(Fowles and Cassiday Chapter 9)

We remove the assumptions that the axis or rotation is either fixed or confined to a plane.

To emphasize that the moment of inertia involves distances which are perpendicular to the axis of rotation, it will be represented as: I = S m_i r^ _i² ,

with r^ _i = r_i sin q _i = | r_i ´ n|

q represents angle between vector r and the axis of rotation,

defined above by the unit vector n, as illustrated in Fig.9.1.1.

Since the rotation axis is not fixed, it is not taken as one of the coordinate axes.

Instead: r_i = i x_i + j y_i + k z_i

Using the above equation and n = i cos a + j cos b + k cos g

gives Eq.(9.1.10), in which:

I_xx = S m_i(y_i² + z_i²) , I_yy = S m_i(z_i² + x_i²) , I_zz = S m_i(x_i² + y_i²)

We can recognise these three quantities as the moments of inertia about the x, y and z axes (previously represented as I_x, I_y and I_z).

I_xy = -S m_i(x_i y_i) is called the xy product of inertia

I_yz = -S m_i(y_i z_i) is called the yz product of inertia

I_zx = -S m_i(z_i x_i) is called the zx product of inertia

In practice, each summation over particles (i) can be evaluated as an integral, as previously.

As shown on p.337, a more compact form of Eq.(9.1.10) is: I = n^~ I n

where I is the moment of inertia tensor, n is a column vector representation of the unit vector n, and n^~ is its transpose, the row vector: (cos a cos b cos g )

Note that the symbols use by Fowles and Cassidy do not distinguish between a vector and a tensor quantity; both are represented in bold type. A less ambiguous notation uses a single underscore for a vector (which is a 1^st-rank tensor) and a double underscore for a 2^nd-rank tensor, with no underscore for a zero-rank tensor (a scalar).

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To see how our tensor notation works, consider Example 9.1.1 on p.340.

The products of inertia are evaluated as integrals, including also the minus sign.

Many books do not include the minus sign in the definitions of I_xy etc., in which case the off-diagonal components of the M of I tensor are represented by - I_xy etc.

Check: I_zz = I + I (from parallel axis theorem and symmetry)

where I = ma²/12 = M of I about the diagonal, as already calculated on p.341,

giving I_zz = ma²/6 (as expected, e.g. mid-term question 4d).

Note: the M of I tensor does depend on the choice of origin of the coordinate system. If we take this origin to be at the center of the square, the I matrix consists of zero terms except for those on the diagonal, which are 1/12 , 1/12 , and 1/6 . But when we evaluate the scalar quantity I for the diagonal axis of rotation, it comes out the same as above.

As in the case of a system of particles, the angular momentum of a rigid body is:

L = S r_i ´ m_iv_i = S r_i ´ m_i [w _i ´ r_i ] = S m_i r_i² w - S m_i r_i(r_i.w _i)

since from Eq.(1.7.3), p.19, the vector triple product A ´ (B ´ C) = B(A.C) – C(A.B)

From this equation (9.1.19a on p.338), we see that in general L is not in the same direction as w (the axis of rotation).

Exception: the distribution of mass within the body makes the second term S m_i (r_i.w _i) r_i equal to zero, as a result of symmetry about the rotation axis.

The two terms in the expression for L can be combined by writing Eq.(9.1.19.a), p. 338, in the form (9.1.19b).

Looking at the second term of Eq.(9.1.19b), we imply that

r_ir_i.w means the same thing as r_i(r_i.w _i)

Or in general that there exists a product ab of two vectors such that ab.c = a(b.c) ,

as in Eq.(9.1.23a). ab is called a dyadic product (or dyad) of the vectors a and b .

We can establish this by writing a(b.c) in terms of its column-vector components, as in the bottom half of Eq.(9.1.23a); (b.c) is just a scalar quantity which multiplies each component.

The resulting column vector can be generated by means of a matrix operating on the column vector c according to rules established on p.21.

The components of this matrix are products of the components of a and b, as indicated by the top half of Eq. (9.1.23b).

Thus we have a matrix representation of the dyadic tensor ab .

We have shown that the second half of vector L in Eq.(9.1.19b) involves a 2^nd-rank tensor.

Going back to Eq.(9.1.19b), the two terms can be bracketed together if we define a unit tensor 1, which (to be consistent with the preceding discussion) must be capable of being written as a dyadic product, as in Eq.(9.1.21).

As demonstrated in Eq.(9.1.27a) and Eq.(9.1.27a), the bracketed term in Eq.(9.1.21) has all of the properties of the moment of inertia tensor whose components are given in Eq. (9.1.6). Therefore the expression for angular momentum becomes, as in Eq.(9.1.28) :

L = I . w

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Example 9.1.2:

1. For rotation about the non-central x-axis, L has a y-component !

This reflects the lack of symmetry about the chosen axis of rotation.

The x-component of L is the value we would calculate from fixed-axis mechanics, neglecting the products of inertia.

(b) For rotation about a diagonal, L has equal x and y components, so it is in the same direction as w . This is a result of symmetery about the axis of rotation.

Using L² = (L _. L) = L^~ L , where the transpose L^~ is a row vector:

1. for x-axis rotation, the magnitude L of the angular momentum is different to that expected from simple (non-tensor) theory
2. for diagonal rotation, the answer is what we would expect from the moment of inertia.

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Note that tensor I (unlike scalar I) completely describes the rotational properties of an object within its 9 components; no need to specify the axis of rotation.

However, L = I . w does depend on the choice of rotation axis.

A general expression for the total KE of a system of particles is (p.261):

T = S (m_i/2) v_i . v_i

where are v_i represents velocity relative to some fixed coordinate system.

If we take the center of coordinates at the center of mass of a rigid body

and consider rotation about the C of M,

we can substitute w ´ r for one of the v_i’s (now measured relative to the C of M) and show (p.343) that T_rot = (1/2) w . L = (1/2) w . I . w

where L is relative to the center of mass.

Similarly, T_trans = S (m_i/2) v_i.v_i = (1/2) v.p

where v is linear velocity of the center of mass.

The total KE is the sum of T_rot and T_trans.

It can be shown that for any rigid body, and any point within it, that there exist three perpendicular coordinates axes, the principal axes , for which all products of inertia are zero. The corresponding moments of inertia are called the principal moments.

Usually, the origin of the principal axes is taken as the centre of mass.

If we employ the notation: I_xx = I₁, I_yy = I₂, I_zz = I₃,

the M of I tensor is represented by a diagonal matrix whose only non-zero terms (those on the diagonal) are I₁, I₂ and I₃, as indicated in Eq.(9.2.2), p.344.

The moment of inertia about any rotation axis is therefore:

I = n^~ I n = I₁ cos^2a + I₂ cos^2b + I₃ cos^2g

as also seen from Eq.(9.1.10) on p.336 since the product of inertia terms are zero.

Expressions for L and T_rot are given on p.345.

If a body is rotated about the 1-axis, w = w ₁ and L = e₁I_{1w 1} = I₁ w

Then the angular momentum coincides with the rotation axis .

Obviously the same argument can be applied to the 2 and 3 principal axes.

If the body has an axis of symmetry, this axis is one of the principal axes.

Calling this axis the z-axis, we can divide the whole object into small elements and identify pairs of elements with the same mass m_I and distance r_i from the axis but equal and opposite y and z coordinates, such that I_zx and I_yz are both zero.

For example, a cylinder whose length axis is z : this axis is one of the principal axes and the other two are any x and y axes perpendicular to z.

For a sphere, any perpendicular (Cartesian) set of axes are principal axes.

Other examples are given on p. 345.

Example 9.2.1 on p.346: uniform rectangular block with dimensions a, b and c.

From symmetry, axes parallel to the sides and passing through the center of mass are 1,2,3 principal axes.

I₁, I₂ and I₃ are given by the expressions for a rectangular lamina (p.315) since a cross-section perpendicular to each axis is rectangular and distance perpendicular to this cross section (along the axis) does not affect the moment of inertia.

For a cubic block with sides of length a, I₁= I₂ = I₃ = (m/6) a²

Then I = I₁ cos^2a + I₂ cos^2b + I₃ cos^2g = I₁ (cos^2a + cos^2b + cos^2g ) = I₁

Therefore any set of Cartesian axes comprise principal axes, as for a sphere.

(The same principle makes many physical properties of cubic crystals isotropic).

For an object with no axis of symmetry, the principal axes are more difficult to identify but must exist. Consider a laminar object (p.346); choosing x and y axes which lie in the plane of the lamina ensures z=0 and therefore I_yz and I_zx are zero. Then, rotating the axes from x,y to x',y' (as shown in Figure 9.2.2) reverses the sign of each y-coordinate and therefore the sign of I_xy. At some intermediate orientation, I_xy must therefore be zero and x,y,z then correspond to the 1,2,3 principal axes.

Dynamic balancing.

For any axis of rotation passing through the center of mass, an object is statically balanced. But unless this axis coincides with one of the principal axes, the object will not be dynamically balanced; the angular momentum L will itself rotate as the object spins, giving rise to an oscillating torque at the bearings which may cause vibration of the whole assembly.

A practical example is the motor-driven fan shown in Figs.9.2.3 and 9.2.4, p.348.

In this case, the axis of symmetry (z- or 3-axis) is not the required axis of rotation but we know that there must be two other principal axes (1,2) which are possible rotation axes which would not cause vibration.

In general, we have L = I . w where I is the M of I tensor

But if rotation is about the 1-axis, L = I_1w also.

Equating these two expressions, in terms of the x- and y- components of L , gives Eqs.(9.2.10) on p.349. In these two equations, I₁, w ₁ and w ₂ are unknowns; however,

w _x and w _y are perpendicular components of w , so w ₁/w ₂ = tan q and the equations can be solved by eliminating I₁, leading to an expression for tan(2q ) in terms of the M of I components: I_xx, I_yy and I_xy : tan(2q ) = 2I_xy / (I_xx - I_yy )

This equation, Eq.(9.2.13) on p.349, is a general equation for the case where one axis (the z-axis) has been identified as a principal axis.

For the case of two fan blades, the value of q will depend on I_xy , which in turn depends on the angle between the blades. But if that angle is 90° , I_xx = I_yy (one of the x- and y-axes being within the plane of each blade) and tan(2q ) = ¥ giving q = 45 ° .

Example: square lamina rotating about a corner (as on pp. 340 - 342).

The z-axis, perpendicular to the lamina, is a principal axis (z-dependent products of inertia vanish because of the zero thickness).

Also, I_xx = I_yy (from symmetry), so 2q = 90 deg or 270 deg, q = 45° or 135° .

So the other principal axes are along the diagonal and perpendicular to the diagonal.

If the rotation axis passes through the center of the square (the C of M), all products of inertia ( including I_xy ) would be zero but I_xx = I_yy so tan(2q ) = 0 / 0 = any value.

So any axes in the plane are principal axes, as in the previous case of a cube rotated about its C or M.

Example 9.2.2: automobile wheel, approximated as a thin disk (radius a) which is not quite perpendicular to the rotation axis (it deviates by an angle q ).

Let x be the symmetry axis of the disk, lying in the yz plane.

The wheel can be balanced by adding two weights (m') in the xy plane, a distance x = b from the disk, as shown in Fig. 9.2.5.

The z-axis would be a principal axis for the unweighted wheel and adding the weights causes no contribution to I_xz and I_zy (since z=0 for each weight).

We can then use: tan(2q ) = 2I_xy / (I_xx - I_yy )

with I_xx = (1/2)ma² + 2m'a²

and I_yy = (1/2)ma² + 2m'b²

and I_xy = -[b(-a)m' + (-b)am'] = 2abm'

We need to diagonalize the tensor I.

Parallel L and w implies: L = I.w = l w where l is a scalar quantity, which must correspond to one of the principal moments of inertia.

(There will actually be three values of l , corresponding to the 3 principal axes.)

Dividing through by the scalar quantity w , we get: I.e = l e

where e is a unit vector directed along any one of the principal axes.

In other words, (I-l ).e = 0 whose solution is given by ½ (I-l )½ = 0

which can be written out in terms of the components of the determinant:

½ ½

½ I_yx I_yy-l I_yx ½ = 0

½ ½

½ (I-l )½ = 0 therefore implies a cubic equation (9.2.18, p.351) with three solutions, the three principal moments.

The direction cosines of each principal axis are found by writing (I-l ).e = 0 in terms of matrix components:

moments of inertia are found from ½ (I-l )½ = 0 .

In general, the angular momentum L is not parallel to the axis of rotation w

In addition, its direction changes as the body rotates.

Our previous discussion (of L) implicitly assumed a rotating (non-inertial) coordinate system, rather than a fixed (inertial) one. From Chapter 5 (p.176),

(dL/dt)_fixed = (dL/dt)_rot + w ´ L as in Eq.(9.3.2)

The rotational version of Newton’s second law becomes:

torque N = (dL/dt)_rot + (w ´ L) = I . (dw /dt) + w ´ (I.w ), as in Eq.(9.3.4)

The last term in Eq.(9.3.4) can be written as a determinant, as in Eq.(9.3.4c) [compare with the vector triple product of Eq.(1.7.1), p.19]. This leads to three simultaneous equations (known as the Euler equations) for the components of N along the three principal axes, which are written in vector form as Eq.(9.3.5) .

For a body caused to rotate at uniform angular speed, the acceleration terms in the Euler equations are zero. If this rotation occurs about a principal axis, two of the w components are zero, which implies N₁ = N₂ = N₃ = 0 i.e. no rotational torque required (excluding air resistance and friction, not included here) and no torque on the bearings, consistent with earlier statements.

Section 9.4: Free rotation implies no applied torque. This includes an object in free space or in free fall (under gravitational force) or suspended at its center of mass by a frictionless pivot, in which case a uniform gravitational field will have no effect.

Angular momentum is conserved in magnitude and direction with respect to fixed axes.

With respect to rotating axes, the magnitude remains fixed but not the direction.

Therefore: L² = (I_{1w 1})² + (I_{2w 2})² +(I_{3w 3})² = constant

Rotational kinetic energy is also conserved. Since T_rot = ½(w .I.w ),

2T_rot ² = I_{1w 1}² + I_{2w 2}² + I_{3w 3}² = constant

These two equations describes two ellipsoids of inertia, whose intersection represents the behavior of the angular velocity vector w ; see Fig.9.4.1.

Under certain circumstances, this intersection occurs only at two points, which define the axis of rotation, and simple rotation is stable about this axis. This is where rotation occurs about a principal axis (3 or 1) having the largest or smallest moment of inertia (see p. 357 and p. 358). In these two cases, rotational momentum and kinetic energy are simultaneously maximized or minimized.

Examples: rocket or bullet given spin to stabilize the forward motion.

Rotation about the intermediate principal axis (2) or about an arbitary axis causes precession, as illustrated on p.358.

Section 9.5 analyses the rotation of a body with an axis of complete rotational symmetry (I₃ = I_s , I₁=I₂=I) such as prolate spheroid (football). The Euler equations indicate that the components of angular motion (w ₁ and w ₂) perpendicular to the symmetry axis are modulated harmonically by an angular frequency: W = w ₃ (I_s/I – 1) = (w cos a ) (I_s/I – 1),

the rate of spin precession. [Note that W ® 0 for a sphere or a cube]

This is equivalent to the rotation axis w moving in a cone (the body cone) of semi-angle a at angular frequency W about the 3-axis (see Figure 9.5.2) which is the direction of the total angular momentum L.

In the case of an arbitary-shaped object (I_1¹ I_2¹ I₃) such as a general ellipsoid (Fig. 9.5.3), there is a double-precession (wobbling) motion in which w ₃ is also modulated (Fig. 9.5.4) , as discussed on pp.360-364.

Section 9.6: Rotation of an arbitary object is best discussed by introducing a new set of axes (x’,y’,z’), in addition to the fixed (x,y,z) and rotating principal (1,2,3) axes; see Fig. 9.6.1. The fixed z-axis defines the total angular momentum L .

z’ is parallel to the 3-axis and rotates with the object

x’ is the intersection of the rotating 1-2 plane with the fixed x-y plane (horizontal in Fig. 9.6.1) and is known as the line of nodes. Its instantaneous angle relative to the x-axis is f . f , q and y (see Fig. 9.6.1) are the Euler angles, which suffice to completely specify the rotation.

During free rotation, the direction of angular momentum L remains fixed (the z-axis).

Relative to L , the spin w will precess at angular frequency W in a cone of semi-angle a , the space cone. The projection of w on the L-axis (w cos a ) is constant.

Relative to the z’-axis attached to the body, w precesses around the body cone (semi-angle q ) at angular frequency df /dt.

Equivalently, the body cone rolls without slipping around the space cone, the line of contact being w .

Example 9.6.1 (p.369): thin lamina with a symmetry axis e.g. frisbee.

I₁ = I₂ = I ,

I_s = I₃ = I₁ + I₂ (from parallel-axis theorem) = 2 I

If the disk is set rotating at an angle a relative to its symmetry (z’ or 3) axis,

W = (w cos a ) (I_s/I – 1) = w cos a » w for small a .

From Eq.(9.6.12), df /dt = w [1+ 3 cos^2a ]^1/2 ~ 2w for small a

Example 9.6.2: free precession of the earth.

I_s/I = 1.00327 and a very small (0.2 arc sec), so W = (w cos a ) (I_s/I – 1) = 0.00327 w giving the period of precession (seen from Earth) = 1/0.00327 = 305 day

Seen from space, df /dt = w [1+ (1.00327² – 1) cos^2a ]^1/2 ~ 1.00327w ,

period of "wobble" = 0.997 day. This is in addition to gyroscopic precession, due to external torque exerted by Moon and Sun, with period 26000 years.

Section 9.7: Gyroscope or top.

Here there is an external torque, of magnitude N_x’ = mgl sinq (where l = distance between pivot and C of M) and direction along the horizontal (but rotating) x’ axis.

The angular momentum L about the fixed z-axis remains constant.

If there is no friction, the rotational kinetic energy, the spin angular velocity S about the z’ axis, and the angular momentum I_sS, remain constant.

Transforming to rotating (x’,y’,z’) coordinates gives Eqs.(9.7)

If the spin axis is horizontal (q = 90 deg), N_x’ = mg l sinq = I_sS (df /dt) sinq

and the precession rate is df /dt = mg l / I_sS

For other values of q , two values of df /dt are possible, given by Eq.(9.7.10),

provided that I_s²S² > 4mg l I cosq . This provides a condition for stable motion (sleeping).

Note that I is the moment of inertia about the x’ axis (see Fig. 9.7.1) which is perpendicular to the spin S and passes through the point of contact O.

If the gyroscope is suddenly released, the spin axis starts to falls but then recovers. In the vertical direction, the axis executes simple harmonic motion (damped because of friction at the pivot) in addition to precession. This behavior, called nutation, results in a cycloidal motion; see Fig.9.8.4. It can be analysed in terms of kinetic energy (p.376) to find the turning points (q ₁ and q ₂) for the elevation angle q .

Gyrocompass : if the angle q held at 90 deg. but the gyro is supported at both ends (on a gimball) so that there is no net torque, the rotation axis oscillates like a simple pendulum (Eq. 9.9.9) about a direction parallel to the horizontal component (w _e cosl ) of the earth’s angular momentum. In other words, allowing for frictional or air damping, the gyroscope will eventually point north, similar to a compass. Note, however, that the gyro is seeking the earth’s rotational axis w _e rather than the magnetic North pole.

The horizontal frictional force F_P causes rolling. Motion of the center of mass arises from F_P and from the weight mg, acting vertically. So rolling and C of M motion can be linked, by eliminating F_P from the vector equations of motion (p.383). In addition, v_cm = w ´ r, where r is the radius vector joining the C of M and point of contact with the plane.

A general solution is complicated, but of practical interest is the case of steady rolling, where the angle between the vertical and rotation axis q » 90 deg.

The deviation c from q = 90° obeys a SHM equation, Eq.(9.10.11) and (in the presence of damping) settles down to zero provided the square-bracket term in Eq.(9.10.11) is positive. Therefore the condition for stability of steady rolling is Eq.(9.10.12):

S² > I mga / [I_s(I_s + ma²)]

I is the M of I about the vertical axis (perpendicular to the spin and through the point of contact, as for the spinning top) = (1/4) ma² for a thin disk

I_s is the M of I about the rotation axis = (1/2) ma² for thin disk, a is the disk radius

This gives S² > (1/3)(g/a), v_cm = aS > 17.6 cm/s for a rolling penny.

For a bicycle wheel, v_cm » 1 m/s (mainly due to M of I of the rim).