3
FORMULATION Chapter OF ROTATIONAL SYSTEMS and review of second moments
Many engineering problems involve rotation, and it is not realistic to speak of a single "point" mass. But if we idealize the mass system as rigid, we can write similar differential equations for rotation as we did for translation x, by formulating a rotational equivalent mass called "moment of inertia."
3.1
Newton's Law Revisited
Let us begin by examining a very simple rotational system. If a concentrated "point" mass m is placed at the end of a rigid, weightless lever of length r (Figure 3.1) and a torque T is applied to the lever, we observe that a small displacement x of the mass can also be expressed as r, a velocity x as r, an acceleration x as r, and the force F at the mass as T /r. Substituting these into Equation ?? F = ma/gc , we obtain T/r = (mr/gc ) , which we can solve
Figure 3.1: Rotational System 23
24
Chapter 3. FORMULATION OF ROTATIONAL SYSTEMS
Figure 3.2: Simple Pendulum with Torsion Spring for T to obtain Newton's law in units of torque: (mr2 ) T = gc
(3.1)
We identify (mr2 ) as the rotational equivalent mass or "moment of inertia"; it is a function of both the magnitude of the mass and the distance r from pivot to mass. Since this radius r is squared, this moment is also called the "second moment of the mass" about the pivot; it is designated by the symbol Jpivot .
3.2
Torsional Systems
If we add a torsional spring (a clock spring or a torsion bar), a torsional damper, and torsional excitation to Figure 3.1, we can write a differential equation that sums up torques similar to the way Equation ?? sums up forces:
Jp i v o t gc
+ cT + kT = T (t)
(3.2)
where cT , torsional damping constant, kT , torsional spring constant, and T (t) , any externally applied torque. Example: A pendulum with a concentrated mass is called a "simple pendulum." If we also add a torsion spring kT (Figure 3.2) attached such that its force is zero when the pendulum is vertical, and evaluate the restoring momentforce times lever armdue to gravity, the equations above reduce to mg mL2 + kT + L sin = 0 gc gc (3.3)
We can interpret the last term as either the tangential force component (mg/gc ) sin times the radial length, or else as the vertical force mg/gc times the horizontal lever arm L sin . Unlike our crude analysis in the previous chapter (page ??),
3.3. RIGID BODIES this equation is not yet linearized. Recalling the Maclaurin series expansion sin = - 3 5 + -... 3! 5! 3 , if = 6
25
(3.4)
we recognize that we can linearize the equation by replacing sin by . This leads to the same gravitational term as the crude analysis of Section ??, but now we know the magnitude of the terms we have neglected when we linearized. 2 The fractional error in the last term of the linearized equation is -sin 6 : sin = if remains within 6 degrees (0.1 radians) of the vertical, the error in the gravitational term remains within 0.2%; if increases to 57 degrees (1 radian) from the vertical, the error in the last term increases to 16%. Because it is important to know the limitations of the analysis, it is always best to delay linearization as long as possible.
3.3
Rigid Bodies
In Chapter ??, where we had pure straight-line translation x, a body was treated as a concentrated "point" mass assuming that all parts of it moved as one; i.e., that it was rigid. Thus the translation x of any point in the body was also the translation of any other point. In mathematical terms: if a rigid body is made up of a large number of small masses dm, it can be represented by the RRR single concentrated mass M = dm, where the integration stands for the summation of all the infinitesimal parts of the body. Similarly, in problems where we have pure rotation about a point, we can sum the second moments of all the small masses dm in a rigid body to obtain the total moment of inertia RRR 2 Jpivot = r dm (3.5) where r is the distance of each mass dm from the axis of rotation. Example: For a flywheel of uniform thickness b and outside radius R, the mass is the sum of a series of infinitesimal rings of width b, circumference 2r, and thickness dr: Z
R
M = b
(2r) dr = bR2
0
The second moment about the central axis of the disk is: Z R Jzz = b (2r) r2 dr = b R4 = 0.5R2 M 2 0
In translation, a rigid flywheel behaves like a concentrated mass M located at the center of gravity. In rotation, it behaves as if that mass were an average
26
Chapter 3. FORMULATION OF ROTATIONAL SYSTEMS
distance of 0.5 R2 from the axis. We can define that average distance as the "radius of gyration" RRR 2 r dm J 2 (3.6) = RRR rgyr , dm M which turns out to be 0.707R for any uniform-disk flywheel of radius R. Exercise 3.1 Find the second moment of a disk with a hole in the center. Exercise 3.2 Show that a thin hoop has Jzz 1.0MR2 . What is the order of magnitude of the error due to neglecting the thickness of the hoop? Since sectional second moments are of importance also in the torsion of rods and bending of beams, values for different shapes can be found in many engineering texts1 and handbooks.2 R R 2 Sectional second moments for torsion Jz , r2 dA = x + y 2 dA can be adapted to find the polar moments Jzz of uniform-thickness bodies, whether circular or not. R R 2 Sectional second moments in bending Ix , y 2 dA and Iy , xdA, RRR 2 can be adapted to find the moments of inertia Ixx , y + z 2 dm RRR 2 or Iyy , x + z 2 dm of flat, thin shapes (z x or y) about axes within their own plane. Exercise 3.3 Compare the Jzz = 0.5MR2 obtained above for a uniform circular flywheel with the sectional moment of a circle in a strength-of-materials book. Exercise 3.4 Show that a butterfly valve (a thin circular disk rotating symmetrically about an axis within its own plane) such as is used as an automotiveengine throttle, has a moment of inertia Ixx 0.25MR2 . What is the order of = magnitude of the error due to the thickness of the disk? In sectional moments, we can show from Pythagoras' theorem r2 = y 2 + x2 that Jzz = Ixx + Iyy ; this is approximately true also for flat, thin shapes (z x or y). Since the integrals for one or the other of these may be much easier to obtain, depending on shape boundaries, we can use this relationship for developing hard-to-get second moments. Exercise 3.5 Show that a thin circular hoop has Ixx = Iyy 0.5M R2 . = Exercise 3.6 Find Ixx and Iyy for a thin rectangular plate by integration, and obtain Jzz from them.
1 e.g., Warren C. Young, Roark's Formulas for Stress and Strain, 6th Edition, McGrawHill, New York, 1989, ISBN 0-07-072541-1. 2 e.g., Ray E. Bolz and G.L. Tuve, editors, CRC Handbook of Tables for Applied Engineering Science, 2nd Edition, CRC Press, Boca Raton, Florida, 1973, ISBN 0-8493-0252-8.
3.4. CENTER OF GRAVITY
27
3.4
Center of Gravity
Any rigid-body rotation about some particular pivot can also be represented at any instant as the combination of rotation about some other point, plus a translation of that point. (Conversely, any combination of rotation and translation can be represented as pure rotation about the "instantaneous center of rotation.") In rigid-body dynamics, it is often convenient to represent a motion as the sum of rotation about the center of gravity (c.g.) plus translation of the c.g. The coordinates of the c.g. in Cartesian coordinates can be obtained by taking the first moments of the mass RRR RRR RRR xdm ydm zdm y ; = RRR ; z = RRR (3.7) x = RRR dm dm dm Exercise 3.7 Find the mass, the location of the c.g., and some moments of inertia of a thin plate in the shape of a right triangle. Exercise 3.8 Find the mass, the location of the c.g., and some moments of inertia of a thin plate in the shape of a half-circle. Exercise 3.9 Find the mass, the location of the c.g., and some moments of inertia of a thin plate in the shape of a ninety-degree sector of a circle. Exercise 3.10 Find the mass, the location of the c.g., and some moments of inertia of a thin plate in the shape of a ninety-degree segment of a circle (i.e., a sector minus a right triangle).
3.5
Parallel-Axis Theorem
The mass of a body is the same, no matter what coordinates are chosen; The location of its c.g. is in the same place, no matter in what coordinates it is expressed. But the second moment of a body depends on the location of the axis about which it turns. Steiner's parallel-axis theorem transfers the second moment about the c.g., to the second moment about any other pivot: Jpivot = Jc.g. + M d2 (3.8)
where d is the distance from the axis of rotation to the center of gravity. If we know the second moment about any axis through the c.g., we can find the second moment about any other parallel axis. Conversely, if we know the second moment about any axis, we can find the second moment about the parallel axis through the c.g., and from that the second moment about any other parallel axis.
28
Chapter 3. FORMULATION OF ROTATIONAL SYSTEMS
Figure 3.3: Compound Pendulum Example: The mass of a yardstick of cross-sectional area A is Z L M = A dx = AL
0
The first moment of mass is A Z
L
xdx = 0.5LM
0
so that the c.g. must be at x = 0.5L. The second moment for a thin stick about a pivot at one end is Z L Jpivot = A x2 dx = M L2 /3
0
and the second moment for a thin stick about its center is Z L/2 x2 dx = ML2 /12 Jc.g. = 2A
0
We can verify that for a pivot at the end Jpivot = Jc.g. + M d2 = M L2 /12 + M (L/2)2 = ML2 /3 We can also obtain the second moment about another pivot, for example at the one-foot mark (L/3 from the end or L/6 from the center) by writing Jpivot = Jc.g. + Md2 = ML2 /12 + M (L/6)2 = M L2 /9. Exercise 3.11 In this example, what is the order of magnitude of the error due to the width of the yardstick?
3.6
The Compound Pendulum
Let us study a pendulum which consists of a mass M in the form of a flat, solid disk of radius R, supported by a "weightless" rod measuring length L from
3.7. SUMMARY
29
pivot to the mass' center of gravity (Figure 3.3). The torque in the governing Equation 3.1 is the force due to gravity multiplied by the lever-arm length upon which it acts: Jpivot -M gL sin() = (3.9) gc gc where Jpivot = Jc.g. + M L2 ; for a flat disk, Jc.g. = 0.5MR2 so that we get 0.5M R2 M L2 M gL + + sin = 0 (3.10) gc gc gc Compare this with the solutions for the simple pendulum in the previous chapter (page ??) and in Section 3.2. 1. This is a nonlinear equation. We can linearize it by substituting for sin . We know what kind of error this introduces from the Maclaurin series expansion (Equation 3.4); the error is small if 2 /6 1. 2. The effect of the size of the mass is now quantified. It is negligible if R2 L2 . For a compact mass, this reduces to our Equation ?? for the simple pendulum. However, using the rotational-system approach, we now know what the effects of the linearization and of the point-mass assumption are. On the one hand, the effect of the size of the mass is less than 0.5% if R < 0.1L. In the other extreme casea flywheel mounted slightly eccentricallythe length L is negligible for the moment-of-inertia term (but important in the gravitational term) if L R.
3.7
Summary
Rotational motion of a body is conveniently described in either of two coordinate systems: either as a combination of translation of the c.g. plus rotationthe variables are xc.g. and ; or as pure rotation about a pivot or other instantaneous center of rotation the variables are and the location of the center of rotation. For the second description, a single differential equation can be written by summing up all the torques about the instantaneous center of rotation. The coefficient of is the moment of inertia, which can be computed by means of a second-moment integral about the given pivot axis. Also, if the location of the c.g. is known, the transfer theorem lets us compute moment of inertia about any pivot axis from the moment of inertia about any other paralle...