{"id":958,"date":"2018-02-06T16:53:28","date_gmt":"2018-02-06T16:53:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/?post_type=chapter&#038;p=958"},"modified":"2018-02-28T19:35:27","modified_gmt":"2018-02-28T19:35:27","slug":"10-5-calculating-moments-of-inertia","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/10-5-calculating-moments-of-inertia\/","title":{"raw":"10.5 Calculating Moments of Inertia","rendered":"10.5 Calculating Moments of Inertia"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Calculate the moment of inertia for uniformly shaped, rigid bodies<\/li>\r\n \t<li>Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known<\/li>\r\n \t<li>Calculate the moment of inertia for compound objects<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167133856164\">In the preceding section, we defined the moment of inertia but did not show how to calculate it. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. This section is very useful for seeing how to apply a general equation to complex objects (a skill that is critical for more advanced physics and engineering courses).<\/p>\r\n\r\n<div id=\"fs-id1167133538983\" class=\"bc-section section\">\r\n<h3>Moment of Inertia<\/h3>\r\n<p id=\"fs-id1167133826555\">We defined the moment of inertia <em>I<\/em> of an object to be $$ I=\\sum _{i}{m}_{i}{r}_{i}^{2} $$ for all the point masses that make up the object. Because <em>r<\/em> is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let\u2019s take a simple example of two masses at the end of a massless (negligibly small mass) rod (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Barbell\">(Figure)<\/a>) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.<\/p>\r\nIn the case with the axis in the center of the barbell, each of the two masses <em>m<\/em> is a distance <em>R<\/em> away from the axis, giving a moment of inertia of\r\n<div id=\"fs-id1167133521048\" class=\"unnumbered\">$${I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}.$$<\/div>\r\n<p id=\"fs-id1167133426576\">In the case with the axis at the end of the barbell\u2014passing through one of the masses\u2014the moment of inertia is<\/p>\r\n\r\n<div id=\"fs-id1167133568600\" class=\"unnumbered\">$${I}_{2}=m{(0)}^{2}+m{(2R)}^{2}=4m{R}^{2}.$$<\/div>\r\n<p id=\"fs-id1167133456090\">From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_10_05_Barbell\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"891\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194447\/CNX_UPhysics_10_05_Barbell.jpg\" alt=\"Figure A shows a barbell of the length 2 R with the masses m at the ends. It is rotating through its center. Figure B shows a barbell of the length 2 R with the masses m at the ends. It is rotating through one end.\" width=\"891\" height=\"301\" \/> <strong>Figure 10.23<\/strong> (a) A barbell with an axis of rotation through its center; (b) a barbell with an axis of rotation through one end.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133517824\">In this example, we had two point masses and the sum was simple to calculate. However, to deal with objects that are not point-like, we need to think carefully about each of the terms in the equation. The equation asks us to sum over each \u2018piece of mass\u2019 a certain distance from the axis of rotation. But what exactly does each \u2018piece of mass\u2019 mean? Recall that in our derivation of this equation, each piece of mass had the same magnitude of velocity, which means the whole piece had to have a single distance <em>r<\/em> to the axis of rotation. However, this is not possible unless we take an infinitesimally small piece of mass <em>dm<\/em>, as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Pointmass\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_10_05_Pointmass\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"372\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194450\/CNX_UPhysics_10_05_pointmass.jpg\" alt=\"Figure shows a point dm located on the X axis at distance r from the center.\" width=\"372\" height=\"381\" \/> <strong>Figure 10.24<\/strong> Using an infinitesimally small piece of mass to calculate the contribution to the total moment of inertia.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133534200\">The need to use an infinitesimally small piece of mass <em>dm<\/em> suggests that we can write the moment of inertia by evaluating an integral over infinitesimal masses rather than doing a discrete sum over finite masses:<\/p>\r\n\r\n<div id=\"fs-id1167133335739\" class=\"equation-callout\">\r\n<div id=\"fs-id1167133385098\">$$I=\\sum _{i}{m}_{i}{r}_{i}{}^{2}\\enspace\\text{becomes}I=\\int {r}^{2}dm.$$<\/div>\r\n<\/div>\r\nThis, in fact, is the form we need to generalize the equation for complex shapes. It is best to work out specific examples in detail to get a feel for how to calculate the moment of inertia for specific shapes. This is the focus of most of the rest of this section.\r\n<div id=\"fs-id1167133698940\" class=\"bc-section section\">\r\n<h4>A uniform thin rod with an axis through the center<\/h4>\r\nConsider a uniform (density and shape) thin rod of mass <em>M<\/em> and length <em>L<\/em> as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Thinrod\">(Figure)<\/a>. We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the <em>z<\/em>-axis is the axis of rotation and the <em>x<\/em>-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the <em>x<\/em>-axis.\r\n<div id=\"CNX_UPhysics_10_05_Thinrod\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"433\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194452\/CNX_UPhysics_10_05_thinrod.jpg\" alt=\"Figure shows a thin rod that rotates about an axis through the center. Part of the rod of the length dx has a mass dm.\" width=\"433\" height=\"158\" \/> <strong>Figure 10.25<\/strong> Calculation of the moment of inertia I for a uniform thin rod about an axis through the center of the rod.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133519807\">We define <em>dm<\/em> to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the <strong>linear mass density<\/strong> $$ \\lambda  $$ of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write<\/p>\r\n\r\n<div id=\"fs-id1167133661384\" class=\"unnumbered\">$$\\lambda =\\frac{m}{l}\\enspace\\text{or}m=\\lambda l.$$<\/div>\r\n<p id=\"fs-id1167133410629\">If we take the differential of each side of this equation, we find<\/p>\r\n\r\n<div id=\"fs-id1167133557177\" class=\"unnumbered\">$$dm=d(\\lambda l)=\\lambda (dl)$$<\/div>\r\n<p id=\"fs-id1167133321526\">since $$ \\lambda  $$ is constant. We chose to orient the rod along the <em>x<\/em>-axis for convenience\u2014this is where that choice becomes very helpful. Note that a piece of the rod <em>dl<\/em> lies completely along the <em>x<\/em>-axis and has a length <em>dx<\/em>; in fact, $$ dl=dx $$ in this situation. We can therefore write $$ dm=\\lambda (dx)$$, giving us an integration variable that we know how to deal with. The distance of each piece of mass <em>dm<\/em> from the axis is given by the variable <em>x<\/em>, as shown in the figure. Putting this all together, we obtain<\/p>\r\n\r\n<div id=\"fs-id1167133467275\" class=\"unnumbered\">$$I=\\int {r}^{2}dm=\\int {x}^{2}dm=\\int {x}^{2}\\lambda dx.$$<\/div>\r\n<p id=\"fs-id1167133558728\">The last step is to be careful about our limits of integration. The rod extends from $$ x=\\text{\u2212}L\\text{\/}2 $$ to $$ x=L\\text{\/}2$$, since the axis is in the middle of the rod at $$ x=0$$. This gives us<\/p>\r\n\r\n<div id=\"fs-id1167133440496\" class=\"unnumbered\">$$\\begin{array}{cc}\\hfill I&amp; =\\underset{\\text{\u2212}L\\text{\/}2}{\\overset{L\\text{\/}2}{\\int }}{x}^{2}\\lambda dx=\\lambda \\frac{{x}^{3}}{3}{|}_{\\text{\u2212}L\\text{\/}2}^{L\\text{\/}2}=\\lambda (\\frac{1}{3})[{(\\frac{L}{2})}^{3}-{(\\frac{\\text{\u2212}L}{2})}^{3}]\\hfill \\\\ &amp; =\\lambda (\\frac{1}{3})\\frac{{L}^{3}}{8}(2)=\\frac{M}{L}(\\frac{1}{3})\\frac{{L}^{3}}{8}(2)=\\frac{1}{12}M{L}^{2}.\\hfill \\end{array}$$<\/div>\r\n<p id=\"fs-id1167133326786\">Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133319345\" class=\"bc-section section\">\r\n<h4>A uniform thin rod with axis at the end<\/h4>\r\n<p id=\"fs-id1167133465541\">Now consider the same uniform thin rod of mass <em>M<\/em> and length <em>L<\/em>, but this time we move the axis of rotation to the end of the rod. We wish to \ufb01nd the moment of inertia about this new axis (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Thinrod2\">(Figure)<\/a>). The quantity <em>dm<\/em> is again defined to be a small element of mass making up the rod. Just as before, we obtain<\/p>\r\n\r\n<div id=\"fs-id1167133795151\" class=\"unnumbered\">$$I=\\int {r}^{2}dm=\\int {x}^{2}dm=\\int {x}^{2}\\lambda dx.$$<\/div>\r\n<p id=\"fs-id1167133528548\">However, this time we have different limits of integration. The rod extends from $$ x=0 $$ to $$ x=L$$, since the axis is at the end of the rod at $$ x=0$$. Therefore we find<\/p>\r\n\r\n<div id=\"fs-id1167133377543\" class=\"unnumbered\">$$\\begin{array}{cc}\\hfill I&amp; =\\underset{0}{\\overset{L}{\\int }}{x}^{2}\\lambda dx=\\lambda \\frac{{x}^{3}}{3}{|}_{0}^{L}=\\lambda (\\frac{1}{3})[{(L)}^{3}-{(0)}^{3}]\\hfill \\\\ &amp; =\\lambda (\\frac{1}{3}){L}^{3}=\\frac{M}{L}(\\frac{1}{3}){L}^{3}=\\frac{1}{3}M{L}^{2}.\\hfill \\end{array}$$<\/div>\r\n<div id=\"CNX_UPhysics_10_05_Thinrod2\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"438\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194454\/CNX_UPhysics_10_05_thinrod2.jpg\" alt=\"Figure shows a thin rod that rotates about an axis through the end. Part of the rod of the length dx has a mass dm.\" width=\"438\" height=\"151\" \/> <strong>Figure 10.26<\/strong> Calculation of the moment of inertia I for a uniform thin rod about an axis through the end of the rod.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133677178\">Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center (consistent with the barbell example) by a factor of four.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133784487\" class=\"bc-section section\">\r\n<h3>The Parallel-Axis Theorem<\/h3>\r\n<p id=\"fs-id1167133497854\">The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Such an axis is called a <strong>parallel axis<\/strong>. There is a theorem for this, called the <strong>parallel-axis theorem<\/strong>, which we state here but do not derive in this text.<\/p>\r\n\r\n<div id=\"fs-id1167133829900\">\r\n<h4>Parallel-Axis Theorem<\/h4>\r\n<p id=\"fs-id1167133544524\">Let <em>m<\/em> be the mass of an object and let <em>d<\/em> be the distance from an axis through the object\u2019s center of mass to a new axis. Then we have<\/p>\r\n\r\n<div id=\"fs-id1167133536034\">$${I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133803000\">Let\u2019s apply this to the rod examples solved above:<\/p>\r\n\r\n<div id=\"fs-id1167133539624\" class=\"unnumbered\">$${I}_{\\text{end}}={I}_{\\text{center of mass}}+m{d}^{2}=\\frac{1}{12}m{L}^{2}+m{(\\frac{L}{2})}^{2}=(\\frac{1}{12}+\\frac{1}{4})m{L}^{2}=\\frac{1}{3}m{L}^{2}.$$<\/div>\r\n<p id=\"fs-id1167133858580\">This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems.<\/p>\r\n\r\n<div id=\"fs-id1167133472470\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3><div id=\"fs-id1167133375616\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133550293\">\r\n\r\n<p id=\"fs-id1167133792700\">What is the moment of inertia of a cylinder of radius <em>R<\/em> and mass <em>m<\/em> about an axis through a point on the surface, as shown below?<\/p>\r\n<span id=\"fs-id1167133523823\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194457\/CNX_UPhysics_10_05_Cyl_img.jpg\" alt=\"Figure shows a cylinder of radius R that rotates about an axis through a point on the surface.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133824053\">\r\n<p id=\"fs-id1167132303828\">[reveal-answer q=\"530534\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"530534\"]$${I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2}=m{R}^{2}+m{R}^{2}=2m{R}^{2}$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133374515\" class=\"bc-section section\">\r\n<h4>A uniform thin disk about an axis through the center<\/h4>\r\n<p id=\"fs-id1167133698444\">Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study\u2014a uniform thin disk about an axis through its center (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Disk\">(Figure)<\/a>).<\/p>\r\n\r\n<div id=\"CNX_UPhysics_10_05_Disk\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"674\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194500\/CNX_UPhysics_10_05_disk.jpg\" alt=\"Figure shows a uniform thin disk of radius r that rotates about a Z axis that passes through its center.\" width=\"674\" height=\"304\" \/> <strong>Figure 10.27<\/strong> Calculating the moment of inertia for a thin disk about an axis through its center.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\nSince the disk is thin, we can take the mass as distributed entirely in the <em>xy<\/em>-plane. We again start with the relationship for the <strong>surface mass density<\/strong>, which is the mass per unit surface area. Since it is uniform, the surface mass density $$ \\sigma  $$ is constant:\r\n<div id=\"fs-id1167133560777\" class=\"unnumbered\">$$\\sigma =\\frac{m}{A}\\quad \\text{or}\\quad \\sigma A=m,\\,\\text{so}\\,dm=\\sigma (dA).$$<\/div>\r\n<p id=\"fs-id1167133377088\">Now we use a simplification for the area. The area can be thought of as made up of a series of thin rings, where each ring is a mass increment <em>dm<\/em> of radius <em>r<\/em> equidistanct from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring <em>dA<\/em> is therefore given by the length of each ring ($$2\\pi r$$) times the infinitesimmal width of each ring <em>dr<\/em>:<\/p>\r\n\r\n<div id=\"fs-id1167133335221\" class=\"unnumbered\">$$A=\\pi {r}^{2},dA=d(\\pi {r}^{2})=\\pi d{r}^{2}=2\\pi rdr.$$<\/div>\r\n<p id=\"fs-id1167133826443\">The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to <em>R<\/em>. This radius range then becomes our limits of integration for <em>dr<\/em>, that is, we integrate from $$ r=0 $$ to $$ r=R$$. Putting this all together, we have<\/p>\r\n\r\n<div id=\"fs-id1167133773688\" class=\"unnumbered\">$$\\begin{array}{cc}\\hfill I&amp; =\\underset{0}{\\overset{R}{\\int }}{r}^{2}\\sigma (2\\pi r)dr=2\\pi \\sigma \\underset{0}{\\overset{R}{\\int }}{r}^{3}dr=2\\pi \\sigma \\frac{{r}^{4}}{4}{|}_{0}^{R}=2\\pi \\sigma (\\frac{{R}^{4}}{4}-0)\\hfill \\\\ &amp; =2\\pi \\frac{m}{A}(\\frac{{R}^{4}}{4})=2\\pi \\frac{m}{\\pi {R}^{2}}(\\frac{{R}^{4}}{4})=\\frac{1}{2}m{R}^{2}.\\hfill \\end{array}$$<\/div>\r\n<p id=\"fs-id1167133525890\">Note that this agrees with the value given in <a class=\"autogenerated-content\" href=\"\/contents\/1b0f645a-e293-4925-9c9b-1dfc7a29ab03#CNX_UPhysics_10_04_RotInertia\">(Figure)<\/a>.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133353900\" class=\"bc-section section\">\r\n<h4>Calculating the moment of inertia for compound objects<\/h4>\r\n<p id=\"fs-id1167133847080\">Now consider a compound object such as that in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_RodDisk\">(Figure)<\/a>, which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound object\u2019s moment of inertia can be found from the sum of each part of the object:<\/p>\r\n\r\n<div id=\"fs-id1167133828814\" class=\"equation-callout\">\r\n<div id=\"fs-id1167133811543\">$${I}_{\\text{total}}=\\sum _{i}{I}_{i}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133553885\">It is important to note that the moments of inertia of the objects in <a class=\"autogenerated-content\" href=\"#fs-id1167133811543\">(Figure)<\/a> are <em>about a common axis<\/em>. In the case of this object, that would be a rod of length <em>L<\/em> rotating about its end, and a thin disk of radius <em>R<\/em> rotating about an axis shifted off of the center by a distance $$ L+R$$, where <em>R<\/em> is the radius of the disk. Let\u2019s define the mass of the rod to be $$ {m}_{\\text{r}} $$ and the mass of the disk to be $$ {m}_{\\text{d}}.$$<\/p>\r\n\r\n<div id=\"CNX_UPhysics_10_05_RodDisk\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"290\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194505\/CNX_UPhysics_10_05_RodDisk.jpg\" alt=\"Figure shows a disk with radius R connected to a rod with length L.\" width=\"290\" height=\"254\" \/> <strong>Figure 10.28<\/strong> Compound object consisting of a disk at the end of a rod. The axis of rotation is located at A.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<p id=\"fs-id1167133797122\">The moment of inertia of the rod is simply $$ \\frac{1}{3}{m}_{\\text{r}}{L}^{2}$$, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is $$ \\frac{1}{2}{m}_{\\text{d}}{R}^{2} $$ and we apply the parallel-axis theorem $$ {I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2} $$ to find<\/p>\r\n\r\n<div id=\"fs-id1167133774220\" class=\"unnumbered\">$${I}_{\\text{parallel-axis}}=\\frac{1}{2}{m}_{\\text{d}}{R}^{2}+{m}_{\\text{d}}{(L+R)}^{2}.$$<\/div>\r\n<p id=\"fs-id1167133447451\">Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be<\/p>\r\n\r\n<div id=\"fs-id1167133803182\" class=\"unnumbered\">$${I}_{\\text{total}}=\\frac{1}{3}{m}_{\\text{r}}{L}^{2}+\\frac{1}{2}{m}_{\\text{d}}{R}^{2}+{m}_{\\text{d}}{(L+R)}^{2}.$$<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133333570\" class=\"bc-section section\">\r\n<h4>Applying moment of inertia calculations to solve problems<\/h4>\r\n<p id=\"fs-id1167133697400\">Now let\u2019s examine some practical applications of moment of inertia calculations.<\/p>\r\n\r\n<div id=\"fs-id1167133824110\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Person on a Merry-Go-Round<\/h4>\r\nA 25-kg child stands at a distance $$ r=1.0\\,\\text{m} $$ from the axis of a rotating merry-go-round (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Merrygor\">(Figure)<\/a>). The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system.\r\n<div id=\"CNX_UPhysics_10_05_Merrygor\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"431\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194508\/CNX_UPhysics_10_05_merrygor.jpg\" alt=\"Figure is a drawing of a child on a merry-go-round. Merry\u2013go-round has a 2 meter radius. Child is standing one meter from the center.\" width=\"431\" height=\"254\" \/> <strong>Figure 10.29<\/strong> Calculating the moment of inertia for a child on a merry-go-round.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<h4>Strategy<\/h4>\r\nThis problem involves the calculation of a moment of inertia. We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. The notation we use is $$ {m}_{\\text{c}}=25\\,\\text{kg},{r}_{\\text{c}}=1.0\\,\\text{m},{m}_{\\text{m}}=500\\,\\text{kg},{r}_{\\text{m}}=2.0\\,\\text{m}$$.\r\n<p id=\"fs-id1167133677861\">Our goal is to find $$ {I}_{\\text{total}}=\\sum _{i}{I}_{i}$$.<\/p>\r\n\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167133612679\">For the child, $$ {I}_{\\text{c}}={m}_{\\text{c}}{r}^{2}$$, and for the merry-go-round, $$ {I}_{\\text{m}}=\\frac{1}{2}{m}_{\\text{m}}{r}^{2}$$. Therefore<\/p>\r\n\r\n<div id=\"fs-id1167133549750\" class=\"unnumbered\">$${I}_{\\text{total}}=25{(1)}^{2}+\\frac{1}{2}(500){(2)}^{2}=25+1000=1025\\,\\text{kg}\u00b7{\\text{m}}^{2}.$$<\/div>\r\n<h4>Significance<\/h4>\r\nThe value should be close to the moment of inertia of the merry-go-round by itself because it has much more mass distributed away from the axis than the child does.\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133345942\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Rod and Solid Sphere<\/h4>\r\nFind the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg.\r\n\r\n<span id=\"fs-id1167133318645\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194511\/CNX_UPhysics_10_05_RodSphere_img.jpg\" alt=\"Figure A shows a disk with radius R connected to a rod with length L. The point A is at the end of the rod opposite to the disk. Figure B shows a disk with radius R connected to a rod with length L. The point B is at the end of the rod connected to the disk.\" \/><\/span>\r\n<h4>Strategy<\/h4>\r\nSince we have a compound object in both cases, we can use the parallel-axis theorem to find the moment of inertia about each axis. In (a), the center of mass of the sphere is located at a distance $$ L+R $$ from the axis of rotation. In (b), the center of mass of the sphere is located a distance <em>R<\/em> from the axis of rotation. In both cases, the moment of inertia of the rod is about an axis at one end. Refer to <a class=\"autogenerated-content\" href=\"\/contents\/1b0f645a-e293-4925-9c9b-1dfc7a29ab03#fs-id1167133375236\">(Figure)<\/a> for the moments of inertia for the individual objects.\r\n<ol id=\"fs-id1167133324770\" type=\"a\">\r\n \t<li>$${I}_{\\text{total}}=\\sum _{i}{I}_{i}={I}_{\\text{Rod}}+{I}_{\\text{Sphere}}$$;$${I}_{\\text{Sphere}}={I}_{\\text{center of mass}}+{m}_{\\text{Sphere}}{(L+R)}^{2}=\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{(L+R)}^{2}$$;$${I}_{\\text{total}}={I}_{\\text{Rod}}+{I}_{\\text{Sphere}}=\\frac{1}{3}{m}_{\\text{Rod}}{L}^{2}+\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{(L+R)}^{2};$$$${I}_{\\text{total}}=\\frac{1}{3}(2.0\\,\\text{kg}){(0.5\\,\\text{m})}^{2}+\\frac{2}{5}(1.0\\,\\text{kg})(0.2\\,{\\text{m})}^{2}+(1.0\\,\\text{kg}){(0.5\\,\\text{m}+0.2\\,\\text{m})}^{2};$$$${I}_{\\text{total}}=(0.167+0.016+0.490)\\,\\text{kg}\u00b7{\\text{m}}^{2}=0.673\\,\\text{kg}\u00b7{\\text{m}}^{2}.$$<\/li>\r\n \t<li>$${I}_{\\text{Sphere}}=\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{R}^{2}$$;$${I}_{\\text{total}}={I}_{\\text{Rod}}+{I}_{\\text{Sphere}}=\\frac{1}{3}{m}_{\\text{Rod}}{L}^{2}+\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{R}^{2}$$;$${I}_{\\text{total}}=\\frac{1}{3}(2.0\\,\\text{kg}){(0.5\\,\\text{m})}^{2}+\\frac{2}{5}(1.0\\,\\text{kg})(0.2\\,{\\text{m})}^{2}+(1.0\\,\\text{kg}){(0.2\\,\\text{m})}^{2}$$;$${I}_{\\text{total}}=(0.167+0.016+0.04)\\,\\text{kg}\u00b7{\\text{m}}^{2}=0.223\\,\\text{kg}\u00b7{\\text{m}}^{2}.$$<\/li>\r\n<\/ol>\r\n<h4>Significance<\/h4>\r\nUsing the parallel-axis theorem eases the computation of the moment of inertia of compound objects. We see that the moment of inertia is greater in (a) than (b). This is because the axis of rotation is closer to the center of mass of the system in (b). The simple analogy is that of a rod. The moment of inertia about one end is $$ \\frac{1}{3}m{L}^{2}$$, but the moment of inertia through the center of mass along its length is $$ \\frac{1}{12}m{L}^{2}$$.\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133775662\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Angular Velocity of a Pendulum<\/h4>\r\nA pendulum in the shape of a rod (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Pend\">(Figure)<\/a>) is released from rest at an angle of $$ 30\\text{\u00b0}$$. It has a length 30 cm and mass 300 g. What is its angular velocity at its lowest point?\r\n<div id=\"CNX_UPhysics_10_05_Pend\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"317\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194514\/CNX_UPhysics_10_05_Pend.jpg\" alt=\"Figure shows a pendulum in the form of a rod with a mass of 300 grams and length of 30 centimeters. Pendulum is released from rest at an angle of 30 degrees.\" width=\"317\" height=\"308\" \/> <strong>Figure 10.30<\/strong> A pendulum in the form of a rod is released from rest at an angle of $$ 30\\text{\u00b0}.$$[\/caption]\r\n<div class=\"wp-caption-text\"><\/div>\r\n<\/div>\r\n<h4>Strategy<\/h4>\r\nUse conservation of energy to solve the problem. At the point of release, the pendulum has gravitational potential energy, which is determined from the height of the center of mass above its lowest point in the swing. At the bottom of the swing, all of the gravitational potential energy is converted into rotational kinetic energy.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1167132279386\">The change in potential energy is equal to the change in rotational kinetic energy, $$ \\Delta U+\\Delta K=0$$.<\/p>\r\n<p id=\"fs-id1167133851770\">At the top of the swing: $$ U=mg{h}_{\\text{cm}}=mg\\frac{L}{2}(\\text{cos}\\,\\theta )$$. At the bottom of the swing, $$ U=mg\\frac{L}{2}.$$<\/p>\r\n<p id=\"fs-id1167133686379\">At the top of the swing, the rotational kinetic energy is $$ K=0$$. At the bottom of the swing, $$ K=\\frac{1}{2}I{\\omega }^{2}$$. Therefore:<\/p>\r\n\r\n<div id=\"fs-id1167132206607\" class=\"unnumbered\">$$\\text{\u0394}U+\\text{\u0394}K=0\u21d2(mg\\frac{L}{2}(1-\\text{cos}\\,\\theta )-0)+(0-\\frac{1}{2}I{\\omega }^{2})=0$$<\/div>\r\n<p id=\"fs-id1167132202151\">or<\/p>\r\n\r\n<div id=\"fs-id1167132202154\" class=\"unnumbered\">$$\\frac{1}{2}I{\\omega }^{2}=mg\\frac{L}{2}(1-\\text{cos}\\,\\theta ).$$<\/div>\r\n<p id=\"fs-id1167132202203\">Solving for $$ \\omega $$, we have<\/p>\r\n\r\n<div id=\"fs-id1167133795236\" class=\"unnumbered\">$$\\omega =\\sqrt{mg\\frac{L}{I}(1-\\text{cos}\\,\\theta )}=\\sqrt{mg\\frac{L}{1\\text{\/}3m{L}^{2}}(1-\\text{cos}\\,\\theta )}=\\sqrt{g\\frac{3}{L}(1-\\text{cos}\\,\\theta )}.$$<\/div>\r\n<p id=\"fs-id1167132303820\">Inserting numerical values, we have<\/p>\r\n\r\n<div id=\"fs-id1167132303823\" class=\"unnumbered\">$$\\omega =\\sqrt{9.8\\,\\text{m}\\text{\/}{\\text{s}}^{2}\\frac{3}{0.3\\,\\text{m}}(1-\\text{cos}\\,30)}=3.6\\,\\text{rad}\\text{\/}\\text{s}.$$<\/div>\r\n<h4>Significance<\/h4>\r\nNote that the angular velocity of the pendulum does not depend on its mass.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133349420\" class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"fs-id1167133566402\">\r\n \t<li>Moments of inertia can be found by summing or integrating over every \u2018piece of mass\u2019 that makes up an object, multiplied by the square of the distance of each \u2018piece of mass\u2019 to the axis. In integral form the moment of inertia is $$ I=\\int {r}^{2}dm$$.<\/li>\r\n \t<li>Moment of inertia is larger when an object\u2019s mass is farther from the axis of rotation.<\/li>\r\n \t<li>It is possible to find the moment of inertia of an object about a new axis of rotation once it is known for a parallel axis. This is called the parallel axis theorem given by $$ {I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2}$$, where <em>d<\/em> is the distance from the initial axis to the parallel axis.<\/li>\r\n \t<li>Moment of inertia for a compound object is simply the sum of the moments of inertia for each individual object that makes up the compound object.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1167133549678\" class=\"review-conceptual-questions\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"fs-id1167133357710\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133357712\">\r\n<p id=\"fs-id1167133863061\">If a child walks toward the center of a merry-go-round, does the moment of inertia increase or decrease?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133359289\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133359292\">\r\n<p id=\"fs-id1167133359294\">A discus thrower rotates with a discus in his hand before letting it go. (a) How does his moment of inertia change after releasing the discus? (b) What would be a good approximation to use in calculating the moment of inertia of the discus thrower and discus?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133845959\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167133845959\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167133845959\"]\r\n<p id=\"fs-id1167133328460\">a. It decreases. b. The arms could be approximated with rods and the discus with a disk. The torso is near the axis of rotation so it doesn\u2019t contribute much to the moment of inertia.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133845992\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133845994\">\r\n<p id=\"fs-id1167133353348\">Does increasing the number of blades on a propeller increase or decrease its moment of inertia, and why?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133357558\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133357560\">\r\n<p id=\"fs-id1167133357562\">The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is $$ m{L}^{2}\\text{\/}3$$. Why is this moment of inertia greater than it would be if you spun a point mass <em>m<\/em> at the location of the center of mass of the rod (at <em>L<\/em>\/2) (that would be $$ m{L}^{2}\\text{\/}4$$)?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133354662\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1167133354662\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1167133354662\"]\r\n<p id=\"fs-id1167133354664\">Because the moment of inertia varies as the square of the distance to the axis of rotation. The mass of the rod located at distances greater than <em>L<\/em>\/2 would provide the larger contribution to make its moment of inertia greater than the point mass at <em>L<\/em>\/2.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133773940\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133773942\">\r\n<p id=\"fs-id1167133773944\">Why is the moment of inertia of a hoop that has a mass <em>M<\/em> and a radius <em>R<\/em> greater than the moment of inertia of a disk that has the same mass and radius?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133871657\" class=\"review-problems textbox exercises\">\r\n<h3>Problems<\/h3>\r\n<div id=\"fs-id1167133407930\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133407932\">\r\n<p id=\"fs-id1167133407934\">While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is $$ 3.75{\\,\\text{kg-m}}^{2} $$ and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter\u2019s shoe if it is 1.05 m from the hip joint?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133357822\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133357824\">\r\n<p id=\"fs-id1167133859006\">Using the parallel axis theorem, what is the moment of inertia of the rod of mass <em>m<\/em> about the axis shown below?<\/p>\r\n<span id=\"fs-id1167133325838\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194516\/CNX_UPhysics_10_05_Prob8_img.jpg\" alt=\"Figure shows a rod that rotates around the axis that passes through it at 1\/6 of length from one end and 5\/6 of length from the opposite end.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133497626\">\r\n<p id=\"fs-id1167133497628\">[reveal-answer q=\"96051\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"96051\"]$$I=\\frac{7}{36}m{L}^{2}$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133775336\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133775338\">\r\n<p id=\"fs-id1167133775340\">Find the moment of inertia of the rod in the previous problem by direct integration.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133346254\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133792663\">\r\n<p id=\"fs-id1167133792665\">A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one end (see the following figure). If the rod is released from rest at an angle of $$ 60\\text{\u00b0} $$ with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?<\/p>\r\n<span id=\"fs-id1167133359187\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194518\/CNX_UPhysics_10_05_RodAng_img.jpg\" alt=\"Figure shows a rod that is released from rest at an angle of 60 degrees with respect to the horizontal.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167133344789\">\r\n<p id=\"fs-id1167133344791\">[reveal-answer q=\"71295\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"71295\"]$$v=7.14\\,\\text{m}\\text{\/}\\text{s}.$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133365711\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133365713\">\r\n<p id=\"fs-id1167133319885\">A pendulum consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm (see the following figure). If the pendulum is released from rest at an angle of $$ 30\\text{\u00b0}$$, what is the angular velocity at the lowest point?<\/p>\r\n<span id=\"fs-id1167133325585\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194521\/CNX_UPhysics_10_05_PendSph_img.jpg\" alt=\"Figure shows a pendulum that consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm. The pendulum is released from rest at an angle of 30 degrees.\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133447838\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133357875\">\r\n<p id=\"fs-id1167133357877\">A solid sphere of radius 10 cm is allowed to rotate freely about an axis. The sphere is given a sharp blow so that its center of mass starts from the position shown in the following figure with speed 15 cm\/s. What is the maximum angle that the diameter makes with the vertical?<\/p>\r\n<span id=\"fs-id1167133345349\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194523\/CNX_UPhysics_10_05_SphPiv_img.jpg\" alt=\"Left figure shows a solid sphere of radius 10 cm that first rotates freely about an axis and then received a sharp blow in its center of mass. Right figure is the image of the same sphere after the blow. An angle that the diameter makes with the vertical is marked as theta.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167132305511\">\r\n<p id=\"fs-id1167132305513\">[reveal-answer q=\"602090\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"602090\"]$$\\theta =10.2\\text{\u00b0}$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167133846985\" class=\"problem textbox\">\r\n<div id=\"fs-id1167133846987\">\r\n<p id=\"fs-id1167133851633\">Calculate the moment of inertia by direct integration of a thin rod of mass <em>M<\/em> and length <em>L<\/em> about an axis through the rod at <em>L<\/em>\/3, as shown below. Check your answer with the parallel-axis theorem.<\/p>\r\n<span id=\"fs-id1167132202884\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194525\/CNX_UPhysics_10_05_RodMomen_img.jpg\" alt=\"Figure shows a rod that rotates around the axis that passes through it at 1\/3 of length from one end and 2\/3 of length from the opposite end.\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1167133686238\">\r\n \t<dt><strong>linear mass density<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167133686243\">the mass per unit length $$ \\lambda  $$ of a one dimensional object<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167133822794\">\r\n \t<dt><strong>parallel axis<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167133802923\">axis of rotation that is parallel to an axis about which the moment of inertia of an object is known<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167132202132\">\r\n \t<dt><strong>parallel-axis theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167132202137\">if the moment of inertia is known for a given axis, it can be found for any axis parallel to it<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1167133399142\">\r\n \t<dt><strong>surface mass density<\/strong><\/dt>\r\n \t<dd id=\"fs-id1167133858665\">mass per unit area $$ \\sigma  $$ of a two dimensional object<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Calculate the moment of inertia for uniformly shaped, rigid bodies<\/li>\n<li>Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known<\/li>\n<li>Calculate the moment of inertia for compound objects<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167133856164\">In the preceding section, we defined the moment of inertia but did not show how to calculate it. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. This section is very useful for seeing how to apply a general equation to complex objects (a skill that is critical for more advanced physics and engineering courses).<\/p>\n<div id=\"fs-id1167133538983\" class=\"bc-section section\">\n<h3>Moment of Inertia<\/h3>\n<p id=\"fs-id1167133826555\">We defined the moment of inertia <em>I<\/em> of an object to be $$ I=\\sum _{i}{m}_{i}{r}_{i}^{2} $$ for all the point masses that make up the object. Because <em>r<\/em> is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let\u2019s take a simple example of two masses at the end of a massless (negligibly small mass) rod (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Barbell\">(Figure)<\/a>) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.<\/p>\n<p>In the case with the axis in the center of the barbell, each of the two masses <em>m<\/em> is a distance <em>R<\/em> away from the axis, giving a moment of inertia of<\/p>\n<div id=\"fs-id1167133521048\" class=\"unnumbered\">$${I}_{1}=m{R}^{2}+m{R}^{2}=2m{R}^{2}.$$<\/div>\n<p id=\"fs-id1167133426576\">In the case with the axis at the end of the barbell\u2014passing through one of the masses\u2014the moment of inertia is<\/p>\n<div id=\"fs-id1167133568600\" class=\"unnumbered\">$${I}_{2}=m{(0)}^{2}+m{(2R)}^{2}=4m{R}^{2}.$$<\/div>\n<p id=\"fs-id1167133456090\">From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center.<\/p>\n<div id=\"CNX_UPhysics_10_05_Barbell\" class=\"wp-caption aligncenter\">\n<div style=\"width: 901px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194447\/CNX_UPhysics_10_05_Barbell.jpg\" alt=\"Figure A shows a barbell of the length 2 R with the masses m at the ends. It is rotating through its center. Figure B shows a barbell of the length 2 R with the masses m at the ends. It is rotating through one end.\" width=\"891\" height=\"301\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.23<\/strong> (a) A barbell with an axis of rotation through its center; (b) a barbell with an axis of rotation through one end.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id1167133517824\">In this example, we had two point masses and the sum was simple to calculate. However, to deal with objects that are not point-like, we need to think carefully about each of the terms in the equation. The equation asks us to sum over each \u2018piece of mass\u2019 a certain distance from the axis of rotation. But what exactly does each \u2018piece of mass\u2019 mean? Recall that in our derivation of this equation, each piece of mass had the same magnitude of velocity, which means the whole piece had to have a single distance <em>r<\/em> to the axis of rotation. However, this is not possible unless we take an infinitesimally small piece of mass <em>dm<\/em>, as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Pointmass\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_UPhysics_10_05_Pointmass\" class=\"wp-caption aligncenter\">\n<div style=\"width: 382px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194450\/CNX_UPhysics_10_05_pointmass.jpg\" alt=\"Figure shows a point dm located on the X axis at distance r from the center.\" width=\"372\" height=\"381\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.24<\/strong> Using an infinitesimally small piece of mass to calculate the contribution to the total moment of inertia.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id1167133534200\">The need to use an infinitesimally small piece of mass <em>dm<\/em> suggests that we can write the moment of inertia by evaluating an integral over infinitesimal masses rather than doing a discrete sum over finite masses:<\/p>\n<div id=\"fs-id1167133335739\" class=\"equation-callout\">\n<div id=\"fs-id1167133385098\">$$I=\\sum _{i}{m}_{i}{r}_{i}{}^{2}\\enspace\\text{becomes}I=\\int {r}^{2}dm.$$<\/div>\n<\/div>\n<p>This, in fact, is the form we need to generalize the equation for complex shapes. It is best to work out specific examples in detail to get a feel for how to calculate the moment of inertia for specific shapes. This is the focus of most of the rest of this section.<\/p>\n<div id=\"fs-id1167133698940\" class=\"bc-section section\">\n<h4>A uniform thin rod with an axis through the center<\/h4>\n<p>Consider a uniform (density and shape) thin rod of mass <em>M<\/em> and length <em>L<\/em> as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Thinrod\">(Figure)<\/a>. We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the <em>z<\/em>-axis is the axis of rotation and the <em>x<\/em>-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the <em>x<\/em>-axis.<\/p>\n<div id=\"CNX_UPhysics_10_05_Thinrod\" class=\"wp-caption aligncenter\">\n<div style=\"width: 443px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194452\/CNX_UPhysics_10_05_thinrod.jpg\" alt=\"Figure shows a thin rod that rotates about an axis through the center. Part of the rod of the length dx has a mass dm.\" width=\"433\" height=\"158\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.25<\/strong> Calculation of the moment of inertia I for a uniform thin rod about an axis through the center of the rod.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id1167133519807\">We define <em>dm<\/em> to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the <strong>linear mass density<\/strong> $$ \\lambda  $$ of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write<\/p>\n<div id=\"fs-id1167133661384\" class=\"unnumbered\">$$\\lambda =\\frac{m}{l}\\enspace\\text{or}m=\\lambda l.$$<\/div>\n<p id=\"fs-id1167133410629\">If we take the differential of each side of this equation, we find<\/p>\n<div id=\"fs-id1167133557177\" class=\"unnumbered\">$$dm=d(\\lambda l)=\\lambda (dl)$$<\/div>\n<p id=\"fs-id1167133321526\">since $$ \\lambda  $$ is constant. We chose to orient the rod along the <em>x<\/em>-axis for convenience\u2014this is where that choice becomes very helpful. Note that a piece of the rod <em>dl<\/em> lies completely along the <em>x<\/em>-axis and has a length <em>dx<\/em>; in fact, $$ dl=dx $$ in this situation. We can therefore write $$ dm=\\lambda (dx)$$, giving us an integration variable that we know how to deal with. The distance of each piece of mass <em>dm<\/em> from the axis is given by the variable <em>x<\/em>, as shown in the figure. Putting this all together, we obtain<\/p>\n<div id=\"fs-id1167133467275\" class=\"unnumbered\">$$I=\\int {r}^{2}dm=\\int {x}^{2}dm=\\int {x}^{2}\\lambda dx.$$<\/div>\n<p id=\"fs-id1167133558728\">The last step is to be careful about our limits of integration. The rod extends from $$ x=\\text{\u2212}L\\text{\/}2 $$ to $$ x=L\\text{\/}2$$, since the axis is in the middle of the rod at $$ x=0$$. This gives us<\/p>\n<div id=\"fs-id1167133440496\" class=\"unnumbered\">$$\\begin{array}{cc}\\hfill I&amp; =\\underset{\\text{\u2212}L\\text{\/}2}{\\overset{L\\text{\/}2}{\\int }}{x}^{2}\\lambda dx=\\lambda \\frac{{x}^{3}}{3}{|}_{\\text{\u2212}L\\text{\/}2}^{L\\text{\/}2}=\\lambda (\\frac{1}{3})[{(\\frac{L}{2})}^{3}-{(\\frac{\\text{\u2212}L}{2})}^{3}]\\hfill \\\\ &amp; =\\lambda (\\frac{1}{3})\\frac{{L}^{3}}{8}(2)=\\frac{M}{L}(\\frac{1}{3})\\frac{{L}^{3}}{8}(2)=\\frac{1}{12}M{L}^{2}.\\hfill \\end{array}$$<\/div>\n<p id=\"fs-id1167133326786\">Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation.<\/p>\n<\/div>\n<div id=\"fs-id1167133319345\" class=\"bc-section section\">\n<h4>A uniform thin rod with axis at the end<\/h4>\n<p id=\"fs-id1167133465541\">Now consider the same uniform thin rod of mass <em>M<\/em> and length <em>L<\/em>, but this time we move the axis of rotation to the end of the rod. We wish to \ufb01nd the moment of inertia about this new axis (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Thinrod2\">(Figure)<\/a>). The quantity <em>dm<\/em> is again defined to be a small element of mass making up the rod. Just as before, we obtain<\/p>\n<div id=\"fs-id1167133795151\" class=\"unnumbered\">$$I=\\int {r}^{2}dm=\\int {x}^{2}dm=\\int {x}^{2}\\lambda dx.$$<\/div>\n<p id=\"fs-id1167133528548\">However, this time we have different limits of integration. The rod extends from $$ x=0 $$ to $$ x=L$$, since the axis is at the end of the rod at $$ x=0$$. Therefore we find<\/p>\n<div id=\"fs-id1167133377543\" class=\"unnumbered\">$$\\begin{array}{cc}\\hfill I&amp; =\\underset{0}{\\overset{L}{\\int }}{x}^{2}\\lambda dx=\\lambda \\frac{{x}^{3}}{3}{|}_{0}^{L}=\\lambda (\\frac{1}{3})[{(L)}^{3}-{(0)}^{3}]\\hfill \\\\ &amp; =\\lambda (\\frac{1}{3}){L}^{3}=\\frac{M}{L}(\\frac{1}{3}){L}^{3}=\\frac{1}{3}M{L}^{2}.\\hfill \\end{array}$$<\/div>\n<div id=\"CNX_UPhysics_10_05_Thinrod2\" class=\"wp-caption aligncenter\">\n<div style=\"width: 448px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194454\/CNX_UPhysics_10_05_thinrod2.jpg\" alt=\"Figure shows a thin rod that rotates about an axis through the end. Part of the rod of the length dx has a mass dm.\" width=\"438\" height=\"151\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.26<\/strong> Calculation of the moment of inertia I for a uniform thin rod about an axis through the end of the rod.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id1167133677178\">Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center (consistent with the barbell example) by a factor of four.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133784487\" class=\"bc-section section\">\n<h3>The Parallel-Axis Theorem<\/h3>\n<p id=\"fs-id1167133497854\">The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Such an axis is called a <strong>parallel axis<\/strong>. There is a theorem for this, called the <strong>parallel-axis theorem<\/strong>, which we state here but do not derive in this text.<\/p>\n<div id=\"fs-id1167133829900\">\n<h4>Parallel-Axis Theorem<\/h4>\n<p id=\"fs-id1167133544524\">Let <em>m<\/em> be the mass of an object and let <em>d<\/em> be the distance from an axis through the object\u2019s center of mass to a new axis. Then we have<\/p>\n<div id=\"fs-id1167133536034\">$${I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167133803000\">Let\u2019s apply this to the rod examples solved above:<\/p>\n<div id=\"fs-id1167133539624\" class=\"unnumbered\">$${I}_{\\text{end}}={I}_{\\text{center of mass}}+m{d}^{2}=\\frac{1}{12}m{L}^{2}+m{(\\frac{L}{2})}^{2}=(\\frac{1}{12}+\\frac{1}{4})m{L}^{2}=\\frac{1}{3}m{L}^{2}.$$<\/div>\n<p id=\"fs-id1167133858580\">This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems.<\/p>\n<div id=\"fs-id1167133472470\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1167133375616\" class=\"problem textbox\">\n<div id=\"fs-id1167133550293\">\n<p id=\"fs-id1167133792700\">What is the moment of inertia of a cylinder of radius <em>R<\/em> and mass <em>m<\/em> about an axis through a point on the surface, as shown below?<\/p>\n<p><span id=\"fs-id1167133523823\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194457\/CNX_UPhysics_10_05_Cyl_img.jpg\" alt=\"Figure shows a cylinder of radius R that rotates about an axis through a point on the surface.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1167133824053\">\n<p id=\"fs-id1167132303828\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530534\">Show Answer<\/span><\/p>\n<div id=\"q530534\" class=\"hidden-answer\" style=\"display: none\">$${I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2}=m{R}^{2}+m{R}^{2}=2m{R}^{2}$$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133374515\" class=\"bc-section section\">\n<h4>A uniform thin disk about an axis through the center<\/h4>\n<p id=\"fs-id1167133698444\">Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study\u2014a uniform thin disk about an axis through its center (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Disk\">(Figure)<\/a>).<\/p>\n<div id=\"CNX_UPhysics_10_05_Disk\" class=\"wp-caption aligncenter\">\n<div style=\"width: 684px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194500\/CNX_UPhysics_10_05_disk.jpg\" alt=\"Figure shows a uniform thin disk of radius r that rotates about a Z axis that passes through its center.\" width=\"674\" height=\"304\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.27<\/strong> Calculating the moment of inertia for a thin disk about an axis through its center.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p>Since the disk is thin, we can take the mass as distributed entirely in the <em>xy<\/em>-plane. We again start with the relationship for the <strong>surface mass density<\/strong>, which is the mass per unit surface area. Since it is uniform, the surface mass density $$ \\sigma  $$ is constant:<\/p>\n<div id=\"fs-id1167133560777\" class=\"unnumbered\">$$\\sigma =\\frac{m}{A}\\quad \\text{or}\\quad \\sigma A=m,\\,\\text{so}\\,dm=\\sigma (dA).$$<\/div>\n<p id=\"fs-id1167133377088\">Now we use a simplification for the area. The area can be thought of as made up of a series of thin rings, where each ring is a mass increment <em>dm<\/em> of radius <em>r<\/em> equidistanct from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring <em>dA<\/em> is therefore given by the length of each ring ($$2\\pi r$$) times the infinitesimmal width of each ring <em>dr<\/em>:<\/p>\n<div id=\"fs-id1167133335221\" class=\"unnumbered\">$$A=\\pi {r}^{2},dA=d(\\pi {r}^{2})=\\pi d{r}^{2}=2\\pi rdr.$$<\/div>\n<p id=\"fs-id1167133826443\">The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to <em>R<\/em>. This radius range then becomes our limits of integration for <em>dr<\/em>, that is, we integrate from $$ r=0 $$ to $$ r=R$$. Putting this all together, we have<\/p>\n<div id=\"fs-id1167133773688\" class=\"unnumbered\">$$\\begin{array}{cc}\\hfill I&amp; =\\underset{0}{\\overset{R}{\\int }}{r}^{2}\\sigma (2\\pi r)dr=2\\pi \\sigma \\underset{0}{\\overset{R}{\\int }}{r}^{3}dr=2\\pi \\sigma \\frac{{r}^{4}}{4}{|}_{0}^{R}=2\\pi \\sigma (\\frac{{R}^{4}}{4}-0)\\hfill \\\\ &amp; =2\\pi \\frac{m}{A}(\\frac{{R}^{4}}{4})=2\\pi \\frac{m}{\\pi {R}^{2}}(\\frac{{R}^{4}}{4})=\\frac{1}{2}m{R}^{2}.\\hfill \\end{array}$$<\/div>\n<p id=\"fs-id1167133525890\">Note that this agrees with the value given in <a class=\"autogenerated-content\" href=\"\/contents\/1b0f645a-e293-4925-9c9b-1dfc7a29ab03#CNX_UPhysics_10_04_RotInertia\">(Figure)<\/a>.<\/p>\n<\/div>\n<div id=\"fs-id1167133353900\" class=\"bc-section section\">\n<h4>Calculating the moment of inertia for compound objects<\/h4>\n<p id=\"fs-id1167133847080\">Now consider a compound object such as that in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_RodDisk\">(Figure)<\/a>, which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound object\u2019s moment of inertia can be found from the sum of each part of the object:<\/p>\n<div id=\"fs-id1167133828814\" class=\"equation-callout\">\n<div id=\"fs-id1167133811543\">$${I}_{\\text{total}}=\\sum _{i}{I}_{i}.$$<\/div>\n<\/div>\n<p id=\"fs-id1167133553885\">It is important to note that the moments of inertia of the objects in <a class=\"autogenerated-content\" href=\"#fs-id1167133811543\">(Figure)<\/a> are <em>about a common axis<\/em>. In the case of this object, that would be a rod of length <em>L<\/em> rotating about its end, and a thin disk of radius <em>R<\/em> rotating about an axis shifted off of the center by a distance $$ L+R$$, where <em>R<\/em> is the radius of the disk. Let\u2019s define the mass of the rod to be $$ {m}_{\\text{r}} $$ and the mass of the disk to be $$ {m}_{\\text{d}}.$$<\/p>\n<div id=\"CNX_UPhysics_10_05_RodDisk\" class=\"wp-caption aligncenter\">\n<div style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194505\/CNX_UPhysics_10_05_RodDisk.jpg\" alt=\"Figure shows a disk with radius R connected to a rod with length L.\" width=\"290\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.28<\/strong> Compound object consisting of a disk at the end of a rod. The axis of rotation is located at A.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id1167133797122\">The moment of inertia of the rod is simply $$ \\frac{1}{3}{m}_{\\text{r}}{L}^{2}$$, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is $$ \\frac{1}{2}{m}_{\\text{d}}{R}^{2} $$ and we apply the parallel-axis theorem $$ {I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2} $$ to find<\/p>\n<div id=\"fs-id1167133774220\" class=\"unnumbered\">$${I}_{\\text{parallel-axis}}=\\frac{1}{2}{m}_{\\text{d}}{R}^{2}+{m}_{\\text{d}}{(L+R)}^{2}.$$<\/div>\n<p id=\"fs-id1167133447451\">Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be<\/p>\n<div id=\"fs-id1167133803182\" class=\"unnumbered\">$${I}_{\\text{total}}=\\frac{1}{3}{m}_{\\text{r}}{L}^{2}+\\frac{1}{2}{m}_{\\text{d}}{R}^{2}+{m}_{\\text{d}}{(L+R)}^{2}.$$<\/div>\n<\/div>\n<div id=\"fs-id1167133333570\" class=\"bc-section section\">\n<h4>Applying moment of inertia calculations to solve problems<\/h4>\n<p id=\"fs-id1167133697400\">Now let\u2019s examine some practical applications of moment of inertia calculations.<\/p>\n<div id=\"fs-id1167133824110\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Person on a Merry-Go-Round<\/h4>\n<p>A 25-kg child stands at a distance $$ r=1.0\\,\\text{m} $$ from the axis of a rotating merry-go-round (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Merrygor\">(Figure)<\/a>). The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system.<\/p>\n<div id=\"CNX_UPhysics_10_05_Merrygor\" class=\"wp-caption aligncenter\">\n<div style=\"width: 441px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194508\/CNX_UPhysics_10_05_merrygor.jpg\" alt=\"Figure is a drawing of a child on a merry-go-round. Merry\u2013go-round has a 2 meter radius. Child is standing one meter from the center.\" width=\"431\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.29<\/strong> Calculating the moment of inertia for a child on a merry-go-round.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<h4>Strategy<\/h4>\n<p>This problem involves the calculation of a moment of inertia. We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. The notation we use is $$ {m}_{\\text{c}}=25\\,\\text{kg},{r}_{\\text{c}}=1.0\\,\\text{m},{m}_{\\text{m}}=500\\,\\text{kg},{r}_{\\text{m}}=2.0\\,\\text{m}$$.<\/p>\n<p id=\"fs-id1167133677861\">Our goal is to find $$ {I}_{\\text{total}}=\\sum _{i}{I}_{i}$$.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167133612679\">For the child, $$ {I}_{\\text{c}}={m}_{\\text{c}}{r}^{2}$$, and for the merry-go-round, $$ {I}_{\\text{m}}=\\frac{1}{2}{m}_{\\text{m}}{r}^{2}$$. Therefore<\/p>\n<div id=\"fs-id1167133549750\" class=\"unnumbered\">$${I}_{\\text{total}}=25{(1)}^{2}+\\frac{1}{2}(500){(2)}^{2}=25+1000=1025\\,\\text{kg}\u00b7{\\text{m}}^{2}.$$<\/div>\n<h4>Significance<\/h4>\n<p>The value should be close to the moment of inertia of the merry-go-round by itself because it has much more mass distributed away from the axis than the child does.<\/p>\n<\/div>\n<div id=\"fs-id1167133345942\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Rod and Solid Sphere<\/h4>\n<p>Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg.<\/p>\n<p><span id=\"fs-id1167133318645\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194511\/CNX_UPhysics_10_05_RodSphere_img.jpg\" alt=\"Figure A shows a disk with radius R connected to a rod with length L. The point A is at the end of the rod opposite to the disk. Figure B shows a disk with radius R connected to a rod with length L. The point B is at the end of the rod connected to the disk.\" \/><\/span><\/p>\n<h4>Strategy<\/h4>\n<p>Since we have a compound object in both cases, we can use the parallel-axis theorem to find the moment of inertia about each axis. In (a), the center of mass of the sphere is located at a distance $$ L+R $$ from the axis of rotation. In (b), the center of mass of the sphere is located a distance <em>R<\/em> from the axis of rotation. In both cases, the moment of inertia of the rod is about an axis at one end. Refer to <a class=\"autogenerated-content\" href=\"\/contents\/1b0f645a-e293-4925-9c9b-1dfc7a29ab03#fs-id1167133375236\">(Figure)<\/a> for the moments of inertia for the individual objects.<\/p>\n<ol id=\"fs-id1167133324770\" type=\"a\">\n<li>$${I}_{\\text{total}}=\\sum _{i}{I}_{i}={I}_{\\text{Rod}}+{I}_{\\text{Sphere}}$$;$${I}_{\\text{Sphere}}={I}_{\\text{center of mass}}+{m}_{\\text{Sphere}}{(L+R)}^{2}=\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{(L+R)}^{2}$$;$${I}_{\\text{total}}={I}_{\\text{Rod}}+{I}_{\\text{Sphere}}=\\frac{1}{3}{m}_{\\text{Rod}}{L}^{2}+\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{(L+R)}^{2};$$$${I}_{\\text{total}}=\\frac{1}{3}(2.0\\,\\text{kg}){(0.5\\,\\text{m})}^{2}+\\frac{2}{5}(1.0\\,\\text{kg})(0.2\\,{\\text{m})}^{2}+(1.0\\,\\text{kg}){(0.5\\,\\text{m}+0.2\\,\\text{m})}^{2};$$$${I}_{\\text{total}}=(0.167+0.016+0.490)\\,\\text{kg}\u00b7{\\text{m}}^{2}=0.673\\,\\text{kg}\u00b7{\\text{m}}^{2}.$$<\/li>\n<li>$${I}_{\\text{Sphere}}=\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{R}^{2}$$;$${I}_{\\text{total}}={I}_{\\text{Rod}}+{I}_{\\text{Sphere}}=\\frac{1}{3}{m}_{\\text{Rod}}{L}^{2}+\\frac{2}{5}{m}_{\\text{Sphere}}{R}^{2}+{m}_{\\text{Sphere}}{R}^{2}$$;$${I}_{\\text{total}}=\\frac{1}{3}(2.0\\,\\text{kg}){(0.5\\,\\text{m})}^{2}+\\frac{2}{5}(1.0\\,\\text{kg})(0.2\\,{\\text{m})}^{2}+(1.0\\,\\text{kg}){(0.2\\,\\text{m})}^{2}$$;$${I}_{\\text{total}}=(0.167+0.016+0.04)\\,\\text{kg}\u00b7{\\text{m}}^{2}=0.223\\,\\text{kg}\u00b7{\\text{m}}^{2}.$$<\/li>\n<\/ol>\n<h4>Significance<\/h4>\n<p>Using the parallel-axis theorem eases the computation of the moment of inertia of compound objects. We see that the moment of inertia is greater in (a) than (b). This is because the axis of rotation is closer to the center of mass of the system in (b). The simple analogy is that of a rod. The moment of inertia about one end is $$ \\frac{1}{3}m{L}^{2}$$, but the moment of inertia through the center of mass along its length is $$ \\frac{1}{12}m{L}^{2}$$.<\/p>\n<\/div>\n<div id=\"fs-id1167133775662\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Angular Velocity of a Pendulum<\/h4>\n<p>A pendulum in the shape of a rod (<a class=\"autogenerated-content\" href=\"#CNX_UPhysics_10_05_Pend\">(Figure)<\/a>) is released from rest at an angle of $$ 30\\text{\u00b0}$$. It has a length 30 cm and mass 300 g. What is its angular velocity at its lowest point?<\/p>\n<div id=\"CNX_UPhysics_10_05_Pend\" class=\"wp-caption aligncenter\">\n<div style=\"width: 327px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194514\/CNX_UPhysics_10_05_Pend.jpg\" alt=\"Figure shows a pendulum in the form of a rod with a mass of 300 grams and length of 30 centimeters. Pendulum is released from rest at an angle of 30 degrees.\" width=\"317\" height=\"308\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.30<\/strong> A pendulum in the form of a rod is released from rest at an angle of $$ 30\\text{\u00b0}.$$<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<h4>Strategy<\/h4>\n<p>Use conservation of energy to solve the problem. At the point of release, the pendulum has gravitational potential energy, which is determined from the height of the center of mass above its lowest point in the swing. At the bottom of the swing, all of the gravitational potential energy is converted into rotational kinetic energy.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1167132279386\">The change in potential energy is equal to the change in rotational kinetic energy, $$ \\Delta U+\\Delta K=0$$.<\/p>\n<p id=\"fs-id1167133851770\">At the top of the swing: $$ U=mg{h}_{\\text{cm}}=mg\\frac{L}{2}(\\text{cos}\\,\\theta )$$. At the bottom of the swing, $$ U=mg\\frac{L}{2}.$$<\/p>\n<p id=\"fs-id1167133686379\">At the top of the swing, the rotational kinetic energy is $$ K=0$$. At the bottom of the swing, $$ K=\\frac{1}{2}I{\\omega }^{2}$$. Therefore:<\/p>\n<div id=\"fs-id1167132206607\" class=\"unnumbered\">$$\\text{\u0394}U+\\text{\u0394}K=0\u21d2(mg\\frac{L}{2}(1-\\text{cos}\\,\\theta )-0)+(0-\\frac{1}{2}I{\\omega }^{2})=0$$<\/div>\n<p id=\"fs-id1167132202151\">or<\/p>\n<div id=\"fs-id1167132202154\" class=\"unnumbered\">$$\\frac{1}{2}I{\\omega }^{2}=mg\\frac{L}{2}(1-\\text{cos}\\,\\theta ).$$<\/div>\n<p id=\"fs-id1167132202203\">Solving for $$ \\omega $$, we have<\/p>\n<div id=\"fs-id1167133795236\" class=\"unnumbered\">$$\\omega =\\sqrt{mg\\frac{L}{I}(1-\\text{cos}\\,\\theta )}=\\sqrt{mg\\frac{L}{1\\text{\/}3m{L}^{2}}(1-\\text{cos}\\,\\theta )}=\\sqrt{g\\frac{3}{L}(1-\\text{cos}\\,\\theta )}.$$<\/div>\n<p id=\"fs-id1167132303820\">Inserting numerical values, we have<\/p>\n<div id=\"fs-id1167132303823\" class=\"unnumbered\">$$\\omega =\\sqrt{9.8\\,\\text{m}\\text{\/}{\\text{s}}^{2}\\frac{3}{0.3\\,\\text{m}}(1-\\text{cos}\\,30)}=3.6\\,\\text{rad}\\text{\/}\\text{s}.$$<\/div>\n<h4>Significance<\/h4>\n<p>Note that the angular velocity of the pendulum does not depend on its mass.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133349420\" class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"fs-id1167133566402\">\n<li>Moments of inertia can be found by summing or integrating over every \u2018piece of mass\u2019 that makes up an object, multiplied by the square of the distance of each \u2018piece of mass\u2019 to the axis. In integral form the moment of inertia is $$ I=\\int {r}^{2}dm$$.<\/li>\n<li>Moment of inertia is larger when an object\u2019s mass is farther from the axis of rotation.<\/li>\n<li>It is possible to find the moment of inertia of an object about a new axis of rotation once it is known for a parallel axis. This is called the parallel axis theorem given by $$ {I}_{\\text{parallel-axis}}={I}_{\\text{center of mass}}+m{d}^{2}$$, where <em>d<\/em> is the distance from the initial axis to the parallel axis.<\/li>\n<li>Moment of inertia for a compound object is simply the sum of the moments of inertia for each individual object that makes up the compound object.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1167133549678\" class=\"review-conceptual-questions\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1167133357710\" class=\"problem textbox\">\n<div id=\"fs-id1167133357712\">\n<p id=\"fs-id1167133863061\">If a child walks toward the center of a merry-go-round, does the moment of inertia increase or decrease?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133359289\" class=\"problem textbox\">\n<div id=\"fs-id1167133359292\">\n<p id=\"fs-id1167133359294\">A discus thrower rotates with a discus in his hand before letting it go. (a) How does his moment of inertia change after releasing the discus? (b) What would be a good approximation to use in calculating the moment of inertia of the discus thrower and discus?<\/p>\n<\/div>\n<div id=\"fs-id1167133845959\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167133845959\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167133845959\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167133328460\">a. It decreases. b. The arms could be approximated with rods and the discus with a disk. The torso is near the axis of rotation so it doesn\u2019t contribute much to the moment of inertia.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133845992\" class=\"problem textbox\">\n<div id=\"fs-id1167133845994\">\n<p id=\"fs-id1167133353348\">Does increasing the number of blades on a propeller increase or decrease its moment of inertia, and why?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133357558\" class=\"problem textbox\">\n<div id=\"fs-id1167133357560\">\n<p id=\"fs-id1167133357562\">The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is $$ m{L}^{2}\\text{\/}3$$. Why is this moment of inertia greater than it would be if you spun a point mass <em>m<\/em> at the location of the center of mass of the rod (at <em>L<\/em>\/2) (that would be $$ m{L}^{2}\\text{\/}4$$)?<\/p>\n<\/div>\n<div id=\"fs-id1167133354662\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167133354662\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167133354662\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167133354664\">Because the moment of inertia varies as the square of the distance to the axis of rotation. The mass of the rod located at distances greater than <em>L<\/em>\/2 would provide the larger contribution to make its moment of inertia greater than the point mass at <em>L<\/em>\/2.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133773940\" class=\"problem textbox\">\n<div id=\"fs-id1167133773942\">\n<p id=\"fs-id1167133773944\">Why is the moment of inertia of a hoop that has a mass <em>M<\/em> and a radius <em>R<\/em> greater than the moment of inertia of a disk that has the same mass and radius?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133871657\" class=\"review-problems textbox exercises\">\n<h3>Problems<\/h3>\n<div id=\"fs-id1167133407930\" class=\"problem textbox\">\n<div id=\"fs-id1167133407932\">\n<p id=\"fs-id1167133407934\">While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is $$ 3.75{\\,\\text{kg-m}}^{2} $$ and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter\u2019s shoe if it is 1.05 m from the hip joint?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133357822\" class=\"problem textbox\">\n<div id=\"fs-id1167133357824\">\n<p id=\"fs-id1167133859006\">Using the parallel axis theorem, what is the moment of inertia of the rod of mass <em>m<\/em> about the axis shown below?<\/p>\n<p><span id=\"fs-id1167133325838\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194516\/CNX_UPhysics_10_05_Prob8_img.jpg\" alt=\"Figure shows a rod that rotates around the axis that passes through it at 1\/6 of length from one end and 5\/6 of length from the opposite end.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1167133497626\">\n<p id=\"fs-id1167133497628\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q96051\">Show Answer<\/span><\/p>\n<div id=\"q96051\" class=\"hidden-answer\" style=\"display: none\">$$I=\\frac{7}{36}m{L}^{2}$$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133775336\" class=\"problem textbox\">\n<div id=\"fs-id1167133775338\">\n<p id=\"fs-id1167133775340\">Find the moment of inertia of the rod in the previous problem by direct integration.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133346254\" class=\"problem textbox\">\n<div id=\"fs-id1167133792663\">\n<p id=\"fs-id1167133792665\">A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one end (see the following figure). If the rod is released from rest at an angle of $$ 60\\text{\u00b0} $$ with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?<\/p>\n<p><span id=\"fs-id1167133359187\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194518\/CNX_UPhysics_10_05_RodAng_img.jpg\" alt=\"Figure shows a rod that is released from rest at an angle of 60 degrees with respect to the horizontal.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1167133344789\">\n<p id=\"fs-id1167133344791\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q71295\">Show Answer<\/span><\/p>\n<div id=\"q71295\" class=\"hidden-answer\" style=\"display: none\">$$v=7.14\\,\\text{m}\\text{\/}\\text{s}.$$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133365711\" class=\"problem textbox\">\n<div id=\"fs-id1167133365713\">\n<p id=\"fs-id1167133319885\">A pendulum consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm (see the following figure). If the pendulum is released from rest at an angle of $$ 30\\text{\u00b0}$$, what is the angular velocity at the lowest point?<\/p>\n<p><span id=\"fs-id1167133325585\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194521\/CNX_UPhysics_10_05_PendSph_img.jpg\" alt=\"Figure shows a pendulum that consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm. The pendulum is released from rest at an angle of 30 degrees.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133447838\" class=\"problem textbox\">\n<div id=\"fs-id1167133357875\">\n<p id=\"fs-id1167133357877\">A solid sphere of radius 10 cm is allowed to rotate freely about an axis. The sphere is given a sharp blow so that its center of mass starts from the position shown in the following figure with speed 15 cm\/s. What is the maximum angle that the diameter makes with the vertical?<\/p>\n<p><span id=\"fs-id1167133345349\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194523\/CNX_UPhysics_10_05_SphPiv_img.jpg\" alt=\"Left figure shows a solid sphere of radius 10 cm that first rotates freely about an axis and then received a sharp blow in its center of mass. Right figure is the image of the same sphere after the blow. An angle that the diameter makes with the vertical is marked as theta.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1167132305511\">\n<p id=\"fs-id1167132305513\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q602090\">Show Answer<\/span><\/p>\n<div id=\"q602090\" class=\"hidden-answer\" style=\"display: none\">$$\\theta =10.2\\text{\u00b0}$$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167133846985\" class=\"problem textbox\">\n<div id=\"fs-id1167133846987\">\n<p id=\"fs-id1167133851633\">Calculate the moment of inertia by direct integration of a thin rod of mass <em>M<\/em> and length <em>L<\/em> about an axis through the rod at <em>L<\/em>\/3, as shown below. Check your answer with the parallel-axis theorem.<\/p>\n<p><span id=\"fs-id1167132202884\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31194525\/CNX_UPhysics_10_05_RodMomen_img.jpg\" alt=\"Figure shows a rod that rotates around the axis that passes through it at 1\/3 of length from one end and 2\/3 of length from the opposite end.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1167133686238\">\n<dt><strong>linear mass density<\/strong><\/dt>\n<dd id=\"fs-id1167133686243\">the mass per unit length $$ \\lambda  $$ of a one dimensional object<\/dd>\n<\/dl>\n<dl id=\"fs-id1167133822794\">\n<dt><strong>parallel axis<\/strong><\/dt>\n<dd id=\"fs-id1167133802923\">axis of rotation that is parallel to an axis about which the moment of inertia of an object is known<\/dd>\n<\/dl>\n<dl id=\"fs-id1167132202132\">\n<dt><strong>parallel-axis theorem<\/strong><\/dt>\n<dd id=\"fs-id1167132202137\">if the moment of inertia is known for a given axis, it can be found for any axis parallel to it<\/dd>\n<\/dl>\n<dl id=\"fs-id1167133399142\">\n<dt><strong>surface mass density<\/strong><\/dt>\n<dd id=\"fs-id1167133858665\">mass per unit area $$ \\sigma  $$ of a two dimensional object<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-958\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>OpenStax University Physics. <strong>Authored by<\/strong>: OpenStax CNX. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\">https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax University Physics\",\"author\":\"OpenStax CNX\",\"organization\":\"\",\"url\":\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"all-rights-reserved"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-958","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":908,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/958","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/958\/revisions"}],"predecessor-version":[{"id":2154,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/958\/revisions\/2154"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/parts\/908"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/958\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/media?parent=958"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=958"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/contributor?post=958"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/license?post=958"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}