72)
pre
fast-rotation
L
L sin
L
I
xy
3
rot
limit
Thus, rather counter-intuitively, the fast-rotating top does not follow the external, vertical force and, in addition to fast spinning about the symmetry axis n3, performs a revolution, called the torque-induced precession, about the vertical axis.18 Note that, similarly to the free-precession frequency (59), the torque-induced precession frequency (72) does not depend on the initial (and sustained) angle .
However, the torque-induced precession frequency is inversely (rather than directly) proportional to rot .
This fact makes the above simple theory valid in many practical cases. Indeed, Eq. (71) is quantitatively valid if the contribution of the precession into L is relatively small: Ipre << I 3rot, where I is a certain effective moment of inertia for the precession – to be calculated below. Using Eq. (72), this condition may be rewritten as
1/ 2
MglI
.
(4.73)
rot
2
I 3
According to Eq. (16), for a body of not too extreme proportions, i.e. with all linear dimensions of the same length scale l, all inertia moments are of the order of Ml 2, so that the right-hand side of Eq. (73) is of the order of ( g/ l)1/2, i.e. comparable with the frequency of small oscillations of the same body as the physical pendulum at the absence of its fast rotation.
To develop a quantitative theory that would be valid beyond such approximate treatment, the Euler equations (66) may be used, but are not very convenient. A better approach, suggested by the same L. Euler, is to introduce a set of three independent angles between the principal axes {n1, n2, n3}
bound to the rigid body, and the axes {n x, n y, n z} of an inertial reference frame (Fig. 10), and then express the basic equation (33) of rotation, via these angles. There are several possible options for the definition of such angles; Fig. 10 shows the set of Euler angles, most convenient for analyses of fast rotation.19 As one can see, the first Euler angle, , is the usual polar angle measured from the n z-axis to the n3-axis. The second one is the azimuthal angle , measured from the n x-axis to the so-called line of nodes formed by the intersection of planes [n x, n y] and [n1, n2]. The last Euler angle, , is measured 18 A semi-quantitative interpretation of this effect is a very useful exercise, highly recommended to the reader.
19 Of the several choices more convenient in the absence of fast rotation, the most common is the set of so-called Tait-Brian angles (called the yaw, pitch, and roll), which are broadly used for aircraft and maritime navigation.
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CM: Classical Mechanics
Euler
angles
within the plane [n1, n2], from the line of nodes to the n1-axis. For example, in the simple picture of slow force-induced precession of a symmetric top, that was discussed above, the angle is constant, the angle changes rapidly, with the rotation velocity rot, while the angle evolves with the precession frequency pre (72).
n z
n3
n2
plane [n
“line of
1, n2]
nodes"
plane [n x, n y]
O
n
y
n1
Fig. 4.10. Definition of
n x
the Euler angles.
Now we can express the principal-axes components of the instantaneous angular velocity vector,
1, 2, and 3, as measured in the lab reference frame, in terms of the Euler angles. This may be readily done by calculating, from Fig. 10, the contributions of the Euler angles’ evolution to the rotation about each principal axis, and then adding them up:
sin sin cos ,
1
via
sin cos sin ,
(4.74) Euler
2
angles
cos .
3
These relations enable the expression of the kinetic energy of rotation (25) and the angular momentum components (26) via the generalized coordinates , , and and their time derivatives (i.e.
the corresponding generalized velocities), and then using the powerful Lagrangian formalism to derive their equations of motion. This is especially simple to do in the case of symmetric tops (with I 1 = I 2), because plugging Eqs. (74) into Eq. (25) we get an expression,
I
I
1
T
sin
cos ,
(4.75)
rot
2 2 2 3
2
2
2
which does not include explicitly either or . (This reflects the fact that for a symmetric top we can always select the n1-axis to coincide with the line of nodes, and hence take = 0 at the considered moment of time. Note that this trick does not mean we can take 0 , because the n1-axis, as observed from an inertial reference frame, moves!) Now we should not forget that at the torque-induced precession, the center of mass moves as well (see, e.g., Fig. 9), so that according to Eq. (14), the total kinetic energy of the body is the sum of two terms,
M
M
2
2
T T T ,
T
V
l sin
,
(4.76)
rot
tran
tran
2 2 2
2
2
while its potential energy is just
U Mgl cos const .
(4.77)
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Now we could readily use Eqs. (2.19) to write the Lagrange equations of motion for the Euler angles, but it is simpler to immediately notice that according to Eqs. (75)-(77), the Lagrangian function, T – U, does not depend explicitly on the “cyclic” coordinates and , so that the corresponding generalized momenta (2.31) are conserved:
T
p
I sin2 I ( cos )cos const,
(4.78)
A
3
T
p
I ( cos ) const,
(4.79)
3
where I A I 1 + Ml 2. (According to Eq. (29), I A is just the body’s moment of inertia for rotation about a horizontal axis passing through the support point A . ) According to the last of Eqs. (74), p is just L 3, i.e.
the angular momentum’s component along the precessing axis n3. On the other hand, by its very definition (78), p is Lz, i.e. the same vector L’s component along the stationary axis z. (Actually, we could foresee in advance the conservation of both these components of L for our system, because the vector (69) of the external torque is perpendicular to both n3 and nz.) Using this notation, and solving the simple system of two linear equations (78)-(79) for the angle derivatives, we get
L L cos
L
L L cos
z
3
,
3
z
3
cos .
(4.80)
I sin 2
I
I sin 2
A
3
A
One more conserved quantity in this problem is the full mechanical energy20
I
I
A
E T U
2 2
sin2 3
cos 2 Mgl cos.
(4.81)
2
2
Plugging Eqs. (80) into Eq. (81), we get a first-order differential equation for the angle , which may be represented in the following physically transparent form:
I
L L
L
A
2
(
cos )2
2
U ( ) E,
U ( )
z
3
3
Mgl cos const .
(4.82)
2
ef
ef
2 I sin 2
2 I
A
3
Thus, similarly to the planetary problems considered in Sec. 3.4, the torque-induced precession of a symmetric top has been reduced (without any approximations!) to a 1D problem of the motion of just one of its degrees of freedom, the polar angle , in the effective potential U ef(). According to Eq.
(82), very similar to Eq. (3.44) for the planetary problem, this potential is the sum of the actual potential energy U given by Eq. (77), and a contribution from the kinetic energy of motion along two other angles. In the absence of rotation about the axes nz and n3 (i.e., L z = L 3 = 0), Eq. (82) is reduced to the first integral of the equation (40) of motion of a physical pendulum, with I’ = I A. If the rotation is present, then (besides the case of very special initial conditions when (0) = 0 and Lz = L 3),21 the first contribution to U ef() diverges at 0 and , so that the effective potential energy has a minimum at some non-zero value 0 of the polar angle – see Fig. 11.
20 Indeed, since the Lagrangian does not depend on time explicitly, H = const, and since the full kinetic energy T
(75)-(76) is a quadratic-homogeneous function of the generalized velocities, we have E = H.
21 In that simple case, the body continues to rotate about the vertical symmetry axis: ( t) = 0. Note, however, that such motion is stable only if the spinning speed is sufficiently high – see Eq. (85) below.
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2
2
th
1.5
2
.
1
0
.
1
U
ef
8
.
0
Fig. 4.11. The effective potential energy
const 1
Mgl
U ef of the symmetric top, given by Eq.
(82), as a function of the polar angle ,
for a particular value (0.95) of the ratio r
0.5
0
Lz/ L 3 (so that at rot >> th, 0 = cos-1 r
cos
0.1011), and several values of the
ratio rot/th – see Eq. (85).
00
0.1
0.2
0.3
0.4
/
If the initial angle (0) is equal to this value 0, i.e. if the initial effective energy is equal to its minimum value U ef(0), the polar angle remains constant through the motion: ( t) = 0. This corresponds to the pure torque-induced precession whose angular velocity is given by the first of Eqs. (80): L L cos
z
3
0
.
(4.83)
pre
I sin 2
A
0
The condition for finding 0, dU ef/ d = 0, is a transcendental algebraic equation that cannot be solved analytically for arbitrary parameters. However, in the high spinning speed limit (73), this is possible.
Indeed, in this limit the Mgl- proportional contribution to U ef is small, and we may analyze its effect by successive approximations. In the 0th approximation, i.e. at Mgl = 0, the minimum of U ef is evidently achieved at cos0 = Lz/ L 3, turning the precession frequency (83) to zero. In the next, 1st approximation, we may require that at = 0, the derivative of the first term of Eq. (82) for U ef over cos, equal to –
Lz( Lz – L 3cos)/ I Asin2,22 is canceled with that of the gravity-induced term, equal to Mgl. This immediately yields pre = ( Lz – L 3cos0) /I Asin20 = Mgl/ L 3, so that by identifying rot with 3 L 3/ I 3
(see Fig. 8), we recover the simple expression (72).
The second important result that may be readily obtained from Eq. (82) is the exact expression for the threshold value of the spinning speed for a vertically rotating top ( = 0, Lz = L 3). Indeed, in the limit 0 this expression may be readily simplified:
2
L
U ( )
Mgl
const
3
2
.
(4.84)
ef
8 I
2
A
This formula shows that if rot L 3/ I 3 is higher than the following threshold value, 1/ 2
MglI
Threshold
2
A
,
(4.85) rotation
th
2
I
speed
3
22 Indeed, the derivative of the fraction 1/2 I Asin2, taken at the point cos = Lz/ L 3, is multiplied by the numerator, ( Lz – L 3cos)2, which turns to zero at this point.
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then the coefficient at 2 in Eq. (84) is positive, so that U ef has a stable minimum at 0 = 0. On the other hand, if 3 is decreased below th, the fixed point becomes unstable, so that the top falls. As the plots in Fig. 11 show, Eq. (85) for the threshold frequency works very well even for non-zero but small values of the precession angle 0. Note that if we take I = I A in the condition (73) of the approximate treatment, it acquires a very simple sense: rot >> th.
Finally, Eqs. (82) give a natural description of one more phenomenon. If the initial energy is larger than U ef(0), the angle oscillates between two classical turning points on both sides of the fixed point 0 – see Fig. 11 again. The law and frequency of these oscillations may be found exactly as in Sec.
3.3 – see Eqs. (3.27) and (3.28). At 3 >> th, this motion is a fast rotation of the symmetry axis n3 of the body about its average position performing the slow torque-induced precession. Historically, these oscillations are called nutations, but their physics is similar to that of the free precession that was analyzed in the previous section, and the order of magnitude of their frequency is given by Eq. (59).
It may be proved that small friction (not taken into account in the above analysis) leads first to decay of these nutations, then to a slower drift of the precession angle 0 to zero, and finally, to a gradual decay of the spinning speed rot until it reaches the threshold (85) and the top falls.
4.6. Non-inertial reference frames
Now let us use the results of our analysis of the rotation kinematics in Sec. 1 to complete the discussion of the transfer between two reference frames, which was started in the introductory Chapter 1. As Fig. 12 (which reproduces Fig. 1.2 in a more convenient notation) shows, even if the “moving”
frame 0 rotates relative to the “lab” frame 0 ’, the radius vectors observed from these two frames are still related, at any moment of time, by the simple Eq. (1.5). In our new notation:
r ' r r .
(4.86)
0
particle
r '
r
“lab”
“moving”
frame
frame
'
0
Fig. 4.12. The general case of transfer
r0
0
between two reference frames.
However, as was mentioned in Sec. 1, the general addition rule for velocities is already more complex. To find it, let us differentiate Eq. (86) over time:
d
d
d
r '
r
.
r
(4.87)
0
dt
dt
dt
The left-hand side of this relation is evidently the particle’s velocity as measured in the lab frame, and the first term on the right-hand side is the velocity v0 of point 0, as measured in the same lab frame. The last term is more complex: due to the possible mutual rotation of the frames 0 and 0 ’, that term may not vanish even if the particle does not move relative to the rotating frame 0 – see Fig. 12.
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Fortunately, we have already derived the general Eq. (8) to analyze situations exactly like this one. Taking A = r in it, we may apply the result to the last term of Eq. (87), to get Transformation
v
v
(v ω r),
(4.88) of
lab
in
0
lab
in
velocity
where is the instantaneous angular velocity of an imaginary rigid body connected to the moving reference frame (or we may say, of this frame as such), as measured in the lab frame 0’, while v is dr/ dt as measured in the moving frame 0 . The relation (88), on one hand, is a natural generalization of Eq.
(10) for v 0; on the other hand, if = 0, it is reduced to simple Eq. (1.8) for the translational motion of the frame 0.
To calculate the particle’s acceleration, we may just repeat the same trick: differentiate Eq. (88) over time, and then use Eq. (8) again, now for the vector A = v + r. The result is d
a
a
(v ω r) ω (v ω r).
(4.89)
lab
in
0
lab
in
dt
Carrying out the differentiation in the second term, we finally get the goal relation,
Transformation
a
a
a ω r 2ω v ω (ω r) , (4.90) of
lab
in
0
lab
in
acceleration
where a is particle’s acceleration as measured in the moving frame. This result is a natural generalization of the simple Eq. (1.9) to the rotating frame case.
Now let the lab frame 0 ’ be inertial; then the 2nd Newton’s law for a particle of mass m is a
m
F ,
(4.91)
lab
in
where F is the vector sum of all forces exerted on the particle. This is simple and clear; however, in many cases it is much more convenient to work in a non-inertial reference frame. For example, when describing most phenomena on the Earth’s surface, it is rather inconvenient to use a reference frame bound to the Sun (or to the galactic center, etc.). In order to understand what we should pay for the convenience of using a moving frame, we may combine Eqs. (90) and (91) to write
Inertial
a
m F a
m
ω
m (ω r) 2 ω
m v ω
m .
r
(4.92) “forces”
0
lab
in
This result means that if we want to use an analog of the 2nd Newton’s law in a non-inertial reference frame, we have to add, to the actual net force F exerted on a particle, four pseudo-force terms, called inertial forces, all proportional to the particle’s mass. Let us analyze them one by one, always remembering that these are just mathematical terms, not actual physical forces. (In particular, it would be futile to seek a 3rd-Newton’s-law counterpart for any inertial force.)
The first term, – ma0in lab, is the only one not related to rotation and is well known from undergraduate mechanics. (Let me hope the reader remembers all these weight-in-the-accelerating-elevator problems.) However, despite its simplicity, this term has more subtle consequences. As an example, let us consider, semi-qualitatively, the motion of a planet, such as our Earth, orbiting a star and also rotating about its own axis – see Fig. 13. The bulk-distributed gravity forces, acting on a planet from its star, are not quite uniform, because they obey the 1/ r 2 gravity law (1.15), and hence are equivalent to a single force applied to a point A slightly offset from the planet’s center of mass 0, toward Chapter 4
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the star. For a spherically-symmetric planet, the direction from 0 to A would be exactly aligned with the direction toward the star. However, real planets are not absolutely rigid, so due to the centrifugal “force”
(to be discussed momentarily), the rotation about their own axis makes them slightly ellipsoidal – see Fig. 13. (For our Earth, this equatorial bulge is about 10 km.) As a result, the net gravity force is slightly offset from the direction toward the center of mass 0. On the other hand, repeating all the arguments of this section for a body (rather than a point), we may see that, in the reference frame moving with the planet, the inertial force – Ma0 (with the magnitude of the total gravity force, but directed from the star) is applied exactly to the center of mass. As a result, this pair of forces creates a torque perpendicular to both the direction toward the star and the vector 0A. (In Fig. 13, the torque vector is perpendicular to the plane of the drawing). If the angle between the planet’s “polar” axis of rotation and the direction towards the star was fixed, then, as we have seen in the previous section, this torque would induce a slow axis precession about that direction.
L ω
I
rot
pre
( t)
direction toward the star
- Ma F
-
0
g
F a
M
A 0
g
0
Fig. 4.13. The axial precession
equator
of a planet (with the equatorial
bulge and the 0A-offset
polar axis
polar axis
strongly exaggerated).
in “summer”
in “winter”
However, as a result of the orbital motion, the angle oscillates in time much faster (once a year) between values (/2 + ) and (/2 – ), where is the axis tilt, i.e. angle between the polar axis (the direction of vectors L and rot) and the normal to the ecliptic plane of the planet’s orbit. (For the Earth, 23.4.) A straightforward averaging over these fast oscillations23 shows that the torque leads to the polar axis’ precession about the axis perpendicular to the ecliptic plane, keeping the angle
constant – see Fig. 13. For the Earth, the period T pre = 2/pre of this precession of the equinoxes, corrected to a substantial effect of the Moon’s gravity, is close to 26,000 years.24
Returning to Eq. (92), the direction of the second term of its right-hand side,
Centrifugal
“force”
F ω
m
,
(4.93)
cf
ωr
called the centrifugal force, is always perpendicular to, and directed out of the instantaneous rotation axis – see Fig. 14. Indeed, the vector r is perpendicular to both and r (in Fig. 14, normal to the drawing plane and directed from the reader) and has the magnitude r sin = , where is the distance of the particle from the rotation axis. Hence the outer vector product, with the account of the minus sign, is normal to the rotation axis , directed from this axis, and is equal to 2 r sin = 2. The centrifugal
“force” is of course just the result of the fact that the centripetal acceleration 2, explicit in the inertial reference frame, disappears in the rotating frame. For a typical location of the Earth ( ~ R E 6106 m), 23 Details of this calculation may be found, e.g., in Sec. 5.8 of the textbook by H. Goldstein et al., Classical Mechanics, 3rd ed., Addison Wesley, 2002.
24 This effect is known from antiquity, apparently discovered by Hipparchus of Rhodes (190-120 BC).
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with its angular velocity E 10-4 s-1, the acceleration is rather considerable, of the order of 3 cm/s2, i.e.
~0.003 g, and is responsible, in particular, for the largest part of the equatorial bulge mentioned above.
ω
m
ω
m (ω r)
r
Fig. 4.14. The centrifugal force.
0
As an example of using the centrifugal force concept, let us return again to our “testbed”
problem on the bead sliding along a rotating ring – see Fig. 2.1. In the non-inertial reference frame attached to the ring, we have to add, to the actual forces mg and N exerted on the bead, the horizontal centrifugal force25 directed from the rotation axis, with the magnitude m2. Its component tangential to the ring equals ( m2)cos = m2 R sincos, and hence the component of Eq. (92) along this direction is ma = – mg sin + m2 R sincos. With a
R , this gives us an equation of motion equivalent to Eq.
(2.25), which had been derived in Sec. 2.2 (in the inertial frame) using the Lagrangian formalism.
The third term on the right-hand side of Eq. (92),
F 2 ω
m v ,
(4.94) Coriolis
C
“force”
is the so-called Coriolis force,26 which is different from zero only if the particle moves in a rotating reference frame. Its physical sense may be understood by considering a projectile fired horizontally, say from the North Pole – see Fig. 15.
E
ω
Fig. 4.15. The trajectory of a projectile fired
horizontally from the North Pole, from the
F 2 mω v
C
point of view of an Earth-bound observer
looking down. The circles show parallels,
v
while the straight lines mark meridians.
d
From the point of view of an Earth-based observer, the projectile will be affected by an additional Coriolis force (94), directed westward, with the magnitude 2 mE v, where v is the main, southward component of the velocity. This force would cause the westward acceleration a = 2E v, and hence the westward deviation growing with time as d = at 2/2 = E vt 2. (This formula is exact only if d is much smaller than the distance r = vt passed by the projectile.) On the other hand, from the point of 25 For this problem, all other inertial “forces”, besides the Coriolis force (see below) vanish, while the latter force is directed normally to the ring and does not affect the bead’s motion along it.
26 Named after G.-G. de Coriolis (already reverently mentioned in Chapter 1) who described its theory and applications in detail in 1835, though the first semi-quantitative analyses of this effect were given by Giovanni Battista Riccioli and Claude François Dechales already in the mid-1600s, and all basic components of the Coriolis theory may be traced to a 1749 work by Leonard Euler.
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view of an inertial-frame observer, the projectile’s trajectory in the horizontal plane is a straight line.
However, during the flight time t, the Earth’s surface slips eastward from under the trajectory by the distance d = r = ( vt)(E t) = E vt 2, where = E t is the azimuthal angle of the Earth’s rotation during the flight). Thus, both approaches give the same result – as they should.
Hence, the Coriolis “force” is just a fancy (but frequently very convenient) way of description of a purely geometric effect pertinent to the rotation, from the point of view of the observer participating in it. This force is responsible, in particular, for the higher right banks of rivers in the Northern hemisphere, regardless of the direction of their flow – see Fig. 16. Despite the smallness of the Coriolis force (for a typical velocity of the water in a river, v ~ 1 m/s, it is equivalent to acceleration a C ~ 10-2
cm/s2 ~ 10-5 g), its multi-century effects may be rather prominent.27
ωE
N
F
F
C
v
C
v
v
0
Fig. 4.16. Coriolis forces due to the
Earth’s rotation, in the Northern
hemisphere.
S
Finally, the last, fourth term of Eq. (92), mω r , exists only when the rotation frequency changes in time, and may be interpreted as a local-position-specific addition to the first term.
The key relation (92), derived above from Newton’s equation (91), may be alternatively obtained from the Lagrangian approach. Indeed, let us use Eq. (88) to represent the kinetic energy of the particle in an inertial “lab” frame in terms of v and r measured in a rotating frame:
m
T
v
(v ω r) ,
(4.95)
0
2
lab
in
2
and use this expression to calculate the Lagrangian function. For the relatively simple case of a particle’s motion in the field of potential forces, measured from a reference frame that performs a pure rotation (so that v0in lab = 0)28 with a constant angular velocity , we get
m
m
m
2
L T U
v mv (ω r)
ωr2
2
U
v mv (ω r) U ,
(4.96a)
ef
2
2
2
where the effective potential energy,29
27 The same force causes the counterclockwise circulation in the “Nor’easter” storms on the US East Coast, with the radial component of the air velocity directed toward the cyclone’s center, due to lower pressure in its middle.
28 A similar analysis of the cases with v0in lab 0, for example, of a translational relative motion of the reference frames, is left for the reader’s exercise.
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m
U U U ,
with U
ω r
(4.96b)
ef
cf
cf
2,
2
is just the sum of the actual potential energy U of the particle and the so-called centrifugal potential energy, associated with the centrifugal “force” (93):
m
F U
ω r
ω
m ω r
(4.97)
cf
cf
2
(
).
2
It is straightforward to verify that the Lagrange equations (2.19), derived from Eqs. (96) considering the Cartesian components of r and v as generalized coordinates and velocities, coincide with Eq. (92) (with a0in lab = 0, ω = 0, and F = – U).
Now it is very informative to have a look at a by-product of this calculation, the generalized momentum (2.31) corresponding to the particle’s coordinate r as measured in the rotating reference frame,30
Canonical
L
momentum
p
m v ω r.
(4.98)
v
at rotation
According to Eq. (88) with v0in lab = 0, the expression in the parentheses is just vin lab. However, from the point of view of the moving frame, i.e. not knowing about the simple physical sense of the vector p, we would have a reason to speak about two different linear momenta of the same particle, the so-called kinetic momentum p = mv and the canonical momentum p = p + mr.31 Let us calculate the Hamiltonian function H defined by Eq. (2.32), and the energy E as functions of the same moving-frame variables:
3
2
L
m
mv
H
v L
p v L m v ω r v
v ω r
, (4.99)
j
2
v
m (
) U
U
ef
ef
j 1
v
2
2
j
m
m
m
2
E T U
v v
m (ω r)
ωr2
2
U
v U v
m (ω r) mω r2 . (4.100)
ef
2
2
2
These expressions clearly show that E and H are not equal.32 In hindsight, this is not surprising, because the kinetic energy (95), expressed in the moving-frame variables, includes a term linear in v, and hence 29 For the attentive reader who has noticed the difference between the negative sign in the expression for U cf, and the positive sign before the similar second term in Eq. (3.44): as was already discussed in Chapter 3, it is due to the difference of assumptions. In the planetary problem, even though the angular momentum L and hence its component Lz are fixed, the corresponding angular velocity is not. On the opposite, in our current discussion, the angular velocity of the reference frame is assumed to be fixed, i.e. is independent of r and v.
30 Here L/v is just a shorthand for a vector with Cartesian components L/ vj. In a more formal language, this is the gradient of the scalar function L in the velocity space.
31 A very similar situation arises at the motion of a particle with electric charge q in magnetic field B. In that case, the role of the additional term p – p = mr is played by the product q A, where A is the vector potential of the field B = A – see, e.g., EM Sec. 9.7, and in particular Eqs. (9.183) and (9.192).
32 Please note the last form of Eq. (99), which shows the physical sense of the Hamiltonian function of a particle in the rotating frame very clearly, as the sum of its kinetic energy (as measured in the moving frame), and the effective potential energy (96b), including that of the centrifugal “force”.
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is not a quadratic-homogeneous function of this generalized velocity. The difference between these functions may be represented as
E H v
m (ω r) mω r2 mv ω r (ω r) v m
(ω r) .
(4.101)
lab
in
Now using the operand rotation rule again, we may transform this expression into a simpler form:33
E – H
E H ω
at rotation
r v
m
ω r p ω L
.
(4.102)
lab
in
lab
in
As a sanity check, let us apply this general expression to the particular case of our testbed problem – see Fig. 2.1. In this case, the vector is aligned with the z- axis, so that of all Cartesian components of the vector L, only the component Lz is important for the scalar product in Eq. (102). This component evidently equals Iz = m2 = m( R sin)2, so that E H 2 2
2
m
R sin ,
(4.103)
i.e. the same result that follows from the subtraction of Eqs. (2.40) and (2.41).
4.7. Exercise problems
4.1. Calculate the principal moments of inertia for the following uniform rigid bodies:
(i)
(ii)
(iii)
(iv)
R
R
R
R
(i) a thin, planar, round hoop, (ii) a flat round disk, (iii) a thin spherical shell, and (iv) a solid sphere.
Compare the results, assuming that all the bodies have the same radius R and mass M, and give an interpretation of their difference.
4.2. Calculate the principal moments of inertia for the rigid bodies shown in the figure below: (i)
(ii)
(iii)
a
a
a
a
a
a
a
a
a
a
a
a
(i) an equilateral triangle made of thin rods with a uniform linear mass density ,
(ii) a thin plate in the shape of an equilateral triangle, with a uniform areal mass density , and 33 Note that by the definition (1.36), the angular momenta L of particles merely add up. As a result, the final form of Eq. (102) is valid for an arbitrary system of particles.
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(iii) a tetrahedron with a uniform bulk mass density .
Assuming that the total mass of the three bodies is the same, compare the results and give an interpretation of their difference.
4.3. Calculate the principal moments of inertia of a thin uniform plate cut in the form of a right triangle with two /4 angles.
4.4. Prove that Eqs. (34)-(36) are valid for rotation of a rigid body about a fixed axis z, even if it does not pass through its center of mass.
f
4.5. Calculate the kinetic energy of a right circular cone with height H,
base radius R, and a constant mass density , that rolls over a horizontal H
surface without slippage, making f turns per second about the vertical axis –
R
see the figure on the right.
4.6. External forces exerted on a rigid body, rotating with an angular velocity , have zero vector sum but a non-vanishing net torque about its
center of mass. Calculate the work of the forces on
the body per unit time, i.e. their instantaneous
power. Prove that the same result is valid for a
body rotating about a fixed axis and the torque’s
component along this axis. Use the last result to
prove that at negligible friction, the differential
gear assembly (see the figure on the right)34
distributes the external torque, applied to its
satellite-carrier axis to rotate it about the common
axis of two axle shafts, equally to both shafts, even
if they rotate with different angular velocities.
m'
4.7. The end of a uniform thin rod of length 2 l and mass m, initially at rest, v 0
is hit by a bullet of mass m' , flying with velocity v
m
0 (see the figure on the right),
which gets stuck in the rod. Use two different approaches to calculate the velocity
of the opposite end of the rod right after the collision.
l
2
4.8. A ball of radius R, initially at rest on a horizontal table, is hit
F t
with a billiard cue in the horizontal direction, at height h above the table –
see the figure on the right. Using the same Coulomb approximation for the
0
friction force between the ball and the table as in Problem 1.4 ( Fmax = N), h
R
calculate the final linear velocity of the rolling ball as a function of h.
Would it matter if the hit point is shifted horizontally (normally to the plane
of the drawing)?
34 Figure from G. Antoni, Sci. World J., 2014, 523281 (2014), adapted with permission. Both satellite gears may rotate freely about their carrier axis.
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Hint: As in most solid body collision problems, during the short time of the cue hit, all other forces exerted on the ball may be considered negligibly small.
4.9. A round cylinder of radius R and mass M may roll, without slippage, over a horizontal surface. The mass density distribution inside the cylinder is not uniform, so that its center of mass is at some distance l 0 from its geometrical axis, and the moment of inertia I (for rotation about the axis parallel to the symmetry axis but passing through the center of mass) is different from MR 2/2, where M
is the cylinder’s mass. Derive the equation of motion of the cylinder under the effect of the uniform vertical gravity field, and use it to calculate the frequency of its small oscillations of the cylinder near its stable equilibrium position.
4.10. A body may rotate about a fixed horizontal axis A – see Fig. 5. Find the frequency of its small oscillations, in a uniform gravity field, as a function of the distance l of the axis from the body’s center of mass 0, and analyze the result.
4.11. Calculate the frequency, and sketch the mode of oscillations of
M , R
a round uniform cylinder of radius R and the mass M, that may roll, without slipping, on a horizontal surface of a block of mass M’. The block, in turn,
M'
may move in the same direction, without friction, on a horizontal surface,
being connected to it with an elastic spring – see the figure on the right.
l'
4.12. A thin uniform bar of mass M and length l is hung on a light thread of length
l’ (like a “chime” bell – see the figure on the right). Derive the equations of motion of the M
system within the plane of the drawing.
l
g
4.13. A solid, uniform, round cylinder of mass M can roll,
R
without slipping, over a concave, round cylindrical surface of a block
R'
of mass M’, in a uniform gravity field – see the figure on the right.
M
g
The block can slide without friction on a horizontal surface. Using the
M'
Lagrangian formalism,
(i) find the frequency of small oscillations of the system near the equilibrium, and
(ii) sketch the oscillation mode for the particular case M’ = M, R’ = 2 R.
4.14. A uniform solid hemisphere of radius R is placed on a
R
horizontal plane – see the figure on the right. Find the frequency of its
'
0
small oscillations within a vertical plane, for two ultimate cases:
(i) there is no friction between the hemisphere and plane surfaces;
g
(ii) the static friction is so strong that there is no slippage between
these surfaces.
4.15. For the “sliding ladder” problem started in Sec. 3 (see Fig. 7), find the critical value c of the angle at that the ladder loses its contact with the vertical wall, assuming that it starts sliding from the vertical position, with a negligible initial velocity.
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a' ?
4.16. Six similar, uniform rods of length l and mass m are connected by
light joints so that they may rotate, without friction, versus each other, forming l, m a planar polygon. Initially, the polygon was at rest, and had the correct hexagon
shape – see the figure on the right. Suddenly, an external force F is applied to
the middle of one rod, in the direction of the hexagon’s symmetry center.
a ?
Calculate the accelerations: of the rod to which the force is applied ( a), and of the opposite rod ( a’), immediately after the application of the force.
F
ω
4.17. A rectangular cuboid (parallelepiped) with sides a
a
1, a 2, and a 3,
1
a
made of a material with a constant mass density , is rotated, with a constant
2
angular velocity , about one of its space diagonals – see the figure on the a
right. Calculate the torque necessary to sustain such rotation.
3
4.18. A uniform round ball moves, without slippage, over a “turntable”: a horizontal plane rotated about a vertical axis with a time-independent angular velocity . Derive a self-consistent equation of motion of the ball’s center of mass, and discuss its solutions.
4.19. Calculate the free precession frequency of a flat round disk rotating with an angular velocity about a direction very close to its symmetry axis, from the point of view of: (i) an observer rotating with the disk, and
(ii) a lab-based observer.
4.20. Use the Euler equations to prove the fact mentioned in Sec. 4: free rotation of an arbitrary body (“asymmetric top”) about its principal axes with the smallest and largest moments of inertia is stable with respect to small variations of initial conditions, while that about the intermediate- Ij axis is not. Illustrate the same fact using the Poinsot construction.
4.21. Give an interpretation of the torque-induced precession, explaining its direction, using any simple system exhibiting this effect, as a model.
4.22. One end of a light shaft of length l is firmly
attached to the center of a thin uniform solid disk of radius R <<
R, M
l and mass M, whose plane is perpendicular to the shaft. Another
end of the shaft is attached to a vertical axis (see the figure on
l
the right) so that the shaft may rotate about the axis without
g
friction. The disk rolls, without slippage, over a horizontal
surface, so that the whole system rotates about the vertical axis with a constant angular velocity .
Calculate the (vertical) supporting force N exerted on the disk by the surface.
4.23. A coin of radius r is rolled over a horizontal surface, without
slippage. Due to its tilt , it rolls around a circle of radius R – see the figure on the right. Modeling the coin as a very thin round disk, calculate the time period
g 2 r
of its motion around the circle.
0
R
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4.24. A symmetric top on point support (as shown see, e.g., Fig. 9), rotating around its symmetry axis with high angular velocity rot, is subjected to not only its weight Mg, but also an additional force also applied to the top’s center of mass, but directed normally to g, with its vector rotating in the horizontal plane with a constant angular velocity << rot. Derive the system of equations describing the top’s motion. Analyze their solution for the simplest case when is exactly equal to the frequency (72) of the torque-induced precession in the gravity field alone.
4.25. Analyze the effect of small friction on the fast rotation of a symmetric top around its symmetry axis, using a simple model in that the lower end of the body is a right cylinder of radius R.
4.26. An air-filled balloon is placed inside a water-filled container, which moves by inertia in free space, at negligible gravity. Suddenly, force F is applied to the container, pointing in a certain direction. What direction does the balloon move relative to the container?
4.27. Two planets are in a circular orbit around their common center of mass. Calculate the effective potential energy of a much lighter body (say, a spacecraft) rotating with the same angular velocity, on the line connecting the planets. Sketch the plot of the radial dependence of U ef and find out the number of so-called Lagrange points is which the potential energy has local maxima. Calculate their position explicitly in the limit when one of the planets is much more massive than the other one.
4.28. Besides the three Lagrange points L1, L2, and L3 discussed in the previous problem, which are located on the line connecting two planets on circular orbits about their mutual center of mass, there are two off-line points L4 and L5 – both within the pane of the planets’ rotation. Calculate their positions.
4.29. The following simple problem may give additional clarity to the physics of the Coriolis
“force”. Consider a bead of mass m, which may slide, without friction, along a straight rod that is rotated, within a horizontal plane with a constant angular velocity – see the figure on the right.
Calculate the bead’s linear acceleration and the force N exerted on it by the rod, in: (i) an inertial (“lab”) reference frame, and
(ii) the non-inertial reference frame rotating with the rod (but not moving with the bead), and compare the results.
4.30. Analyze the dynamics of the famous Foucault pendulum, used for spectacular demonstrations of the Earth’s rotation. In particular, calculate the angular velocity of the rotation of its oscillation plane relative to the Earth’s surface, at the location with the polar angle (“colatitude”) .
Assume that the pendulum oscillation amplitude is small enough to neglect nonlinear effects, and that its oscillation period is much shorter than 24 hours.
4.31. A small body is dropped down to the surface of Earth from height h << R E, without initial velocity. Calculate the magnitude and direction of its deviation from the vertical, due to the Earth rotation. Estimate the effect’s magnitude for a body dropped from the Empire State Building.
4.32. Calculate the height of solar tides on a large ocean, using the following simplifying assumptions: the tide period (½ of Earth's day) is much longer than the period of all ocean waves, the Chapter 4
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Earth (of mass M E) is a sphere of radius R E, and its distance r S from the Sun (of mass M S) is constant and much larger than R E.
4.33. A satellite is on a circular orbit of radius R, around the Earth.
(i) Write the equations of motion of a small body as observed from the satellite, and simplify them for the case when the motion is limited to the satellite’s close vicinity.
(ii) Use the equations to prove that the body may be placed on an elliptical trajectory around the satellite’s center of mass, within its plane of rotation about Earth. Calculate the ellipse’s orientation and eccentricity.
4.34. A non-spherical shape of an artificial satellite may ensure its stable angular orientation relative to Earth’s surface, advantageous for many practical goals. Modeling the satellite as a strongly elongated, axially-symmetric body, moving around the Earth on a circular orbit of radius R, find its stable orientation.
4.35.* A rigid, straight, uniform rod of length l, with the lower end on a pivot, falls in a uniform gravity field – see the figure on the right. Neglecting friction, calculate the distribution of the bending torque along its length, and analyze the result.
l
Hint: The bending torque is the net torque of the force F acting between two parts of the rod, mentally separated by its cross-section, about a certain “neutral axis”.35 As will g
be discussed in detail in Sec. 7.5, at the proper definition of this axis, the bending torque’s gradient along the rod’s length is equal to (– F), where F is the rod-normal (“shear”) component of the force exerted by the top part of the rod on its lower part.
4.36. Let r be the radius vector of a particle, as measured in a possibly non-inertial but certainly non-rotating reference frame. Taking its Cartesian components for the generalized coordinates, calculate the corresponding generalized momentum p of the particle, and its Hamiltonian function H. Compare p
with mv, and H with the particle’s energy E. Derive the Lagrangian equation of motion in this approach, and compare it with Eq. (92).
35 Inadequate definitions of this torque are the main reason for numerous wrong solutions of this problem, posted online – readers beware!
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Chapter 5. Oscillations
In this course, oscillations and waves are discussed in detail, because of their importance for fundamental and applied physics. This chapter starts with a discussion of the harmonic oscillator, whose differential equation of motion is linear and hence allows the full analytical solution, and then proceeds to so-called “nonlinear” and “parametric” systems whose dynamics may be only explored by either approximate analytical or numerical methods.
5.1. Free and forced oscillations
In Sec. 3.2 we briefly discussed oscillations in a keystone Hamiltonian system – a 1D harmonic oscillator described by a very simple Lagrangian1
m
L T ( q) U ( q)
2
2
q q ,
(5.1)
2
2
whose Lagrange equation of motion,2
Harmonic
oscillator:
q
m q
,
0
i.e.
2
q q ,
0
with 2
0 ,
(5.2)
0
0
equation
m
is a linear homogeneous differential equation. Its general solution is given by Eq. (3.16), which is frequently recast into another, amplitude-phase form:
Harmonic
oscillator:
q t
( ) u cos t v sin t A cos
,
(5.3a)
0
0
t
0
motion
where A is the amplitude and the phase of the oscillations, which are determined by the initial conditions. Mathematically, it is frequently easier to work with sinusoidal functions as complex exponents, by rewriting the last form of Eq. (3a) in one more form:3
i( t )
0
i t 0
q t
( ) Re Ae
Re ae
,
(5.3b)
Real
and where a is the complex amplitude of the oscillations:
complex
amplitudes
a Aei ,
a ,
A
Re a A cos u,
Im a A sin .
v
(5.4)
For an autonomous, Hamiltonian oscillator, Eqs. (3) give the full classical description of its dynamics.
However, it is important to understand that this free-oscillation solution, with a constant amplitude A, 1 For the notation brevity, in this chapter I will drop indices “ef” in the energy components T and U, and in parameters like m, , etc. However, the reader should still remember that T and U do not necessarily coincide with the actual kinetic and potential energies (even if those energies may be uniquely identified) – see Sec. 3.1.
2 0 is usually called the own frequency of the oscillator. In quantum mechanics, the Germanized version of the same term, eigenfrequency, is used more. In this series, I will use either of the terms, depending on the context.
3 Note that this is the so-called physics convention. Most engineering texts use the opposite sign in the imaginary exponent, exp{- i t} exp{ i t}, with the corresponding sign implications for intermediate formulas, but (of course) similar final results for real variables.
© K. Likharev
CM: Classical Mechanics
means the conservation of the energy E T + U = A 2/2 of the oscillator. If its energy changes for any reason, the description needs to be generalized.
First of all, if the energy leaks out of the oscillator to its environment (the effect usually called the energy dissipation), the free oscillations decay with time. The simplest model of this effect is represented by an additional linear drag (or “kinematic friction”) force, proportional to the generalized velocity and directed opposite to it:
F
q ,
(5.5)
v
where constant is called the drag coefficient.4 The inclusion of this force modifies the equation of motion (2) to become
q
m q q
0 .
(5.6a)
This equation is frequently rewritten in the form
Free
q
oscillator
2 q
2
q ,
0
with
,
(5.6b)
0
2 m
with
damping
where the parameter is called the damping coefficient (or just “damping”). Note that Eq. (6) is still a linear homogeneous second-order differential equation, and its general solution still has the form of the sum (3.13) of two exponents of the type exp{ t}, with arbitrary pre-exponential coefficients. Plugging such an exponent into Eq. (6), we get the following algebraic characteristic equation for : 2
2
2
.
0
(5.7)
0
Solving this quadratic equation, we get
i ' ,
where '
(5.8)
0
0
2 2
0
1/2,
so that for not very high damping ( < 0) we get the following generalization of Eq. (3):5
t
q ( t)
c e
c e t u
't v
