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§37
The asymmetrical top
121
and by formulae (37.14) tan of = M1/M2, cos 0) 2(1 (M3/M) 22
substituting (2), we obtain
tan 4 = V[I(I3-I2)/I2(I3-I1)] cot wt,
(3)
To find , we note that, by the third formula (35.1), we have, for 0 1,
Hence
= lot
(4)
omitting an arbitrary constant of integration.
A clearer idea of the nature of the motion of the top is obtained if we consider the change
in direction of the three axes of inertia. Let n1, n2, n3 be unit vectors along these axes. The
vectors n1 and n2 rotate uniformly in the XY-plane with frequency So, and at the same time
execute small transverse oscillations with frequency w. These oscillations are given by the
Z-components of the vectors:
22 M1/M = av(I3/I2-1) cos wt,
N2Z 22 M2/M = av(I3/I1-1) sin wt.
For the vector n3 we have, to the same accuracy, N3x 22 0 sin , N3y 22 -0 cos , n3z 1.
(The polar angle and azimuth of n3 with respect to the axes X, Y, Z are 0 and -; see
the footnote to 35.) We also write, using formulae (37.13),
naz=0sin(Qot-4)
= Asin Sot cos 4-0 cos lot sin 4
= (M 2/M) sin Dot-(M1/M) cos Sot
sin Sot sin N/1-1) cos Not cos wt
cos(so
Similarly
From this we see that the motion of n3 is a superposition of two rotations about the Z-axis
with frequencies So + w.
PROBLEM 2. Determine the free rotation of a top for which M2 = 2EI2.
SOLUTION. This case corresponds to the movement of the terminus of M along a curve
through the pole on the x2-axis (Fig. 51). Equation (37.7) becomes ds/dr = 1-s2,
= S = I2/20, where So = M/I2 = 2E|M. Integration of
this equation and the use of formulae (37.6) gives
sech T,
}
(1)
sech T.
To describe the absolute motion of the top, we use Eulerian angles, defining 0 as the angle
between the Z-axis (direction of M) and the x2-axis (not the x3-axis as previously). In formulae
(37.14) and (37.16), which relate the components of the vector CA to the Eulerian angles, we
5
122
Motion of a Rigid Body
§38
must cyclically permute the suffixes 1, 2, 3 to 3, 1, 2. Substitution of (1) in these formulae
then gives cos 0 = tanh T, = lot + constant, tan =
It is seen from these formulae that, as t 8, the vector SC asymptotically approaches the
x2-axis, which itself asymptotically approaches the Z-axis.
$38. Rigid bodies in contact
The equations of motion (34.1) and (34.3) show that the conditions of
equilibrium for a rigid body can be written as the vanishing of the total force
and total torque on the body:
F = f = 0 ,
K ==~rxf=0. =
(38.1)
Here the summation is over all the external forces acting on the body, and r
is the radius vector of the "point of application"; the origin with respect to
which the torque is defined may be chosen arbitrarily, since if F = 0 the
value of K does not depend on this choice (see (34.5)).
If we have a system of rigid bodies in contact, the conditions (38.1) for
each body separately must hold in equilibrium. The forces considered must
include those exerted on each body by those with which it is in contact. These
forces at the points of contact are called reactions. It is obvious that the mutual
reactions of any two bodies are equal in magnitude and opposite in direction.
In general, both the magnitudes and the directions of the reactions are
found by solving simultaneously the equations of equilibrium (38.1) for all the
bodies. In some cases, however, their directions are given by the conditions
of the problem. For example, if two bodies can slide freely on each other, the
reaction between them is normal to the surface.
If two bodies in contact are in relative motion, dissipative forces of friction
arise, in addition to the reaction.
There are two possible types of motion of bodies in contact-sliding and
rolling. In sliding, the reaction is perpendicular to the surfaces in contact,
and the friction is tangential. Pure rolling, on the other hand, is characterised
by the fact that there is no relative motion of the bodies at the point of
contact; that is, a rolling body is at every instant as it were fixed to the point
of contact. The reaction may be in any direction, i.e. it need not be normal
to the surfaces in contact. The friction in rolling appears as an additional
torque which opposes rolling.
If the friction in sliding is negligibly small, the surfaces concerned are
said to be perfectly smooth. If, on the other hand, only pure rolling without
sliding is possible, and the friction in rolling can be neglected, the surfaces
are said to be perfectly rough.
In both these cases the frictional forces do not appear explicitly in the pro-
blem, which is therefore purely one of mechanics. If, on the other hand, the
properties of the friction play an essential part in determining the motion,
then the latter is not a purely mechanical process (cf. $25).
Contact between two bodies reduces the number of their degrees of freedom
as compared with the case of free motion. Hitherto, in discussing such
§38
Rigid bodies in contact
123
problems, we have taken this reduction into account by using co-ordinates
which correspond directly to the actual number of degrees of freedom. In
rolling, however, such a choice of co-ordinates may be impossible.
The condition imposed on the motion of rolling bodies is that the velocities
of the points in contact should be equal; for example, when a body rolls on a
fixed surface, the velocity of the point of contact must be zero. In the general
case, this condition is expressed by the equations of constraint, of the form
E caide = 0,
(38.2)
where the Cai are functions of the co-ordinates only, and the suffix a denumer-
ates the equations. If the left-hand sides of these equations are not the total
time derivatives of some functions of the co-ordinates, the equations cannot
be integrated. In other words, they cannot be reduced to relations between the
co-ordinates only, which could be used to express the position of the bodies
in terms of fewer co-ordinates, corresponding to the actual number of degrees
of freedom. Such constraints are said to be non-holonomic, as opposed to
holonomic constraints, which impose relations between the co-ordinates only.
Let us consider, for example, the rolling of a sphere on a plane. As usual,
we denote by V the translational velocity (the velocity of the centre of the
sphere), and by Sa the angular velocity of rotation. The velocity of the point
of contact with the plane is found by putting r = - an in the general formula
V = +SXR; a is the radius of the sphere and n a unit vector along the
normal to the plane. The required condition is that there should be no sliding
at the point of contact, i.e.
V-aSxxn = 0.
(38.3)
This cannot be integrated: although the velocity V is the total time derivative
of the radius vector of the centre of the sphere, the angular velocity is not in
general the total time derivative of any co-ordinate. The constraint (38.3) is
therefore non-holonomic.t
Since the equations of non-holonomic constraints cannot be used to reduce
the number of co-ordinates, when such constraints are present it is necessary
to use co-ordinates which are not all independent. To derive the correspond-
ing Lagrange's equations, we return to the principle of least action.
The existence of the constraints (38.2) places certain restrictions on the
possible values of the variations of the co-ordinates: multiplying equations
(38.2) by St, we find that the variations dqi are not independent, but are
related by
(38.4)
t It may be noted that the similar constraint in the rolling of a cylinder is holonomic. In
that case the axis of rotation has a fixed direction in space, and hence la = do/dt is the total
derivative of the angle of rotation of the cylinder about its axis. The condition (38.3) can
therefore be integrated, and gives a relation between the angle and the co-ordinate of the
centre of mass.
124
Motion of a Rigid Body
§38
This must be taken into account in varying the action. According to
Lagrange's method of finding conditional extrema, we must add to the inte-
grand in the variation of the action
=
the left-hand sides of equations (38.4) multiplied by undetermined coeffici-
ents da (functions of the co-ordinates), and then equate the integral to zero.
In SO doing the variations dqi are regarded as entirely independent, and the
result is
(38.5)
These equations, together with the constraint equations (38.2), form a com-
plete set of equations for the unknowns qi and da.
The reaction forces do not appear in this treatment, and the contact of
the bodies is fully allowed for by means of the constraint equations. There
is, however, another method of deriving the equations of motion for bodies in
contact, in which the reactions are introduced explicitly. The essential feature
of this method, which is sometimes called d'Alembert's principle, is to write
for each of the bodies in contact the equations.
dP/dt==f,
(38.6)
wherein the forces f acting on each body include the reactions. The latter
are initially unknown and are determined, together with the motion of the
body, by solving the equations. This method is equally applicable for both
holonomic and non-holonomic constraints.
PROBLEMS
PROBLEM 1. Using d'Alembert's principle, find the equations of motion of a homogeneous
sphere rolling on a plane under an external force F and torque K.
SOLUTION. The constraint equation is (38.3). Denoting the reaction force at the point of
contact between the sphere and the plane by R, we have equations (38.6) in the form
u dV/dt = F+R,
(1)
dSu/dt = K-an xR,
(2)
where we have used the facts that P = V and, for a spherical top, M = ISE. Differentiating
the constraint equation (38.3) with respect to time, we have V = aS2xn. Substituting in
equation (1) and eliminating S by means of (2), we obtain (I/au)(F+R) = Kxn-aR+
+an(n . R), which relates R, F and K. Writing this equation in components and substitut-
ing I = zua2 (§32, Problem 2(b)), we have
R2 = -F2,
where the plane is taken as the xy-plane. Finally, substituting these expressions in (1), we
§38
Rigid bodies in contact
125
obtain the equations of motion involving only the given external force and torque:
dVx dt 7u 5 Ky
dt
The components Ox, Q2 y of the angular velocity are given in terms of Vx, Vy by the constraint
equation (38.3); for S2 we have the equation 2 dQ2/dt = K2, the z-component of equa-
tion (2).
PROBLEM 2. A uniform rod BD of weight P and length l rests against a wall as shown in
Fig. 52 and its lower end B is held by a string AB. Find the reaction of the wall and the ten-
sion in the string.
Rc
h
P
RB
T
A
B
FIG. 52
SOLUTION. The weight of the rod can be represented by a force P vertically downwards,
applied at its midpoint. The reactions RB and Rc are respectively vertically upwards and
perpendicular to the rod; the tension T in the string is directed from B to A. The solution
of the equations of equilibrium gives Rc = (Pl/4h) sin 2a, RB = P-Rcsin x, T = Rc cos a.
PROBLEM 3. A rod of weight P has one end A on a vertical plane and the other end B on
a horizontal plane (Fig. 53), and is held in position by two horizontal strings AD and BC,
RB
TA
A
RA
C
FIG. 53
126
Motion of a Rigid Body
§39
the latter being in the same vertical plane as AB. Determine the reactions of the planes and
the tensions in the strings.
SOLUTION. The tensions TA and TB are from A to D and from B to C respectively. The
reactions RA and RB are perpendicular to the corresponding planes. The solution of the
equations of equilibrium gives RB = P, TB = 1P cot a, RA = TB sin B, TA = TB cos B.
PROBLEM 4. Two rods of length l and negligible weight are hinged together, and their ends
are connected by a string AB (Fig. 54). They stand on a plane, and a force F is applied
at the midpoint of one rod. Determine the reactions.
RC
C
PA
F
1
RB
T
T
A
B
FIG. 54
SOLUTION. The tension T acts at A from A to B, and at B from B to A. The reactions RA
and RB at A and B are perpendicular to the plane. Let Rc be the reaction on the rod AC at
the hinge; then a reaction - -Rc acts on the rod BC. The condition that the sum of the moments
of the forces RB, T and - -Rc acting on the rod BC should be zero shows that Rc acts along
BC. The remaining conditions of equilibrium (for the two rods separately) give RA = 1F,
RB = 1F, Rc = 1F cosec a, T = 1F cot a, where a is the angle CAB.
§39. Motion in a non-inertial frame of reference
Up to this point we have always used inertial frames of reference in discuss-
ing the motion of mechanical systems. For example, the Lagrangian
L = 1mvo2- U,
(39.1)
and the corresponding equation of motion m dvo/dt = - au/dr, for a single
particle in an external field are valid only in an inertial frame. (In this section
the suffix 0 denotes quantities pertaining to an inertial frame.)
Let us now consider what the equations of motion will be in a non-inertial
frame of reference. The basis of the solution of this problem is again the
principle of least action, whose validity does not depend on the frame of
reference chosen. Lagrange's equations
(39.2)
are likewise valid, but the Lagrangian is no longer of the form (39.1), and to
derive it we must carry out the necessary transformation of the function Lo.
§39
Motion in a non-inertial frame of reference
127
This transformation is done in two steps. Let us first consider a frame of
reference K' which moves with a translational velocity V(t) relative to the
inertial frame K0. The velocities V0 and v' of a particle in the frames Ko and
K' respectively are related by
vo = v'+ V(t).
(39.3)
Substitution of this in (39.1) gives the Lagrangian in K':
L' = 1mv2+mv.+1mV2-U
Now V2(t) is a given function of time, and can be written as the total deriva-
tive with respect to t of some other function; the third term in L' can there-
fore be omitted. Next, v' = dr'/dt, where r' is the radius vector of the par-
ticle in the frame K'. Hence
mV(t)+v'= mV.dr/'dt = d(mV.r')/dt-mr'.dV/dt.
Substituting in the Lagrangian and again omitting the total time derivative,
we have finally
L' =
(39.4)
where W = dV/dt is the translational acceleration of the frame K'.
The Lagrange's equation derived from (39.4) is
(39.5)
Thus an accelerated translational motion of a frame of reference is equivalent,
as regards its effect on the equations of motion of a particle, to the application
of a uniform field of force equal to the mass of the particle multiplied by the
acceleration W, in the direction opposite to this acceleration.
Let us now bring in a further frame of reference K, whose origin coincides
with that of K', but which rotates relative to K' with angular velocity Su(t).
Thus K executes both a translational and a rotational motion relative to the
inertial frame Ko.
The velocity v' of the particle relative to K' is composed of its velocity
V
relative to K and the velocity Sxr of its rotation with K: v' = Lxr
(since the radius vectors r and r' in the frames K and K' coincide). Substitut-
ing this in the Lagrangian (39.4), we obtain
L = +mv.Sx+1m(xr)2-mW.r-
(39.6)
This is the general form of the Lagrangian of a particle in an arbitrary, not
necessarily inertial, frame of reference. The rotation of the frame leads to the
appearance in the Lagrangian of a term linear in the velocity of the particle.
To calculate the derivatives appearing in Lagrange's equation, we write
128
Motion of a Rigid Body
§39
the total differential
dL = mv.dv+mdv.Sxr+mv.Sxdr+
=
v.dv+mdv.xr+mdr.vxR+
The terms in dv and dr give
0L/dr X Q-mW-dU/0r. - -
Substitution of these expressions in (39.2) gives the required equation of
motion:
mdv/dt = (39.7)
We see that the "inertia forces" due to the rotation of the frame consist
of three terms. The force mrxo is due to the non-uniformity of the rotation,
but the other two terms appear even if the rotation is uniform. The force
2mvxs is called the Coriolis force; unlike any other (non-dissipative) force
hitherto considered, it depends on the velocity of the particle. The force
mSX(rxS) is called the centrifugal force. It lies in the plane through r and
S, is perpendicular to the axis of rotation (i.e. to S2), and is directed away
from the axis. The magnitude of this force is mpO2, where P is the distance
of the particle from the axis of rotation.
Let us now consider the particular case of a uniformly rotating frame with
no translational acceleration. Putting in (39.6) and (39.7) S = constant,
W = 0, we obtain the Lagrangian
L
=
(39.8)
and the equation of motion
mdv/dt = -
(39.9)
The energy of the particle in this case is obtained by substituting
p =
(39.10)
in E = p.v-L, which gives
E =
(39.11)
It should be noticed that the energy contains no term linear in the velocity.
The rotation of the frame simply adds to the energy a term depending only
on the co-ordinates of the particle and proportional to the square of the
angular velocity. This additional term - 1m(Sxr)2 is called the centrifugal
potential energy.
The velocity V of the particle relative to the uniformly rotating frame of
reference is related to its velocity V0 relative to the inertial frame Ko by
(39.12)
§39
Motion in a non-inertial frame of reference
129
The momentum p (39.10) of the particle in the frame K is therefore the same
as its momentum Po = MVO in the frame K0. The angular momenta
M = rxpo and M = rxp are likewise equal. The energies of the particle
in the two frames are not the same, however. Substituting V from (39.12) in
(39.11), we obtain E = 1mv02-mvo Sxr+U = 1mvo2 + mrxvo S.
The first two terms are the energy E0 in the frame K0. Using the angular
momentum M, we have
E = E0 n-M.S.
(39.13)
This formula gives the law of transformation of energy when we change to a
uniformly rotating frame. Although it has been derived for a single particle,
the derivation can evidently be generalised immediately to any system of
particles, and the same formula (39.13) is obtained.
PROBLEMS
PROBLEM 1. Find the deflection of a freely falling body from the vertical caused by the
Earth's rotation, assuming the angular velocity of this rotation to be small.
SOLUTION. In a gravitational field U = -mg. r, where g is the gravity acceleration
vector; neglecting the centrifugal force in equation (39.9) as containing the square of S, we
have the equation of motion
v = 2vxSu+g.
(1)
This equation may be solved by successive approximations. To do so, we put V = V1+V2,
where V1 is the solution of the equation V1 = g, i.e. V1 = gt+ (Vo being the initial velocity).
Substituting V = V1+v2in (1) and retaining only V1 on the right, we have for V2 the equation
V2 = 2v1xSc = 2tgxSt+2voxS. Integration gives
(2)
where h is the initial radius vector of the particle.
Let the z-axis be vertically upwards, and the x-axis towards the pole; then gx = gy = 0,
n sin 1, where A is the latitude (which for definite-
ness we take to be north). Putting V0 = 0 in (2), we find x = 0, =-1t300 cos A. Substitu-
tion of the time of fall t 22 (2h/g) gives finally x = 0,3 = - 1(2h/g)3/2 cos A, the negative
value indicating an eastward deflection.
PROBLEM 2. Determine the deflection from coplanarity of the path of a particle thrown
from the Earth's surface with velocity Vo.
SOLUTION. Let the xx-plane be such as to contain the velocity Vo. The initial altitude
h = 0. The lateral deviation is given by (2), Problem 1: y =
or, substituting the time of flight t 22 2voz/g, y =
PROBLEM 3. Determine the effect of the Earth's rotation on small oscillations of a pendulum
(the problem of Foucault's pendulum).
SOLUTION. Neglecting the vertical displacement of the pendulum, as being a quantity
of the second order of smallness, we can regard the motion as taking place in the horizontal
xy-plane. Omitting terms in N°, we have the equations of motion x+w2x = 20zy, j+w2y
= -20zx, where w is the frequency of oscillation of the pendulum if the Earth's rotation is
neglected. Multiplying the second equation by i and adding, we obtain a single equation
130
Motion of a Rigid Body
§39
+2i02s+w28 = 0 for the complex quantity $ = xtiy. For I2<<, the solution of this
equation is
$ = exp(-is2t) [A1 exp(iwt) +A2 exp(-iwt)]
or
xtiy = (xo+iyo) exp(-is2zt),
where the functions xo(t), yo(t) give the path of the pendulum when the Earth's rotation is
neglected. The effect of this rotation is therefore to turn the path about the vertical with
angular velocity Qz.