Physics
Parallel Axis Theorem
The Parallel Axis Theorem states that the moment of inertia of a rigid body about any axis parallel to its center of mass axis is equal to the sum of the moment of inertia about the center of mass axis and the product of the body's mass and the square of the distance between the two axes. This theorem is a useful tool for calculating the moment of inertia of complex objects.
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7 Key excerpts on "Parallel Axis Theorem"
- Raymond Serway, John Jewett(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
The Parallel Axis Theorem can be used with the geometry shown to determine the moment of inertia of the original object around the z9 axis. y' y' x', y' dm O D r ' x' x' CM y y' CM x' CM x' CM, y' CM x c y x Example 10.9 Applying the Parallel-Axis Theorem Consider once again the uniform rigid rod of mass M and length L shown in Figure 10.15. Find the moment of inertia of the rod about an axis perpendicular to the rod through one end (the y9 axis in Fig. 10.15). S O L U T I O N Conceptualize Imagine twirling the rod around an endpoint rather than the midpoint. If you have a meterstick handy, try it and notice the degree of difficulty in rotating it around the end compared with rotating it around the center. Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 10.7 Rotational Kinetic Energy 267 10.7 Rotational Kinetic Energy After investigating the role of forces in our study of translational motion, we turned our attention to approaches involving energy in Chapters 7 and 8. We do the same thing in our current study of rotational motion. In Chapter 7, we defined the kinetic energy of an object as the energy associated with its motion through space. An object rotating about a fixed axis remains station- ary in space, so there is no kinetic energy associated with translational motion. The individual particles making up the rotating object, however, are moving through space; they follow circular paths. Consequently, there is kinetic energy associated with rotational motion.
eBook - PDFThe Mechanical Universe
Mechanics and Heat, Advanced Edition
- Steven C. Frautschi, Richard P. Olenick, Tom M. Apostol, David L. Goodstein(Authors)
- 2008(Publication Date)
- Cambridge University Press(Publisher)
14.5 THE PARALLEL-AXIS THEOREM Table 14.1 Moments of Inertia for Uniform Bodies 379 Body Rod (length L) Thin ring (radius R) Circular cylinder Thin disk Solid sphere Thin spherical shell Rectangular plate (length a, height b) Axis Perpendicular axis through center Perpendicular axis through center Axis of cylinder Transverse axis through center Any axis through center Any axis through center Axis through center perpendicular to the plate jfttf MR 2 ! * -{MR 1 ;»> !** 12; f b 2 ) first sight the calculation may look complicated, but in fact the answer can be written down immediately with the aid of the parallel-axis theorem, which we'l! now prove. Figure 14.8 represents a section through the rigid body whose moment of inertia we wish to calculate. The axis of rotation, which is perpendicular to the plane of the figure, pierces this plane at point P, and C is the point where an axis passing through the center of mass and parallel to the axis of rotation pierces this plane. The distance between P and C is r; for convenience we choose the x axis so that it passes through both P and C. The moment of inertia of the body about the axis through P is From Fig. 14.8 , the law of cosines gives rj = r'p- + r 2 + 2rr' t cos fl,-, (14.16) and we have r cos 6 ; = x, the x coordinate of m, relative to the center of mass. Hence 'p = E m i r l 2 + 2 V 2 + 2r 2 m : x -(14.17) mass m f P r C Figure 14.8 The parallel-axis theorem. 380 ROTATIONAL DYNAMICS FOR RIGID BODIES The first sum on the right is / c , the moment of inertia about the center of mass; the second is M, the mass of the body, times r 2 , the square of the distance between the two parallel axes. The last sum vanishes because 2 m^JM represents the distance of the x component of the center of mass from C, which is zero. Thus we obtain the parallel-axis theorem: / P = / c + Mr 1 .
- Andrei D. Polyanin, Alexei Chernoutsan(Authors)
- 2010(Publication Date)
- CRC Press(Publisher)
It can be seen from equations (P1.9.1.3) and (P1.9.1.4) that the moment of inertia is a measure of the body’s inertia with respect to rotational motion; moment of inertia plays the same role as mass in translational motion. Example. Compute the moment of inertia of a thin disk of mass m and radius R about its axis of symmetry. Solution. Dividing the disk into thin circular rings and integrating, we obtain I = integraldisplay r 2 dm = integraldisplay R 0 r 2 parenleftBig m πR 2 parenrightBig 2 πrdr = mR 2 2 . The same result holds also for a homogeneous solid cylinder. ◮ Parallel Axis Theorem. The Parallel Axis Theorem (also known as the Huygens–Steiner theorem ) relates the moment of inertia I about an arbitrary axis to the moment of inertia I 0 about a parallel axis passing through the center of mass of the body: I = I 0 + ma 2 , (P1. 9 . 2 . 1 ) where m is mass of the body and a is the distance between the axes. For example, the moment of inertia of a disk about an axis perpendicular to its plane and passing through its edge equals 1 2 mR 2 + mR 2 = 3 2 mR 2 . The minimum moment of inertia among all parallel axes is for the axis passing the center of mass. ◮ Perpendicular axis theorem. The perpendicular axis theorem asserts that the moment of inertia of a plane body about an arbitrary axis z perpendicular to the body plane equals the sum of the moments of inertia about mutually perpendicular axes x and y lying in the body plane and intersecting z : I z = I x + I y . For example, the moment of inertia of a thin disk about an axis of symmetry lying in its plane equals I x = I y = 1 2 I z = 1 4 mR 2 . P1.9. D YNAMICS OF R IGID B ODY 431 ◮ Moments of inertia of some bodies. Listed below are moments of inertia of some bodies of various shape. 1) Thin ring of radius R (about its axis of symmetry): I = mR 2 . The same result holds for a thin hollow cylinder (without end caps).
eBook - PDF- Nivaldo A. Lemos(Author)
- 2018(Publication Date)
- Cambridge University Press(Publisher)
The expression (4.56) coincides with the definition of moment inertia found in elementary n ^ d k q k r k Fig. 4.3 Moment of inertia: d k is the distance of the kth particle from the rotation axis. 121 Kinetic Energy and Parallel Axis Theorem . r k r ´ k m k CM n O ^ R a Fig. 4.4 Parallel Axis Theorem. physics textbooks. Likewise, (4.54) is the usual formula for the rotational kinetic energy of a rigid body. In general, however, the angular velocity changes direction as time passes, so I varies with time. The moment of inertia I is constant if the body is constrained to rotate about a fixed axis, the case normally treated in elementary physics. The moment of inertia depends on the choice of the origin of the coordinate system used to calculate it. There is, however, a simple relation between moments of inertia with respect to parallel axes when one of them goes through the centre of mass. Theorem 4.4.1 (Parallel Axis Theorem) The moment of inertia with respect to a given axis is equal to the moment of inertia with respect to a parallel axis through the centre of mass plus the moment of inertia of the body with respect to the given axis computed as if the entire mass of the body were concentrated at the centre of mass. 2 Proof From Fig. 4.4 it follows that r k = R + r k , where R is the position vector of the centre of mass from origin O on the given axis. The moment of inertia with respect to the given axis is, by (4.56) and (4.57), I a = k m k |ˆ n × r k | 2 = k m k |ˆ n × R + ˆ n × r k | 2 = k m k |ˆ n × R| 2 + k m k |ˆ n × r k | 2 + 2( ˆ n × R)· ˆ n × k m k r k . (4.58) With the use of (1.19) this last equation reduces to I a = I CM + M|ˆ n × R| 2 , (4.59) completing the proof, since |ˆ n × R| 2 is the square of the distance of the centre of mass from the given axis. 2 This result is also known as Steiner’s theorem. 122 Dynamics of Rigid Bodies The inertia tensor itself admits a decomposition analogous to (4.59).
eBook - PDF- Richard C. Hill, Kirstie Plantenberg(Authors)
- 2013(Publication Date)
- SDC Publications(Publisher)
In this case, Equation 6.4-2 can be expressed using integral notation as shown in Equation 6.4-3. Each incremental piece of the rigid body has mass dm and perpendicular distance from the reference axis r. Mass moment of inertia for a rigid body: 2 O I r dm (6.4-3) I O = mass moment of inertia about the axis through O r = perpendicular distance of each incremental mass element dm from an axis through O Mass moments of inertia for many simple rigid body shapes are given in Appendix A. Many of these inertias are given with respect to the axis that passes through the body's center of mass (G). In order to analyze rigid-body motion, we may need the mass moment of inertia of the body with respect to an arbitrary axis as shown in Figure 6.4-4. The parallel - axis theorem (Equation 6.4-4) may be used to calculate the mass moment of inertia of the body with respect to an axis parallel to the axis passing through the body's mass center. To illustrate how Equation 6.4-4 is applied, consider the body shown in Figure 6.4-4. If we want to know the mass moment of inertia about the x'-axis, but are only given the mass moment of inertia about the x G axis (the axis through the body's center of mass) and the distance between the two axes, the equation would be 2 ' G x x x I I md . Parallel-axis theorem: 2 G I I md (6.4-4) I' = mass moment of inertia with respect to an axis parallel to the axis passing through G I G = mass moment of inertia with respect to an axis passing through G m = mass of the body d = the perpendicular distance from the parallel axis to mass center G Conceptual Dynamics Kinetics: Chapter 6 – Rigid Body Newtonian Mechanics 6 - 12 Notice that the mass moment of inertia tables in Appendix A give values of the inertia with respect to a specific axis.
eBook - PDF- Daniel Kleppner, Robert Kolenkow(Authors)
- 2013(Publication Date)
- Cambridge University Press(Publisher)
ANGULAR MOMENTUM AND FIXED AXIS ROTATION 7 7.1 Introduction 240 7.2 Angular Momentum of a Particle 241 7.3 Fixed Axis Rotation 245 7.3.1 Moment of Inertia 246 7.3.2 The Parallel Axis Theorem 249 7.4 Torque 250 7.5 Torque and Angular Momentum 252 7.5.1 Conservation of Angular Momentum 253 7.6 Dynamics of Fixed Axis Rotation 260 7.7 Pendulum Motion and Fixed Axis Rotation 262 7.7.1 The Simple Pendulum 262 7.7.2 The Physical Pendulum 263 7.8 Motion Involving Translation and Rotation 267 7.8.1 Torque on a Moving Body 269 7.9 The Work–Energy Theorem and Rotational Motion 273 7.10 The Bohr Atom 277 Note 7.1 Chasles’ Theorem 280 Note 7.2 A Summary of the Dynamics of Fixed Axis Rotation 282 Problems 282 240 ANGULAR MOMENTUM AND FIXED AXIS ROTATION 7.1 Introduction Our discussion of the principles of mechanics has so far neglected an im-portant issue—the rotational motion of solid bodies. For example, con-sider the common yo-yo running up and down its string as the spool winds and unwinds. In principle we can predict the motion because each particle of the yo-yo moves according to Newton’s laws. The motion is simple but attempting to analyze it on a particle-by-particle basis would quickly prove to be hopeless. To treat the rotational motion of extended bodies as a whole, we need to develop a simple method, which is the goal of this chapter. In attacking the problem of translational motion, we introduced the concepts of force, linear momentum, and center of mass. In this chapter we introduce precisely analogous concepts for rotational motion: torque, angular momentum, and moment of inertia. Our goal, of course, is much more ambitious than merely to under-stand yo-yo motion; our goal is to find a general way to analyze the motion of rigid bodies under any combination of applied forces.
eBook - PDF- H. W. Harkness(Author)
- 2014(Publication Date)
- Academic Press(Publisher)
G is the axis, normal to the plane through the center of gravity of the body and O is a parallel axis. Choosing an element FIG. 1-10 of mass dM we write the expression for the moments of inertia for the two axes. Ig = j> r 2 dM; I 0 = fx 2 dM By using the cosine law, we obtain #2 = r 2 + d 2 _ 2r d cos Θ FIG. 1-9 14 I. RIGID BODY MOTION AROUND A FIXED AXIS and the second of these becomes / = < j > r 2 dM + d* j>dM - 2d < j > r cos Θ · dM The first term is the moment of inertia of the body for the axis through its center of gravity. The second term is the mass by the square of the distance between the axes, and the third term is zero since it is the sum, over the whole body, of the products of elements of mass and their distance from the axis through the center of gravity. Hence, I 0 =I g + Md* (1-12) The Moment of Inertia of a Thin Rectangular Plate for a Polar Axis through Its Center of Gravity Using the above theorems and the result we have obtained for the moment of inertia of a thin rectangular plate for an axis in one edge, we may easily calculate its moment of inertia for a polar axis through its center of gravity. Using the theorem for the shift of axis, we obtain the moment of inertia for an axis in the plane of the plate and through it center of gravity. Thus (see Figure 1-11) I y = £M/ 2 - M/2/4 = T ^M/ 2 0 — / — -1 b I Y FIG. M l INERTIA MOMENT OF A CYLINDER 15 From symmetry the moment of inertia for the XX axis is I x = -^Mb* and hence by theorem I above, the polar moment of inertia is written at once: / - ^ M ( * 2 + / 2 ) (1-13) The Moment of Inertia of a Rectangular Parallelepiped for an Axis Normal to One of Its Faces and through Its Center of Gravity The last result may now be extended, by the principle already established, and the moment of inertia of a rectangular parallel-epiped written at once. For, to the plate in the above analysis, we may add similar plates one after the other until a three-dimensional body is built up.
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