22 Center of Mass, Moment of Inertia

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1 Chapter Center of Mass, Moent of Inertia Center of Mass, Moent of Inertia A istake that crops up in the calculation of oents of inertia, involves the Parallel Ais Theore. The istake is to interchange the oent of inertia of the ais through the center of ass, with the one parallel to that, when applying the Parallel Ais Theore. Recognizing that the subscript CM in the parallel ais theore stands for center of ass will help one avoid this istake. Also, a check on the answer, to ake sure that the value of the oent of inertia with respect to the ais through the center of ass is saller than the other oent of inertia, will catch the istake. Center of Mass Consider two particles, having one and the sae ass, each of which is at a different position on the ais of a Cartesian coordinate syste. y #1 # Coon sense tells you that the average position of the aterial aking up the two particles is idway between the two particles. Coon sense is right. We give the nae center of ass to the average position of the aterial aking up a distribution, and the center of ass of a pair of sae-ass particles is indeed idway between the two particles. How about if one of the particles is ore assive than the other? One would epect the center of ass to be closer to the ore assive particle, and again, one would be right. To deterine the position of the center of ass of the distribution of atter in such a case, we copute a weighted su of the positions of the particles in the distribution, where the weighting factor for a given particle is that fraction, of the total ass, that the particle s own ass is. Thus, for two particles on the ais, one of ass 1, at 1, and the other of ass, at, y ( 1, ) (, ) 1 1

2 Chapter Center of Mass, Moent of Inertia the position of the center of ass is given by = + (-1) Note that each weighting factor is a proper fraction and that the su of the weighting factors is always 1. Also note that if, for instance, 1 is greater than, then the position 1 of particle 1 will count ore in the su, thus ensuring that the center of ass is found to be closer to the ore assive particle (as we know it ust be). Further note that if 1 =, each weighting factor is 1, as is evident when we substitute for both 1 and in equation -1: = = = 1 + The center of ass is found to be idway between the two particles, right where coon sense tells us it has to be. The Center of Mass of a Thin Rod Quite often, when the finding of the position of the center of ass of a distribution of particles is called for, the distribution of particles is the set of particles aking up a rigid body. The easiest rigid body for which to calculate the center of ass is the thin rod because it etends in only one diension. (Here, we discuss an ideal thin rod. A physical thin rod ust have soe nonzero diaeter. The ideal thin rod, however, is a good approiation to the physical thin rod as long as the diaeter of the rod is sall copared to its length.) In the siplest case, the calculation of the position of the center of ass is trivial. The siplest case involves a unifor thin rod. A unifor thin rod is one for which the linear ass density µ, the ass-per-length of the rod, has one and the sae value at all points on the rod. The center of ass of a unifor rod is at the center of the rod. So, for instance, the center of ass of a unifor rod that etends along the ais fro = to = is at ( /, ). The linear ass density µ, typically called linear density when the contet is clear, is a easure of how closely packed the eleentary particles aking up the rod are. Where the linear density is high, the particles are close together. 1

3 Chapter Center of Mass, Moent of Inertia To picture what is eant by a non-unifor rod, a rod whose linear density is a function of position, iagine a thin rod ade of an alloy consisting of lead and aluinu. Further iagine that the percentage of lead in the rod varies soothly fro % at one end of the rod to 1% at the other. The linear density of such a rod would be a function of the position along the length of the rod. A one-illieter segent of the rod at one position would have a different ass than that of a one-illieter segent of the rod at a different position. People with soe eposure to calculus have an easier tie understanding what linear density is than calculus-deprived individuals do because linear density is just the ratio of the aount of ass in a rod segent to the length of the segent, in the liit as the length of the segent goes to zero. Consider a rod that etends fro to along the ais. Now suppose that s () is the ass of that segent of the rod etending fro to where but <. Then, the linear d density of the rod at any point along the rod, is just s evaluated at the value of in d question. Now that you have a good idea of what we ean by linear ass density, we are going to illustrate how one deterines the position of the center of ass of a non-unifor thin rod by eans of an eaple. Eaple -1 Find the position of the center of ass of a thin rod that etends fro to.89 along the ais of a Cartesian coordinate syste and has a linear density given by µ ( ) =. 6. In order to be able to deterine the position of the center of ass of a rod with a given length and a given linear density as a function of position, you first need to be able to find the ass of such a rod. To do that, one ight be tepted to use a ethod that works only for the special case of a unifor rod, naely, to try using = µ with being the length of the rod. The proble with this is, that µ varies along the entire length of the rod. What value would one use for µ? One ight be tepted to evaluate the given µ at = and use that, but that would be acting as if the linear density were constant at µ = µ (). It is not. In fact, in the case at hand, µ () is the aiu linear density of the rod, it only has that value at one point on the rod. What we can do is to say that the infinitesial aount of ass d in a segent d of the rod is µ d. Here we are saying that at soe position on the rod, the aount of ass in the infinitesial length d of the rod is the value of µ at that value, ties the infinitesial length d. Here we don t have to worry about the fact that µ changes with position since the segent d is infinitesially long, eaning, essentially, that it has zero length, so the whole segent is essentially at one position and hence the value of µ at that is good for the whole segent d. d = µ () d (-) 1

4 Chapter Center of Mass, Moent of Inertia y z d (, ) d = µ d Now this is true for any value of, but it just covers an infinitesial segent of the rod at. To get the ass of the whole rod, we need to add up all such contributions to the ass. Of course, since each d corresponds to an infinitesial length of the rod, we will have an infinite nuber of ters in the su of all the d s. An infinite su of infinitesial ters, is an integral. d = µ ( ) d (-) where the values of have to run fro to to cover the length of the rod, hence the liits on the right. Now the atheaticians have provided us with a rich set of algoriths for evaluating integrals, and indeed we will have to reach into that toolbo to evaluate the integral on the right, but to evaluate the integral on the left, we cannot, should not, and will not turn to such an algorith. Instead, we use coon sense and our conceptual understanding of what the integral on the left eans. In the contet of the proble at hand, d eans the su of all the infinitesial bits of ass aking up the rod. Now, if you add up all the infinitesial bits of ass aking up the rod, you get the ass of the rod. So d is just the ass of the rod, which we will call. Equation - then becoes = µ ( ) d (-) Replacing µ () with the given epression for the linear density µ =. 6 which I choose to write as µ = b with b being defined by b.6 we obtain Factoring out the constant yields = b d 1

5 Chapter Center of Mass, Moent of Inertia = b d When integrating the variable of integration raised to a power all we have to do is increase the power by one and divide by the new power. This gives = b Evaluating this at the lower and upper liits yields = b = b b = The value of is given as.89 and we defined b to be the constant. 6 in the given epression for µ, µ =. 6, so. 6 (. 89 ) = =.17 That s a value that will coe in handy when we calculate the position of the center of ass. Now, when we calculated the center of ass of a set of discrete particles (where a discrete particle is one that is by itself, as opposed, for instance, to being part of a rigid body) we just carried out a weighted su in which each ter was the position of a particle ties its weighting factor and the weighting factor was that fraction, of the total ass, represented by the ass of the particle. We carry out a siilar procedure for a continuous distribution of ass such as that which akes up the rod in question. et s start by writing one single ter of the su. We ll consider an infinitesial length d of the rod at a position along the length of the rod. The position, as just stated, is, and the weighting factor is that fraction of the total ass of the rod 16

6 Chapter Center of Mass, Moent of Inertia that the ass d of the infinitesial length d represents. That eans the weighting factor is d, so, a ter in our weighted su of positions looks like: d Now, d can be epressed as µ d so our epression for the ter in the weighted su can be written as µ d That s one ter in the weighted su of positions, the su that yields the position of the center of ass. The thing is, because the value of is unspecified, that one ter is good for any infinitesial segent of the bar. Every ter in the su looks just like that one. So we have an epression for every ter in the su. Of course, because the epression is for an infinitesial length d of the rod, there will be an infinite nuber of ters in the su. So, again we have an infinite su of infinitesial ters. That is, again we have an integral. Our epression for the position of the center of ass is: µ d = Substituting the given epression µ ( ) =. 6 for µ, which we again write as µ = b with b being defined by b.6, yields = b d Rearranging and factoring the constants out gives Net we carry out the integration. = b d = b 17

7 Chapter Center of Mass, Moent of Inertia = b b = Now we substitute values with units; the ass of the rod that we found earlier, the constant b that we defined to siplify the appearance of the linear density function, and the given length of the rod:. 6 ( 89 ). = (. 17 ) =.668 This is our final answer for the position of the center of ass. Note that it is closer to the denser end of the rod, as we would epect. The reader ay also be interested to note that had we substituted the epression b = that we derived for the ass, rather than the value we obtained when we evaluated that epression, our epression for would have siplified to which evaluates to =.668, the sae result as the one above. Moent of Inertia a.k.a. Rotational Inertia You already know that the oent of inertia of a rigid object, with respect to a specified ais of rotation, depends on the ass of that object, and how that ass is distributed relative to the ais of rotation. In fact, you know that if the ass is packed in close to the ais of rotation, the object will have a saller oent of inertia than it would if the sae ass was ore spread out relative to the ais of rotation. et s quantify these ideas. (Quantify, in this contet, eans to put into equation for.) We start by constructing, in our inds, an idealized object for which the ass is all concentrated at a single location which is not on the ais of rotation: Iagine a assless disk rotating with angular velocity w about an ais through the center of the disk and perpendicular to its faces. et there be a particle of ass ebedded in the disk at a distance r fro the ais of rotation. Here s what it looks like fro a viewpoint on the ais of rotation, soe distance away fro the disk: 18

8 Chapter Center of Mass, Moent of Inertia w O r Particle of ass Massless Disk where the ais of rotation is arked with an O. Because the disk is assless, we call the oent of inertia of the construction, the oent of inertia of a particle, with respect to rotation about an ais fro which the particle is a distance r. Knowing that the velocity of the particle can be epressed as v = r w you can show yourself 1 how I ust be defined in order for the kinetic energy epression K = Iw for the object, 1 viewed as a spinning rigid body, to be the sae as the kinetic energy epression K = v for the particle oving through space in a circle. Either point of view is valid so both viewpoints ust yield the sae kinetic energy. Please go ahead and derive what I ust be and then coe back and read the derivation below. Here is the derivation: 1 Given that K = v, we replace v with r w. This gives K = which can be written as 1 K = ( r ) w For this to be equivalent to we ust have 1 K = Iw I = r 1 ( rw) (-) This is our result for the oent of inertia of a particle of ass, with respect to an ais of rotation fro which the particle is a distance r. 19

9 Chapter Center of Mass, Moent of Inertia Now suppose we have two particles ebedded in our assless disk, one of ass 1 at a distance r 1 fro the ais of rotation and another of ass at a distance r fro the ais of rotation. w O r 1 1 r Massless Disk The oent of inertia of the first one by itself would be I 1 = 1 r 1 and the oent of inertia of the second particle by itself would be I = r The total oent of inertia of the two particles ebedded in the assless disk is siply the su of the two individual oents of inertial. I = I 1 + I 1r1 r I = + This concept can be etended to include any nuber of particles. For each additional particle, one siply includes another i r i ter in the su where i is the ass of the additional particle and r i is the distance that the additional particle is fro the ais of rotation. In the case of a rigid object, we subdivide the object up into an infinite set of infinitesial ass eleents d. Each ass eleent contributes an aount of oent of inertia di =r d (-6) to the oent of inertia of the object, where r is the distance that the particular ass eleent is fro the ais of rotation. 1

10 Chapter Center of Mass, Moent of Inertia Eaple - Find the oent of inertia of the rod in Eaple -1 with respect to rotation about the z ais. In Eaple -1, the linear density of the rod was given as µ =. 6. To reduce the nuber of ties we have to write the value in that epression, we will write it as µ = b with b being defined as b.6. The total oent of inertia of the rod is the infinite su of the infinitesial contributions fro each and every ass eleent d aking up the rod. di =r d (-6) y z d (, ) d = µ d In the diagra, we have indicated an infinitesial eleent d of the rod at an arbitrary position on the rod. The z ais, the ais of rotation, looks like a dot in the diagra and the distance r in di =r d, the distance that the bit of ass under consideration is fro the ais of rotation, is siply the abscissa of the position of the ass eleent. Hence, equation -6 for the case at hand can be written as di = d 11

11 Chapter Center of Mass, Moent of Inertia which we copy here di = d By definition of the linear ass density µ, the infinitesial ass d can be epressed as d = µ d. Substituting this into our epression for d I yields d I = µ d Now µ was given as b (with b actually being the sybol that I chose to use to represent the given constant. 6 ). Substituting b in for µ in our epression for d I yields d I = ( b ) d di = b This epression for the contribution of an eleent d of the rod to the total oent of inertia of the rod is good for every eleent d of the rod. The infinite su of all such infinitesial contributions is thus the integral di = Again, as with our last integration, on the left, we have not bothered with liits of integration the infinite su of all the infinitesial contributions to the oent of inertia is siply the total oent of inertia. I = On the right we use the liits of integration to to include every eleent of the rod which etends fro = to =, with given as.89. Factoring out the constant b gives us Now we carry out the integration: I = b b b d d d d I = b I = b 1

12 Chapter Center of Mass, Moent of Inertia I = b Substituting the given values of b and yields: I =. 6 (. 89 ) I =. 76 The Parallel Ais Theore We state, without proof, the parallel ais theore: in which: I =ICM + d (-7) I is the oent of inertia of an object with respect to an ais fro which the center of ass of the object is a distance d. I CM is the oent of inertia of the object with respect to an ais that is parallel to the first ais and passes through the center of ass. is the ass of the object. d is the distance between the two aes. The parallel ais theore relates the oent of inertia I CM of an object, with respect to an ais through the center of ass of the object, to the oent of inertia I of the sae object, with respect to an ais that is parallel to the ais through the center of ass and is at a distance d fro the ais through the center of ass. A conceptual stateent ade by the parallel ais theore is one that you probably could have arrived at by eans of coon sense, naely that the oent of inertia of an object with respect to an ais through the center of ass is saller than the oent of inertia about any ais parallel to that one. As you know, the closer the ass is packed to the ais of rotation, the saller the oent of inertia; and; for a given object, per definition of the center of ass, the ass is packed ost closely to the ais of rotation when the ais of rotation passes through the center of ass. 1

13 Chapter Center of Mass, Moent of Inertia Eaple - Find the oent of inertia of the rod fro eaples -1 and -, with respect to an ais that is perpendicular to the rod and passes through the center of ass of the rod. Recall that the rod in question etends along the ais fro = to = with =.89 and that the rod has a linear density given by µ = b with b. 6. The ais in question can be chosen to be one that is parallel to the z ais, the ais about which, in solving eaple -, we found the oent of inertia to be I =. 76. In solving eaple -1 we found the ass of the rod to be =.17 and the center of ass of the rod to be at a distance d =.668 away fro the z ais. Here we present the solution to the proble: y z d The center of ass of the rod I I =I I CM CM + d =I d C M = (. 668 ) I CM =. 7 1