{"id":577,"date":"2021-07-21T14:24:12","date_gmt":"2021-07-21T18:24:12","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=577"},"modified":"2025-08-01T18:46:02","modified_gmt":"2025-08-01T22:46:02","slug":"7-7-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/7-7-examples\/","title":{"raw":"7.6 Examples","rendered":"7.6 Examples"},"content":{"raw":"Here are examples from Chapter 7 to help you understand these concepts better. These were taken from the real world and supplied by FSDE students in Summer 2021. If you'd like to submit your own examples, please send them to the author <a href=\"mailto:eosgood@upei.ca\">eosgood@upei.ca<\/a>.\r\n\r\n&nbsp;\r\n<h1>Example 7.6.1: All of Ch 7, Submitted by William Craine<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA person is playing soccer. The ball they are using has a diameter of 20 cm and a mass of 0.45 kg. The person\u2019s leg has a mass of 18 kg, and their foot has a mass of 8 kg. Assume that all the shapes have uniform density.\r\n\r\na) Find the cm for the ball.\r\n\r\nb) Calculate the mass moment of inertia (MMOI) for the ball.\r\n\r\nc) Find the cm for the person\u2019s leg and foot.\r\n\r\nd) Find the MMOI for the person\u2019s leg and foot on the y-axis about A.\r\n\r\n[caption id=\"attachment_1514\" align=\"aligncenter\" width=\"309\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/images.jpg\" alt=\"A person kicking a ball.\" class=\"wp-image-1514 \" width=\"309\" height=\"309\" \/> Source:<a href=\"https:\/\/www.pngwing.com\/en\/free-png-aaujw\"> https:\/\/encrypted-tbn0.gstatic.com\/images?q=tbn:ANd9GcTM4e4xHaRSXBdQMGugm1gISi2Qgn7rQx_K3w&amp;usqp=CAU<\/a>[\/caption]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-1024x779.png\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1515\" width=\"417\" height=\"317\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-300x236.jpg\" alt=\"A FBD of the problem.\" class=\"alignleft wp-image-885 size-medium\" width=\"300\" height=\"236\" \/>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-300x201.jpg\" alt=\"A circular sketch of the ball.\" class=\"aligncenter wp-image-886\" width=\"248\" height=\"166\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>m<sub>b<\/sub> = 0.45 kg<\/li>\r\n \t<li>d<sub>b<\/sub> = 20 cm<\/li>\r\n \t<li>m<sub>L<\/sub> = 18 kg<\/li>\r\n \t<li>m<sub>f<\/sub> = 8 kg<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>x<sub>b<\/sub><\/li>\r\n \t<li>I<sub>b<\/sub><\/li>\r\n \t<li>x<sub>p<\/sub><\/li>\r\n \t<li>I<sub>p x<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nFind the cm for both objects using arbitrary coordinates since no origin is given.\r\n\r\nUse the sphere MMOI formula for the ball.\r\n\r\nCalculate the individual MMOIs for the leg and foot, then use the parallel axis theorem to get each shape's MMOI about the system cm, add them, and then use the parallel axis theorem to get the MMOI about A.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a (find the center of mass of the ball):\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-300x264.jpg\" alt=\"Diagram to find the center of mass for the ball.\" class=\"aligncenter wp-image-887 size-medium\" width=\"300\" height=\"264\" \/>\r\n\r\n[latex]x_b=(10\\underline{\\hat{i}}+10\\underline{\\hat{j}}+10\\underline{\\hat{k}})cm[\/latex]\r\n\r\n&nbsp;\r\n\r\nPart b (find the MMOI of the ball about its center of mass):\r\n\r\n[latex]I_{xx}=I_{yy}=I_{zz}=\\frac{2}{3}mr^2\\\\\\qquad \\quad=\\frac{2}{3}(0.45 kg)(0.1m)^2\\\\I_b=0.003kgm^2[\/latex]\r\n\r\nPart c (find the center of mass for the system of the person's leg):\r\n\r\n<em>Step 1: Find the center of mass of the foot (f)<\/em>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-300x166.jpg\" alt=\"Diagram to find the centroid of the persons foot (represented as a cuboid).\" class=\"aligncenter wp-image-890 size-medium\" width=\"300\" height=\"166\" \/>\r\n\r\n[latex]f=\\frac{15cm}{2}\\underline{\\hat{i}}+\\frac{7cm}{2}\\underline{\\hat{j}}+\\frac{7cm}{2}\\underline{\\hat{k}}\\\\f=(7.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+3.5\\underline{\\hat{k}})cm[\/latex]\r\n\r\n&nbsp;\r\n\r\n<em>Step 2: Find the center of mass of the leg (L)<\/em>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6-232x300.jpg\" alt=\"Diagram to find the centroid of the persons leg (represented as a cylinder).\" class=\"aligncenter wp-image-892\" width=\"189\" height=\"244\" \/>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-300x228.jpg\" alt=\"Diagram to find the centroid of the persons foot and leg combined.\" class=\"aligncenter wp-image-906\" width=\"407\" height=\"309\" \/>\r\n\r\n[latex]L_1=(3.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+15\\underline{\\hat{k}})cm\\\\L_2=(15cm-7cm-1cm)\\underline{\\hat{i}}+(0cm)\\underline{\\hat{j}}+(7cm)\\underline{\\hat{k}}\\\\\\qquad \\quad L_2 = (7 \\underline{\\hat{i}}+(0)\\underline{\\hat{j}}+(7)\\underline{\\hat{k}} )cm[\/latex]\r\n\r\n[latex]\\\\L=L_1+L_2\\\\L=(3.5cm+7cm)\\underline{\\hat{i}}+(3.5cm+0cm)\\underline{\\hat{j}}+(15cm+7cm)\\underline{\\hat{k}}\\\\L=(10.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+22\\underline{\\hat{k}})cm[\/latex]\r\n\r\n<em>Step 3: find the center of mass of the system (P)<\/em>\r\n\r\n[latex]x_p=\\frac{\\sum m_i x_i}{\\sum m_i}\\\\X_p=\\frac{m_f\\cdot x_f+m_L\\cdot x_L}{m_f+m_L}\\\\x_p=\\frac{8kg\\cdot 7.5cm+18kg\\cdot 10.5cm}{8kg+18kg}\\\\x_p=9.58cm[\/latex]\r\n\r\n[latex]y_p=\\frac{\\sum m_i y_i}{\\sum m_i}\\\\y_p=\\frac{8kg\\cdot 3.5cm+18kg\\cdot 3.5cm}{8kg+18kg}\\\\y_p=3.5cm[\/latex]\r\n\r\n[latex]z_p=\\frac{\\sum m_i z_i}{\\sum m_i}\\\\z_p=\\frac{8kg\\cdot 3.5cm+18kg\\cdot 22cm}{8kg+18kg}\\\\z_p=16.3cm[\/latex]\r\n\r\n[latex]\\underline{\\underline{P}=(9.58\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+16.3\\underline{\\hat{k}})cm}[\/latex]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-300x246.jpg\" alt=\"Diagram to find the centroid of the persons foot and leg combined.\" class=\"aligncenter wp-image-898\" width=\"413\" height=\"339\" \/>\r\n\r\nPart d (find the inertia of the person's leg about point A):\r\n\r\n<em>Step 1: Find the MMOI of the foot about f (I<sub>ff<\/sub>)<\/em>\r\n\r\n[latex]x_f=0.15m\\\\z_f=0.07m\\\\m_f=8kg[\/latex]\r\n\r\n[latex]I_{ff}=\\frac{1}{12}\\cdot m\\cdot (x^2+z^2)\\\\I_{ff}=\\frac{1}{12}(8kg)(0.15m^2+0.07m^2)\\\\I_{ff}=0.0183kg\\; m^2[\/latex]\r\n\r\n<em>Step 2: Find the MMOI of the foot about P (I<sub>Pf<\/sub>)<\/em>\r\n\r\n[latex]d^2=(0.0208m)^2+(0.128m)^2\\\\I_{pf}=I_{ff}+m_f(d_f)^2\\\\I_{pf}=0.0182kgm^2+8kg[(0.0208m)^2+(0.128m)^2]\\\\I_{pf}=0.1527kgm^2[\/latex]\r\n\r\n<em>Step 3: Find the MMOI of the leg about L (I<sub>LL<\/sub>)<\/em>\r\n\r\n[latex]r=0.035m\\\\h=0.3m\\\\m_L=18kg\\\\I_{LL}=\\frac{1}{12}\\cdot m\\cdot (3_r^2+h^2)\\\\I_{LL}=\\frac{1}{12}\\cdot 18kg(3(0.035m)^2+(0.3m)^2)\\\\I_{LL}=0.1405kgm^2[\/latex]\r\n\r\n<em>Step 4: Find the MMOI of the leg about P (I<sub>PL<\/sub>)<\/em>\r\n\r\n[latex]r_{LP}=[(9.58-10.5)\\underline{\\hat{i}}+(3.5-3.5)\\underline{\\hat{j}}+(16.3-22)\\underline{\\hat{k}}]cm\\\\r_{LP}=(-0.92\\underline{\\hat{i}}-5.7\\underline{\\hat{k}})cm\\\\I_{PL}=I_{LL}+m(d^2)\\\\I_{PL}=0.1405kgm^2+18kg[(0.0092m)^2+(0.057m)^2]\\\\I_{PL}=0.2005kgm^2[\/latex]\r\n\r\n<em>Step 5: Find the MMOI of the entire system about P (I<sub>P<\/sub>)<\/em>\r\n\r\n[latex]I_p=I_{PL}+I_{pf}\\\\I_p=0.1527kgm^2+0.2005kg\\\\I_p=0.3532kgm^2[\/latex]\r\n\r\n<em>Step 6: Find the MMOI of the entire system about A (I<sub>A<\/sub>)<\/em>\r\n\r\n[latex]A=(10.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+37\\underline{\\hat{k}})cm\\\\P=(9.58\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+16.3\\underline{\\hat{k}})cm\\\\r_{AP}=P-A\\\\r_{AP}=[(9.58-10.5)\\underline{\\hat{i}}+(3.5-3.5)\\underline{\\hat{j}}+(16.3-37)\\underline{\\hat{k}}]cm\\\\r_{AP}=(-0.92\\underline{\\hat{i}}-20.7\\underline{\\hat{k}})cm[\/latex]\r\n\r\n[latex]I_A=I_p+m(d^2)\\\\I_A=0.3532kgm^2+(8kg+18kg)[(0.0092m)^2+(0.207m)^2]\\\\I_A=1.4695kgm^2\\\\\\underline{I_A=1.47kgm^2}[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that the numbers are small, since before the final step, the mass was small, or the distance to the new axis was small.\r\n\r\n<\/div>\r\n<h1>Example 7.6.2 Inertia, Submitted by Luke McCarvill<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA figure skater with a mass of 60 kg is about to perform a spin about her long axis (z). She is 167 cm tall, and her body can be approximated as a circular cylinder of 30 cm diameter while her limbs are at her side, and a circular cylinder of 60 cm diameter while her arms (and one leg) are outstretched.\r\n\r\na) What should she do in order to generate the highest angular acceleration, assuming she can generate a net torque of 200 Nm? Does lowering her height increase or decrease her angular acceleration?\r\n\r\nb) How fast will she be spinning after 0.5 seconds of her maximum vs minimum accelerations, assuming she starts from zero (\u2375<sub>0<\/sub> = 0 rad\/sec)?\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1516\" align=\"alignnone\" width=\"1024\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-1024x817.png\" alt=\"A person ice skating.\" class=\"wp-image-1516 size-large\" width=\"1024\" height=\"817\" \/> Annotated from original source: https:\/\/commons.wikimedia.org\/wiki\/File:2019_Internationaux_de_France_Friday_ladies_SP_group_1_Starr_ANDREWS_8D9A6706.jpg[\/caption]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-300x255.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1145 size-medium\" width=\"300\" height=\"255\" \/> <img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-300x254.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1146 size-medium\" width=\"300\" height=\"254\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>m = 60 kg<\/li>\r\n \t<li>\u03a3M = 200Nm<\/li>\r\n \t<li>h = 167cm<\/li>\r\n \t<li>d<sub>1<\/sub> = 30cm<\/li>\r\n \t<li>d<sub>2<\/sub> = 60cm<\/li>\r\n \t<li>t = 0.5sec<\/li>\r\n \t<li>I<sub>zz<\/sub> = \u00bd mr2<\/li>\r\n \t<li>\u03a3M = I\u221d<\/li>\r\n \t<li>\u2375<sub>0<\/sub> = 0rad\/sec<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>\u221d<sub>1<\/sub> = ?<\/li>\r\n \t<li>\u221d<sub>2<\/sub> = ?<\/li>\r\n \t<li>I<sub>zz1<\/sub> = ?<\/li>\r\n \t<li>I<sub>zz2<\/sub> = ?<\/li>\r\n \t<li>\u2375<sub>1<\/sub> = ?<\/li>\r\n \t<li>\u2375<sub>2<\/sub> = ?<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nI\u2019ll be using MMOI for circular cylinders, as well as the sum of moments\/torque equaling MMOI times angular acceleration, as well as acceleration equaling change in velocity over time.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n<em>Step 1: Find the inertia when the arms are hugged to the body<\/em>\r\n\r\nI<sub>zz1<\/sub> = 0.5 (60kg) (0.15m)<sup>2 <\/sup>\r\n\r\nI<sub>zz1<\/sub>= 0.675 kg m<sup>2<\/sup>\r\n\r\n&nbsp;\r\n\r\n<em>Step 2: Find the inertia when the arms are spread out<\/em>\r\n\r\nI<sub>zz2<\/sub> = 0.5(60kg)(0.3m)<sup>2<\/sup>\r\n\r\nI<sub>zz2<\/sub> = 2.7 kg m<sup>2<\/sup>\r\n\r\n&nbsp;\r\n\r\n<em>Step 3: Find the angular acceleration for both cases<\/em>\r\n\r\n\u03a3M = I\u221d ; therefore, \u03a3M\/I = \u221d\r\n\r\n&nbsp;\r\n\r\n\u221d<sub>1<\/sub> = \u03a3M \/ I<sub>zz1<\/sub>\r\n\r\n\u221d<sub>1<\/sub> = 200 Nm \/ 0.675 kg m<sup>2<\/sup>\r\n\r\n\u221d<sub>1<\/sub> \u2248 296.296 rad\/sec<sup>2<\/sup>\r\n\r\n&nbsp;\r\n\r\n\u221d<sub>2<\/sub> = \u03a3M \/ I<sub>zz2<\/sub>\r\n\r\n\u221d<sub>2<\/sub> = 200Nm\/2.7 kg m<sup>2<\/sup>\r\n\r\n\u221d<sub>2<\/sub> \u2248 74.074 rad\/sec<sup>2 <\/sup>\r\n\r\n&nbsp;\r\n\r\nThe acceleration when the skater had her arms close to her body was about 296 rad\/sec<sup>2<\/sup>, while that when she had her arms spread out was about 74 rad\/sec<sup>2<\/sup>. Therefore, having her limbs closer to her body will give her a much higher angular acceleration!\r\n\r\nAs seen in the equation I<sub>zz<\/sub> = \u00bd m r<sup>2<\/sup>, her height is arbitrary; thus, lowering her height would not change the inertia, nor would it change her angular acceleration.\r\n\r\n&nbsp;\r\n\r\nPart b:\r\n\r\n\u221d = \u2375\/t therefore \u221d*t = \u2375\r\n\r\n&nbsp;\r\n\r\n\u2375<sub>1<\/sub> = 296.296 s<sup>-2<\/sup>(0.5s)\r\n\r\n\u2375<sub>1<\/sub> = 148.148 rad\/sec\r\n\r\n&nbsp;\r\n\r\n\u2375<sub>2<\/sub> = 74.074 s<sup>-2<\/sup>(0.5s)\r\n\r\n\u2375<sub>2<\/sub> = 37.037 rad\/sec\r\n\r\n&nbsp;\r\n\r\nGiven that these are in radians per second, let\u2019s convert this to rotations per second to make it more meaningful. To do so, simply divide by 2\u03c0 since there are 2\u03c0 radians per rotation. Thus, with her arms in, she can achieve about 23.6 rotations per second after 0.5 seconds of acceleration, compared to about 5.9 rotations per second with her limbs out. This is on par with an Olympian, according to <a href=\"http:\/\/www.bsharp.org\/physics\/spins\">this site<\/a>.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that they spin faster when their limbs are hugged to their body; we can try this at home with a swivel chair!\r\n\r\n<\/div>\r\n<h1>Example 7.6.3: All of Ch 7, Submitted by Victoria Keefe<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nTwo rectangular compost containers are tied end to end with a 2 m chain of negligible mass. Container A has a mass of 5 kg with a height of 1 m and a width of 2 m. The container B is of 8kg, height 2m, and width 3m. Consider the point which is the center of the chain as the origin of the coordinate frame to find the center of mass, center of gravity,\u00a0 and centroid of the system and the individual containers.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"372\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefes-metal-containers.jpg\" width=\"372\" height=\"279\" alt=\"Compost containers\" class=\"\" \/> Source:https:\/\/felixwong.com\/2021\/10\/differences-between-homes-in-the-u-s-and-spain\/[\/caption]\r\n\r\n<span style=\"text-align: initial;background-color: initial;font-size: 1em\">The real-life scenario of the problem would look similar to the image.<\/span>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5.jpg\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1814 size-full\" width=\"610\" height=\"242\" \/>\r\n\r\n<\/div>\r\n<strong style=\"text-align: initial;background-color: initial;font-size: 1em\">2. Draw<\/strong>\r\n<div><strong style=\"font-size: 1em;background-color: initial\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1807 size-full\" width=\"721\" height=\"314\" \/><\/strong><\/div>\r\n<div><strong>3. Knowns and Unknowns<\/strong><\/div>\r\n<div><\/div>\r\n<div>Knowns:<\/div>\r\n<ul>\r\n \t<li>m<sub>A<\/sub> = 5 kg<\/li>\r\n \t<li>m<sub>B = <\/sub>8 kg<\/li>\r\n \t<li>h<sub>A\u00a0<\/sub>= 1 m<\/li>\r\n \t<li>h<sub>B<\/sub> = 2 m<\/li>\r\n \t<li>w<sub>A =\u00a0<\/sub>2m<\/li>\r\n \t<li>w<sub>B <\/sub>= 3 m<\/li>\r\n \t<li>d =\u00a0 2 m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>[latex]\\bar{x_A}[\/latex]<\/li>\r\n \t<li>[latex]\\bar{x_B}[\/latex]<\/li>\r\n \t<li>[latex]\\bar{y_a}[\/latex]<\/li>\r\n \t<li>[latex]\\bar{y_b}[\/latex]<\/li>\r\n \t<li>[latex]\\bar{x_s}[\/latex]<\/li>\r\n \t<li>[latex]\\bar{y_s}[\/latex]<\/li>\r\n \t<li>[latex]\\bar{r_s} [\/latex]<\/li>\r\n<\/ul>\r\n<div><\/div>\r\n<div><strong>4. Approach\u00a0<\/strong><\/div>\r\n<div>Use equations for the center of mass, the center of gravity, and the centroid. utilizing the advantage of symmetry whenever possible.<\/div>\r\n<div><\/div>\r\n<div><strong>5. Analysis\u00a0<\/strong><\/div>\r\n<div><\/div>\r\n<div><span style=\"text-decoration: underline\">FINDING CENTER OF MASS\u00a0<\/span><\/div>\r\n<div>Calculate the centre of mass of box A. Because it is symmetrical, you divide the height and width by 2 and add the distance from O.<\/div>\r\n<div>[latex]\\bar{x_A}=\\frac{w_A}{2}+1 = 2 m[\/latex]<\/div>\r\n<div>[latex]\\bar{y_A}=\\frac{h_A}{2} =\u00a0 0.5 m[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Thus, in vector form center of mass of container A is [latex]{cm_A}=\\begin{bmatrix} 2 m\\\\ 0.5 m\\end{bmatrix}[\/latex]<\/div>\r\n<div><span style=\"background-color: initial;font-size: 1em\">Calculate the centre of mass of box B. Because it is symmetrical, you divide the height and width by 2, and add the distance from O.<\/span><\/div>\r\n<div>\r\n<div>[latex]\\bar{x_B}=\\frac{-w_B}{2}-1 \\\\=\\frac{-3}{2}-1 =-2.5 m[\/latex]<\/div>\r\n<div>[latex]\\bar{y_B}=\\frac{h_B}{2} = 1 m[\/latex]<\/div>\r\n<div>In vector form, [latex]{cm_B}=\\begin{bmatrix} -2.5 m\\\\ 1 m\\end{bmatrix}[\/latex]<\/div>\r\n<\/div>\r\n<div>Calculate the system centre of mass in x:<\/div>\r\n<div>[latex]cm_{sx} = \\frac {{m_A} \\cdot \\bar{x}_{ax} +m_B\\cdot \\bar{x}_{bx}} {m_A+m_B}\\\\ = \\frac{5kg \\cdot2m +8kg\\cdot(-2.5)m }{5 kg +8kg}\\\\ = -0.769 m[\/latex]\r\nCalculate the System centre of mass in y:\r\n[latex]cm_{sy} = \\frac {{m_A} \\cdot \\bar{x}_{ay} +m_B\\cdot \\bar{x}_{by}} {m_A+m_B}\\\\ = \\frac{5kg \\cdot 0.5m +8kg\\cdot 1m }{5 kg +8kg}\\\\ = 0.808m [\/latex]<\/div>\r\n<div><\/div>\r\n<div>In vector format center of mass of the system:<\/div>\r\n<div>[latex]{cm_S}=\\begin{bmatrix}-0.769 m \\\\0.808 m \\end{bmatrix}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><span style=\"text-decoration: underline\">FINDING CENTER OF GRAVITY<\/span><\/div>\r\n<div>The center of gravity of container A and container B is the same as the center of mass due to the uniform rectangular shape and orientation.<\/div>\r\n<div>Thus<\/div>\r\n<div>[latex]cg_{A} =\\begin{bmatrix}2 m \\\\ 0.5 m\\end{bmatrix}[\/latex]<\/div>\r\n<div>similarly, [latex]cg_{B} =\\begin{bmatrix}-2.5 m\u00a0 \\\\ 1 m\\end{bmatrix}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Since the system has a uniform gravitational field and constant density, the system's centre of gravity is equal to that of the centre of mass found above.<\/div>\r\n<div>Thus<\/div>\r\n<div style=\"text-align: center\">[latex]{cg_S}=\\begin{bmatrix}-0.769\\\\0.808 m \\end{bmatrix}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><span style=\"text-decoration: underline\">FINDING CENTROID<\/span><\/div>\r\n<div>Similar to the center of gravity, the centroid of individual containers is also the same as the center of mass of the individual containers due to the uniform shape of the containers.<\/div>\r\n<div>In this specific problem, where the containers are of simple uniform rectangular shape, we can consider the symmetry and the dimensions of the system to find the centroid of the system.<\/div>\r\n<div><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16.png\" alt=\"Diagram shown to find the centroid.\" class=\"alignnone wp-image-1809 size-full\" width=\"688\" height=\"358\" \/><\/div>\r\n<div>Since the system has a rectangular shape, the centroid would be the geometrical center. The vector [latex]\\overrightarrow{r_S}[\/latex](blue vector line in the above diagram) can be found from the following equation derived from the above image.<\/div>\r\n<div>[latex]\\overrightarrow{r_{Sx}} = \\frac{t_s}{2}- w_B\\\\= \\frac{7m}{2}-3 m\\\\= 0.5 m[\/latex]; here t<sub>S<\/sub> is the total width of the system.<\/div>\r\n<div>[latex]\\overrightarrow{r_{Sy}} =\\frac{h_B}{2}\\\\ = 1m[\/latex]<\/div>\r\n<div>In vector form, [latex]\\overrightarrow{r_S}=\\begin{bmatrix}0.5m\\\\1 m \\end{bmatrix}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><strong>6. Review<\/strong><\/div>\r\n<div>The centre of gravity being the same as the centre of mass is due to only looking at this problem in 2D. The requirements for the centre of mass and the centre of gravity being equal have been fulfilled.<\/div>\r\n<\/div>\r\n<div>\r\n<h1>Example 7.6.4: All of Ch 7, Submitted by Deanna Malone<\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n<p style=\"text-align: left\">A woman doing yoga weighs 65 kg, and her legs are 85 cm long. The combined mass of both her legs is 11.7 kg, one of which is lifted 90 cm above the ground (as in the image below). Her torso contributes to 50% of her mass and is 60 cm long. Her arms are 4.5 kg each, and they are 90 cm from the fingertips to the shoulder. What is her center of gravity in the warrior 3 pose from the origin?<\/p>\r\n\r\n\r\n[caption id=\"attachment_2197\" align=\"aligncenter\" width=\"591\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back.jpg\" alt=\"A person in the warrior 3 pose. \" width=\"591\" height=\"331\" class=\"wp-image-2197 \" \/> Source:https:\/\/commons.wikimedia.org\/wiki\/File:Virabhadrasana_III_from_back.jpg[\/caption]\r\n\r\n<\/div>\r\n<p style=\"text-align: justify\"><strong>2. Draw<\/strong><\/p>\r\n\r\n\r\n[caption id=\"attachment_2174\" align=\"aligncenter\" width=\"459\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554-300x173.jpg\" alt=\"Warrior pose 3 silhouette, and a FBD of the problem.\" width=\"459\" height=\"265\" class=\"wp-image-2174\" \/> Source:https:\/\/www.vhv.rs\/viewpic\/ihxioJo_warrior-3-yoga-pose-silhouette-hd-png-download\/)[\/caption]\r\n<p style=\"text-align: justify\">3. <strong>Knowns and Unknowns<\/strong><\/p>\r\n<p style=\"text-align: justify\">Knowns:<\/p>\r\n\r\n<ul style=\"text-align: justify\">\r\n \t<li>m= 65 kg<\/li>\r\n \t<li>m<sub>L<\/sub> = 11.7 kg<\/li>\r\n \t<li>m<sub>A<\/sub> = 4.5 kg<\/li>\r\n \t<li>m<sub>T<\/sub> = 50% of 65 kg<\/li>\r\n<\/ul>\r\n<p style=\"text-align: justify\">Unknowns:<\/p>\r\n\r\n<ul style=\"text-align: justify\">\r\n \t<li>[latex]\\overrightarrow{r}[\/latex], vector from origin to center of gravity<\/li>\r\n<\/ul>\r\n<p style=\"text-align: justify\"><span style=\"text-decoration: underline\">Assumptions<\/span><\/p>\r\n<p style=\"text-align: justify\">g= 9.81 m\/s<sup>2<\/sup><\/p>\r\n<p style=\"text-align: justify\">4.\u00a0 <strong>Approach<\/strong><\/p>\r\n<p style=\"text-align: justify\">Calculate mass and distance from origin, keeping in mind that the mass distribution is different for legs, torso, and arms, but assuming that the mass is evenly distributed within each of these parts. Use the center of gravity equation.<\/p>\r\n<p style=\"text-align: justify\">5.<strong> Analysis<\/strong><\/p>\r\n<p style=\"text-align: justify\">Point of mass for leg in air \u2013 11.7 kg at 90 cm up and 42.5 cm left from the origin.<\/p>\r\n<p style=\"text-align: justify\">Point mass for leg on ground \u2013 11.7 kg at 42.5 cm up and 90 cm left.<\/p>\r\n<p style=\"text-align: justify\">Point mass for torso \u2013 32.5 kg at 90 cm up and 115 cm left.<\/p>\r\n<p style=\"text-align: justify\">Point mass for both arms \u2013 9 kg at 90 cm up and 190 cm left.<\/p>\r\n<p style=\"text-align: justify\">[latex]\\bar{x_m}=\\frac{(m_L\\cdot0.425m)+(m_L\\cdot0.90m)+(m_T\\cdot1.15)+(2\\cdot m_A\\cdot1.90m)}{m}[\/latex]<\/p>\r\n<p style=\"text-align: justify\">[latex]\\bar{x_m} = \\frac{(11.7 kg \\cdot 0.425 m ) + (11.7 kg \\cdot 0.90 m )+(32.5 kg \\cdot 1.15 m )+(2 \\cdot 4.5 kg \\cdot 1.90 m) }{65 kg}[\/latex]<\/p>\r\n<p style=\"text-align: justify\">\u00a0[latex]\\bar{x_m }=1.08 m[\/latex]<\/p>\r\n[latex]\\bar{y_m} = \\frac{(m_L\\cdot0. 9m)+(m_L\\cdot0.425m)+(m_T\\cdot0.9)+(2\\cdot m_A\\cdot0.90m)}{m}[\/latex]\r\n\r\n[latex]\\bar{y_m} = \\frac{(11.7 kg \\cdot 0.90 m ) + (11.7 kg \\cdot 0.425 m )+(32.5 kg \\cdot 0.90 m )+(2 \\cdot4.5 kg \\cdot 0.90 m) }{65 kg}[\/latex]\r\n<p style=\"text-align: justify\">[latex]\\bar {y_m} = 0.81 m[\/latex]<\/p>\r\nin vector form, [latex]\\overrightarrow{r}=\\begin{bmatrix}1.08m\\\\0.81m \\end{bmatrix}[\/latex]\r\n<p style=\"text-align: justify\"><span style=\"background-color: initial;font-size: 1em\">6.\u00a0<\/span><strong style=\"background-color: initial;font-size: 1em\">Review<\/strong><\/p>\r\n<p style=\"text-align: justify\">The answer makes sense because it is close to where the torso is, and it has the most mass. It is a little lower than the torso, which makes sense as this pose would lower the center of gravity.<\/p>\r\n\r\n<\/div>\r\n<h1>Example 7.6.5: All of Ch 7, Submitted by Liam Murdock<\/h1>\r\n<div class=\"textbox\">\r\n<div style=\"text-align: justify\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nYou are the engineer for a tech company, and your role is to design a new extended mount for their new phones. Currently, you are trying to figure out what distance the mount can be from the hand \u201cd\u201d.\r\n\r\nYou can only use the equation for the midpoint of the rod Izz = (1\/12)ml2at your disposal, and the pole weighs 15 N. If a hand is held up the pole at the bottom, it can have a mass moment of inertia of 5 kgm2 before breaking (Iz\u2019z\u2019). Figure out what size the extended mount can be (d).\r\n\r\n[caption id=\"attachment_1732\" align=\"aligncenter\" width=\"447\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun_Teotihuacan.jpg\" alt=\"A person taking a selfie with a selfie stick.\" class=\"wp-image-1732\" width=\"447\" height=\"398\" \/> Source:https:\/\/commons.wikimedia.org\/wiki\/File:Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun,_Teotihuac%C3%A1n.jpg[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7.jpg\" alt=\"A FBD of the problem.\" width=\"492\" height=\"393\" class=\"alignnone wp-image-2200 size-full\" \/>\r\n\r\n<strong>3.<\/strong> <strong>Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>W<sub>p<\/sub> =15 N<\/li>\r\n \t<li>g = 9.81 m\/s<sup>2<\/sup><\/li>\r\n \t<li>Izz = (1\/12)ml<sup>2<\/sup><\/li>\r\n \t<li>Max Iz\u2019z\u2019 =5 kgm<sup>2<\/sup><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>L, length of stick<\/li>\r\n<\/ul>\r\n<strong>4.<\/strong> <strong>Approach\u00a0<\/strong>\r\n\r\nNewton\u2019s Second Law, Parallel Axis Theorem, Understanding of Mass Moment of Inertia\r\n\r\n<strong>5.<\/strong> <strong>Analysis\u00a0<\/strong>\r\n\r\n[latex]m = \\frac{w_p}{g}[\/latex]\r\n\r\n[latex]m = \\frac{15N} {9.81 \\: m\/s^2}[\/latex]\r\n\r\n[latex]m = 1.529\\: kg[\/latex]\r\n\r\n[latex]I_g =\\frac{1}{12}\\:md^2[\/latex]\r\n\r\n[latex]Iz'z' = I_g + ml^2[\/latex]\r\n\r\n[latex]Iz'z' = \\frac{1}{12}\\:md^2+m\\times \\left(\\frac{1}{2}d\\right) ^2[\/latex]\r\n\r\nthus,\r\n\r\n[latex]5kgm^2 = \\left (\\frac{1}{12}\\times (1.529kg)\\times d^2 \\right)+\\left(1.529kg\\: \\times \\frac{1}{4}d^2\\right)[\/latex]\r\n\r\n[latex]d =3.13 m[\/latex]\r\n\r\nIn conclusion, the maximum size the pole can be is 3.13 m without the device breaking.\r\n\r\n<strong>6.<\/strong> <strong>Review\u00a0<\/strong>\r\n\r\nApplying the derived value of d in\r\n\r\n[latex]Iz'z' = \\frac{1}{3}md^2[\/latex]\r\n\r\n[latex]Iz'z' =\\frac{1}{3}\\times (1.529kg)\\times(3.13m)^2[\/latex]\r\n\r\n[latex]Iz'z' = 4.99999kgm^2 =5.0kgm^2[\/latex]\r\n\r\nAccording to the equation, the max distance of 3.13 m does make sense\r\n\r\n<\/div>\r\n<h1>Example 7.6.6: Center of Mass, Submitted by Dhruvil Kanani<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">A thin uniform metal disc with a diameter of 18cm and an initial mass of 50.87kg has a circular hole cut out of it, turning it into a crescent shape, shown in the figure below. The radius of the hole is 6cm. Assuming the mass of the disc is evenly distributed, calculate the mass of the crescent formed and its center of mass.\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2.png\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1931\" width=\"258\" height=\"258\" \/><\/div>\r\n<div><strong>2. Draw<\/strong><\/div>\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1.png\" alt=\"A FBD of the problem. \" class=\"alignnone wp-image-1930\" width=\"360\" height=\"359\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Mass of Initial disc: m<sub>1<\/sub> = 50.87kg<\/li>\r\n \t<li>Radius of larger disc: r<sub style=\"text-align: initial;background-color: initial\">1<\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\"> = 18cm\/2 = 9cm<\/span><\/li>\r\n \t<li>Radius of cut: r<sub style=\"text-align: initial;background-color: initial\">2<\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\"> = 6cm<\/span><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Mass of crescent: m<sub>2<\/sub><\/li>\r\n \t<li>Center of mass of crescent: CM<sub style=\"text-align: initial;background-color: initial\">c<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\nUse the mass and radius of the larger disc to find the density of the material. Calculate the area of the crescent\r\nUse the area and density to find the mass of the crescent.\r\nFind the center of mass of the smaller circle.\r\nUse the center of mass of each circle and the mass of each circle to determine the center of mass of the crescent.\r\n\r\n<strong>5. Analysis<\/strong>\r\nDensity of materials:\r\n[latex]\\rho = \\frac{m_1}{2\\pi \\cdot r_1 ^2} \\\\ \\rho = \\frac{50.87kg}{2\\pi \\cdot 9^2} \\\\ \\rho = 0.1kg\/cm^2[\/latex]\r\n\r\nArea of crescent A<sub>c<\/sub> :\r\n[latex]A_c = 2\\pi \\cdot (r_1 ^2 - r_2 ^2) cm^2 \\\\ A_c = 2\\pi \\cdot (9^2 - 6^2) cm^2 \\\\ A_c = 90\\pi cm^2[\/latex]\r\n\r\nMass of crescent:\r\n[latex]m_2 = \\rho \\cdot A_c \\\\ m_2 = 0.1kg\/cm^2 \\cdot 90\\pi cm^2 \\\\ m_2 = 9\\pi kg \\approx 28.27kg[\/latex]\r\n\r\nCenter of mass of the cut circle:\r\n\r\nSince the edge of the cut circle lines incident to the larger circle, the radius of the cut can be subtracted from the radius of the full disc to find the center of mass of the cut, CM<sub>cut<\/sub>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3.png\" alt=\"A FBD of the diagram to find its Center of mass cut.\" class=\"alignnone wp-image-1933\" width=\"292\" height=\"291\" \/>\r\n\r\n[latex]CM_{cut x} = R_1 - R_2 \\\\ CM_{cut x} = 9cm - 6cm \\\\ CM_{cut x} = 3cm[\/latex]\r\n\r\nCM<sub>cut x<\/sub> is 3cm in the negative x direction.\r\n\r\nCM<sub>cut y<\/sub> is equal to 0 as the center of mass of both the disc and the cut is zero in the y-direction.\r\n\r\nThe center of mass of the full disc is at [0cm, 0cm] as it's centred at the origin.\r\n\r\nCenter of mass of the crescent:\r\n\r\nThe center of mass of the full circle (CM<sub>full<\/sub>) is [0cm, 0cm] and its mass m<sub>1<\/sub> = 50.87kg.\r\nThe center of mass of the cut (CM<sub>cut<\/sub>) is [-3cm, 0cm], and its mass (m<sub>cut<\/sub>) is the mass of the crescent minus the mass of the full circle:\u00a0 28.27kg - 50.87kg = -22.6kg.\r\n\r\nThe center of mass of the crescent CM<sub>c<\/sub> is:\r\n\r\n[latex]CM_{cx}\u00a0 = \\frac{(CM_{full x} \\cdot m_1) + (CM_{cut x} \\cdot m_{cut})}{m_2}[\/latex]\r\n\r\n[latex]CM_{cx}\u00a0 = \\frac{(0cm \\cdot 50.87kg) + (-3cm \\cdot -22.6kg)}{28.27kg}[\/latex]\r\n\r\n[latex]CM_{cx} = 2.4cm[\/latex]\r\n\r\nBoth the center of mass of the full circle and the cut in the y-direction are 0cm, giving the center of mass of the crescent in the y-direction also 0cm.\r\n\r\nThe complete center of mass of the crescent is therefore:\r\n<strong>CM<sub>c<\/sub> = [2.4cm, 0cm]<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4.png\" alt=\"Diagram showing the complete center of mass.\" class=\"alignleft wp-image-1941\" width=\"373\" height=\"371\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<strong>6. Review:<\/strong>\r\n\r\nThe center of mass of the crescent leans to the side with more mass, as one would expect. The mass and center of mass found are both in the expected units with a reasonable magnitude.\r\n\r\n<\/div>\r\n<h1>Example 7.6.7: Mass moment of inertia, Submitted by <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Michael Oppong-Ampomah&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:156,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:1}\">Michael Oppong-Ampomah<\/span><\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nAn empty tissue paper roll rolls down an angled floor. If the tissue paper roll weighs 2 lbs, has and inner radius 0.9 inches, total diameter 2 inches, and a length of 12 inches, then calculate the mass moment of inertia of the roll.\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\n<strong style=\"font-size: 1em\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908.png\" alt=\"A FBD of the problem.\" width=\"380\" height=\"318\" class=\"alignnone wp-image-2049 size-full\" \/><\/strong>\r\n\r\n&nbsp;\r\n\r\n<strong>3. K<\/strong><strong style=\"text-align: initial;font-size: 1em\">nowns and Unknown<\/strong><strong style=\"text-align: initial;font-size: 1em\">s\u00a0<\/strong>\r\n\r\nKnowns:\r\n\r\nw = 2 lbs\r\n\r\nInner radius r<sub>1<\/sub> = 0.9 in\r\n\r\nOuter radius r<sub>2<\/sub> = 1 in\r\n\r\nh=12 in\r\n\r\n&nbsp;\r\n\r\nUnknowns:\r\n\r\nMass moment of inertia\r\n\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nFinding the mass of the roll, identifying the axis of rotation, and applying the corresponding mass moment of inertia of the cylinder.\r\n\r\n[latex]I_{x}= \\frac{1}{2} m(r_{1}^2 + r_{2}^2)[\/latex]\r\n\r\n[latex]I_{y}= I_{z}= \\frac{1}{12} m[3(r_{1}^2 + r_{2}^2)+h^2][\/latex]\r\n\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\nweight of the roll = 2 lb\r\n\r\nmass of\u00a0 the roll, [latex]m = \\frac{2 lb}{32.2 ft\/s^2} = 0.062 lb [\/latex]\r\n\r\nSince we used constant of gravity with a unit of [latex]ft\/ s^2[\/latex] it is important to convert other values from inches to ft.\r\n\r\n[latex]0.9 in = \\frac{0.9}{12}=0.075 ft[\/latex]\r\n\r\n[latex]1 in =\\frac{1}{12} = 0.083 ft [\/latex]\r\n\r\n[latex]12 in = 1 ft[\/latex]\r\n\r\nnow, [latex]I_{x}= \\frac{1}{2} m(r_{1}^2 + r_{2}^2)[\/latex]\r\n\r\n[latex]I_{x} = \\frac{1}{2} 0.062(0.075^2 + 0.083^2) = 0.00039 \\quad lb ft^2[\/latex]\r\n\r\n[latex]I_{y}= I_{z}= \\frac{1}{12} m[3(r_{1}^2 + r_{2}^2)+h^2][\/latex]\r\n\r\n[latex]I_{y}= I_{z}= \\frac{1}{12} 0.062[3(0.075^2 + 0.083^2)+1^2] =0.0054 \\quad lb ft^2[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nAlthough the weight of an empty paper roll is exaggerated for easier calculation, the answers make sense. The value of I<sub>y<\/sub> is higher than I<sub>x<\/sub><sub>,<\/sub> which means more force will be needed for rotation in that axis. From this, it is important to note that the roll will rotate along the x-axis. Also note that weight is converted to mass, and units must be the same.\r\n\r\n<\/div>\r\n<\/div>","rendered":"<p>Here are examples from Chapter 7 to help you understand these concepts better. These were taken from the real world and supplied by FSDE students in Summer 2021. If you&#8217;d like to submit your own examples, please send them to the author <a href=\"mailto:eosgood@upei.ca\">eosgood@upei.ca<\/a>.<\/p>\n<p>&nbsp;<\/p>\n<h1>Example 7.6.1: All of Ch 7, Submitted by William Craine<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A person is playing soccer. The ball they are using has a diameter of 20 cm and a mass of 0.45 kg. The person\u2019s leg has a mass of 18 kg, and their foot has a mass of 8 kg. Assume that all the shapes have uniform density.<\/p>\n<p>a) Find the cm for the ball.<\/p>\n<p>b) Calculate the mass moment of inertia (MMOI) for the ball.<\/p>\n<p>c) Find the cm for the person\u2019s leg and foot.<\/p>\n<p>d) Find the MMOI for the person\u2019s leg and foot on the y-axis about A.<\/p>\n<figure id=\"attachment_1514\" aria-describedby=\"caption-attachment-1514\" style=\"width: 309px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/images.jpg\" alt=\"A person kicking a ball.\" class=\"wp-image-1514\" width=\"309\" height=\"309\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/images.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/images-150x150.jpg 150w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/images-65x65.jpg 65w\" sizes=\"auto, (max-width: 309px) 100vw, 309px\" \/><figcaption id=\"caption-attachment-1514\" class=\"wp-caption-text\">Source:<a href=\"https:\/\/www.pngwing.com\/en\/free-png-aaujw\"> https:\/\/encrypted-tbn0.gstatic.com\/images?q=tbn:ANd9GcTM4e4xHaRSXBdQMGugm1gISi2Qgn7rQx_K3w&amp;usqp=CAU<\/a><\/figcaption><\/figure>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-1024x779.png\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1515\" width=\"417\" height=\"317\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-1024x779.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-300x228.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-768x584.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-65x49.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-225x171.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM-350x266.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-8.46.02-PM.png 1314w\" sizes=\"auto, (max-width: 417px) 100vw, 417px\" \/><\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-300x236.jpg\" alt=\"A FBD of the problem.\" class=\"alignleft wp-image-885 size-medium\" width=\"300\" height=\"236\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-300x236.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-1024x805.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-768x604.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-1536x1207.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-65x51.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-225x177.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1-350x275.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-1.jpg 1931w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-300x201.jpg\" alt=\"A circular sketch of the ball.\" class=\"aligncenter wp-image-886\" width=\"248\" height=\"166\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-300x201.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-1024x686.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-768x514.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-65x44.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-225x151.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2-350x234.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-draw-2.jpg 1147w\" sizes=\"auto, (max-width: 248px) 100vw, 248px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>m<sub>b<\/sub> = 0.45 kg<\/li>\n<li>d<sub>b<\/sub> = 20 cm<\/li>\n<li>m<sub>L<\/sub> = 18 kg<\/li>\n<li>m<sub>f<\/sub> = 8 kg<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>x<sub>b<\/sub><\/li>\n<li>I<sub>b<\/sub><\/li>\n<li>x<sub>p<\/sub><\/li>\n<li>I<sub>p x<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Find the cm for both objects using arbitrary coordinates since no origin is given.<\/p>\n<p>Use the sphere MMOI formula for the ball.<\/p>\n<p>Calculate the individual MMOIs for the leg and foot, then use the parallel axis theorem to get each shape&#8217;s MMOI about the system cm, add them, and then use the parallel axis theorem to get the MMOI about A.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a (find the center of mass of the ball):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-300x264.jpg\" alt=\"Diagram to find the center of mass for the ball.\" class=\"aligncenter wp-image-887 size-medium\" width=\"300\" height=\"264\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-300x264.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-768x676.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-65x57.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-225x198.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1-350x308.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Will-1-solve-1.jpg 818w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]x_b=(10\\underline{\\hat{i}}+10\\underline{\\hat{j}}+10\\underline{\\hat{k}})cm[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Part b (find the MMOI of the ball about its center of mass):<\/p>\n<p>[latex]I_{xx}=I_{yy}=I_{zz}=\\frac{2}{3}mr^2\\\\\\qquad \\quad=\\frac{2}{3}(0.45 kg)(0.1m)^2\\\\I_b=0.003kgm^2[\/latex]<\/p>\n<p>Part c (find the center of mass for the system of the person&#8217;s leg):<\/p>\n<p><em>Step 1: Find the center of mass of the foot (f)<\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-300x166.jpg\" alt=\"Diagram to find the centroid of the persons foot (represented as a cuboid).\" class=\"aligncenter wp-image-890 size-medium\" width=\"300\" height=\"166\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-300x166.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-768x425.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-65x36.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-225x124.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4-350x194.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-4.jpg 908w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]f=\\frac{15cm}{2}\\underline{\\hat{i}}+\\frac{7cm}{2}\\underline{\\hat{j}}+\\frac{7cm}{2}\\underline{\\hat{k}}\\\\f=(7.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+3.5\\underline{\\hat{k}})cm[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><em>Step 2: Find the center of mass of the leg (L)<\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6-232x300.jpg\" alt=\"Diagram to find the centroid of the persons leg (represented as a cylinder).\" class=\"aligncenter wp-image-892\" width=\"189\" height=\"244\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6-232x300.jpg 232w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6-65x84.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6-225x291.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6-350x453.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-6.jpg 612w\" sizes=\"auto, (max-width: 189px) 100vw, 189px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-300x228.jpg\" alt=\"Diagram to find the centroid of the persons foot and leg combined.\" class=\"aligncenter wp-image-906\" width=\"407\" height=\"309\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-300x228.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-1024x777.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-768x582.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-1536x1165.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-2048x1553.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-65x49.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-225x171.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-11-1-350x265.jpg 350w\" sizes=\"auto, (max-width: 407px) 100vw, 407px\" \/><\/p>\n<p>[latex]L_1=(3.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+15\\underline{\\hat{k}})cm\\\\L_2=(15cm-7cm-1cm)\\underline{\\hat{i}}+(0cm)\\underline{\\hat{j}}+(7cm)\\underline{\\hat{k}}\\\\\\qquad \\quad L_2 = (7 \\underline{\\hat{i}}+(0)\\underline{\\hat{j}}+(7)\\underline{\\hat{k}} )cm[\/latex]<\/p>\n<p>[latex]\\\\L=L_1+L_2\\\\L=(3.5cm+7cm)\\underline{\\hat{i}}+(3.5cm+0cm)\\underline{\\hat{j}}+(15cm+7cm)\\underline{\\hat{k}}\\\\L=(10.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+22\\underline{\\hat{k}})cm[\/latex]<\/p>\n<p><em>Step 3: find the center of mass of the system (P)<\/em><\/p>\n<p>[latex]x_p=\\frac{\\sum m_i x_i}{\\sum m_i}\\\\X_p=\\frac{m_f\\cdot x_f+m_L\\cdot x_L}{m_f+m_L}\\\\x_p=\\frac{8kg\\cdot 7.5cm+18kg\\cdot 10.5cm}{8kg+18kg}\\\\x_p=9.58cm[\/latex]<\/p>\n<p>[latex]y_p=\\frac{\\sum m_i y_i}{\\sum m_i}\\\\y_p=\\frac{8kg\\cdot 3.5cm+18kg\\cdot 3.5cm}{8kg+18kg}\\\\y_p=3.5cm[\/latex]<\/p>\n<p>[latex]z_p=\\frac{\\sum m_i z_i}{\\sum m_i}\\\\z_p=\\frac{8kg\\cdot 3.5cm+18kg\\cdot 22cm}{8kg+18kg}\\\\z_p=16.3cm[\/latex]<\/p>\n<p>[latex]\\underline{\\underline{P}=(9.58\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+16.3\\underline{\\hat{k}})cm}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-300x246.jpg\" alt=\"Diagram to find the centroid of the persons foot and leg combined.\" class=\"aligncenter wp-image-898\" width=\"413\" height=\"339\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-300x246.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-768x630.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-65x53.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-225x185.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12-350x287.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/will-1-solve-12.jpg 913w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><\/p>\n<p>Part d (find the inertia of the person&#8217;s leg about point A):<\/p>\n<p><em>Step 1: Find the MMOI of the foot about f (I<sub>ff<\/sub>)<\/em><\/p>\n<p>[latex]x_f=0.15m\\\\z_f=0.07m\\\\m_f=8kg[\/latex]<\/p>\n<p>[latex]I_{ff}=\\frac{1}{12}\\cdot m\\cdot (x^2+z^2)\\\\I_{ff}=\\frac{1}{12}(8kg)(0.15m^2+0.07m^2)\\\\I_{ff}=0.0183kg\\; m^2[\/latex]<\/p>\n<p><em>Step 2: Find the MMOI of the foot about P (I<sub>Pf<\/sub>)<\/em><\/p>\n<p>[latex]d^2=(0.0208m)^2+(0.128m)^2\\\\I_{pf}=I_{ff}+m_f(d_f)^2\\\\I_{pf}=0.0182kgm^2+8kg[(0.0208m)^2+(0.128m)^2]\\\\I_{pf}=0.1527kgm^2[\/latex]<\/p>\n<p><em>Step 3: Find the MMOI of the leg about L (I<sub>LL<\/sub>)<\/em><\/p>\n<p>[latex]r=0.035m\\\\h=0.3m\\\\m_L=18kg\\\\I_{LL}=\\frac{1}{12}\\cdot m\\cdot (3_r^2+h^2)\\\\I_{LL}=\\frac{1}{12}\\cdot 18kg(3(0.035m)^2+(0.3m)^2)\\\\I_{LL}=0.1405kgm^2[\/latex]<\/p>\n<p><em>Step 4: Find the MMOI of the leg about P (I<sub>PL<\/sub>)<\/em><\/p>\n<p>[latex]r_{LP}=[(9.58-10.5)\\underline{\\hat{i}}+(3.5-3.5)\\underline{\\hat{j}}+(16.3-22)\\underline{\\hat{k}}]cm\\\\r_{LP}=(-0.92\\underline{\\hat{i}}-5.7\\underline{\\hat{k}})cm\\\\I_{PL}=I_{LL}+m(d^2)\\\\I_{PL}=0.1405kgm^2+18kg[(0.0092m)^2+(0.057m)^2]\\\\I_{PL}=0.2005kgm^2[\/latex]<\/p>\n<p><em>Step 5: Find the MMOI of the entire system about P (I<sub>P<\/sub>)<\/em><\/p>\n<p>[latex]I_p=I_{PL}+I_{pf}\\\\I_p=0.1527kgm^2+0.2005kg\\\\I_p=0.3532kgm^2[\/latex]<\/p>\n<p><em>Step 6: Find the MMOI of the entire system about A (I<sub>A<\/sub>)<\/em><\/p>\n<p>[latex]A=(10.5\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+37\\underline{\\hat{k}})cm\\\\P=(9.58\\underline{\\hat{i}}+3.5\\underline{\\hat{j}}+16.3\\underline{\\hat{k}})cm\\\\r_{AP}=P-A\\\\r_{AP}=[(9.58-10.5)\\underline{\\hat{i}}+(3.5-3.5)\\underline{\\hat{j}}+(16.3-37)\\underline{\\hat{k}}]cm\\\\r_{AP}=(-0.92\\underline{\\hat{i}}-20.7\\underline{\\hat{k}})cm[\/latex]<\/p>\n<p>[latex]I_A=I_p+m(d^2)\\\\I_A=0.3532kgm^2+(8kg+18kg)[(0.0092m)^2+(0.207m)^2]\\\\I_A=1.4695kgm^2\\\\\\underline{I_A=1.47kgm^2}[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that the numbers are small, since before the final step, the mass was small, or the distance to the new axis was small.<\/p>\n<\/div>\n<h1>Example 7.6.2 Inertia, Submitted by Luke McCarvill<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A figure skater with a mass of 60 kg is about to perform a spin about her long axis (z). She is 167 cm tall, and her body can be approximated as a circular cylinder of 30 cm diameter while her limbs are at her side, and a circular cylinder of 60 cm diameter while her arms (and one leg) are outstretched.<\/p>\n<p>a) What should she do in order to generate the highest angular acceleration, assuming she can generate a net torque of 200 Nm? Does lowering her height increase or decrease her angular acceleration?<\/p>\n<p>b) How fast will she be spinning after 0.5 seconds of her maximum vs minimum accelerations, assuming she starts from zero (\u2375<sub>0<\/sub> = 0 rad\/sec)?<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1516\" aria-describedby=\"caption-attachment-1516\" style=\"width: 1024px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-1024x817.png\" alt=\"A person ice skating.\" class=\"wp-image-1516 size-large\" width=\"1024\" height=\"817\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-1024x817.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-300x239.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-768x612.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-65x52.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-225x179.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM-350x279.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screen-Shot-2021-08-29-at-9.20.46-PM.png 1244w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-1516\" class=\"wp-caption-text\">Annotated from original source: https:\/\/commons.wikimedia.org\/wiki\/File:2019_Internationaux_de_France_Friday_ladies_SP_group_1_Starr_ANDREWS_8D9A6706.jpg<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-300x255.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1145 size-medium\" width=\"300\" height=\"255\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-300x255.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-1024x869.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-768x652.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-65x55.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-225x191.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1-350x297.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-1.jpg 1272w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/> <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-300x254.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1146 size-medium\" width=\"300\" height=\"254\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-300x254.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-1024x867.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-768x650.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-65x55.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-225x191.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2-350x296.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-2-2.jpg 1183w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>m = 60 kg<\/li>\n<li>\u03a3M = 200Nm<\/li>\n<li>h = 167cm<\/li>\n<li>d<sub>1<\/sub> = 30cm<\/li>\n<li>d<sub>2<\/sub> = 60cm<\/li>\n<li>t = 0.5sec<\/li>\n<li>I<sub>zz<\/sub> = \u00bd mr2<\/li>\n<li>\u03a3M = I\u221d<\/li>\n<li>\u2375<sub>0<\/sub> = 0rad\/sec<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>\u221d<sub>1<\/sub> = ?<\/li>\n<li>\u221d<sub>2<\/sub> = ?<\/li>\n<li>I<sub>zz1<\/sub> = ?<\/li>\n<li>I<sub>zz2<\/sub> = ?<\/li>\n<li>\u2375<sub>1<\/sub> = ?<\/li>\n<li>\u2375<sub>2<\/sub> = ?<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>I\u2019ll be using MMOI for circular cylinders, as well as the sum of moments\/torque equaling MMOI times angular acceleration, as well as acceleration equaling change in velocity over time.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p><em>Step 1: Find the inertia when the arms are hugged to the body<\/em><\/p>\n<p>I<sub>zz1<\/sub> = 0.5 (60kg) (0.15m)<sup>2 <\/sup><\/p>\n<p>I<sub>zz1<\/sub>= 0.675 kg m<sup>2<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><em>Step 2: Find the inertia when the arms are spread out<\/em><\/p>\n<p>I<sub>zz2<\/sub> = 0.5(60kg)(0.3m)<sup>2<\/sup><\/p>\n<p>I<sub>zz2<\/sub> = 2.7 kg m<sup>2<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><em>Step 3: Find the angular acceleration for both cases<\/em><\/p>\n<p>\u03a3M = I\u221d ; therefore, \u03a3M\/I = \u221d<\/p>\n<p>&nbsp;<\/p>\n<p>\u221d<sub>1<\/sub> = \u03a3M \/ I<sub>zz1<\/sub><\/p>\n<p>\u221d<sub>1<\/sub> = 200 Nm \/ 0.675 kg m<sup>2<\/sup><\/p>\n<p>\u221d<sub>1<\/sub> \u2248 296.296 rad\/sec<sup>2<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p>\u221d<sub>2<\/sub> = \u03a3M \/ I<sub>zz2<\/sub><\/p>\n<p>\u221d<sub>2<\/sub> = 200Nm\/2.7 kg m<sup>2<\/sup><\/p>\n<p>\u221d<sub>2<\/sub> \u2248 74.074 rad\/sec<sup>2 <\/sup><\/p>\n<p>&nbsp;<\/p>\n<p>The acceleration when the skater had her arms close to her body was about 296 rad\/sec<sup>2<\/sup>, while that when she had her arms spread out was about 74 rad\/sec<sup>2<\/sup>. Therefore, having her limbs closer to her body will give her a much higher angular acceleration!<\/p>\n<p>As seen in the equation I<sub>zz<\/sub> = \u00bd m r<sup>2<\/sup>, her height is arbitrary; thus, lowering her height would not change the inertia, nor would it change her angular acceleration.<\/p>\n<p>&nbsp;<\/p>\n<p>Part b:<\/p>\n<p>\u221d = \u2375\/t therefore \u221d*t = \u2375<\/p>\n<p>&nbsp;<\/p>\n<p>\u2375<sub>1<\/sub> = 296.296 s<sup>-2<\/sup>(0.5s)<\/p>\n<p>\u2375<sub>1<\/sub> = 148.148 rad\/sec<\/p>\n<p>&nbsp;<\/p>\n<p>\u2375<sub>2<\/sub> = 74.074 s<sup>-2<\/sup>(0.5s)<\/p>\n<p>\u2375<sub>2<\/sub> = 37.037 rad\/sec<\/p>\n<p>&nbsp;<\/p>\n<p>Given that these are in radians per second, let\u2019s convert this to rotations per second to make it more meaningful. To do so, simply divide by 2\u03c0 since there are 2\u03c0 radians per rotation. Thus, with her arms in, she can achieve about 23.6 rotations per second after 0.5 seconds of acceleration, compared to about 5.9 rotations per second with her limbs out. This is on par with an Olympian, according to <a href=\"http:\/\/www.bsharp.org\/physics\/spins\">this site<\/a>.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that they spin faster when their limbs are hugged to their body; we can try this at home with a swivel chair!<\/p>\n<\/div>\n<h1>Example 7.6.3: All of Ch 7, Submitted by Victoria Keefe<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>Two rectangular compost containers are tied end to end with a 2 m chain of negligible mass. Container A has a mass of 5 kg with a height of 1 m and a width of 2 m. The container B is of 8kg, height 2m, and width 3m. Consider the point which is the center of the chain as the origin of the coordinate frame to find the center of mass, center of gravity,\u00a0 and centroid of the system and the individual containers.<\/p>\n<figure style=\"width: 372px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefes-metal-containers.jpg\" width=\"372\" height=\"279\" alt=\"Compost containers\" class=\"\" \/><figcaption class=\"wp-caption-text\">Source:https:\/\/felixwong.com\/2021\/10\/differences-between-homes-in-the-u-s-and-spain\/<\/figcaption><\/figure>\n<p><span style=\"text-align: initial;background-color: initial;font-size: 1em\">The real-life scenario of the problem would look similar to the image.<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5.jpg\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1814 size-full\" width=\"610\" height=\"242\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5.jpg 610w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5-300x119.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5-65x26.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5-225x89.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/victoria-keefe-compost-bin-5-350x139.jpg 350w\" sizes=\"auto, (max-width: 610px) 100vw, 610px\" \/><\/p>\n<\/div>\n<p><strong style=\"text-align: initial;background-color: initial;font-size: 1em\">2. Draw<\/strong><\/p>\n<div><strong style=\"font-size: 1em;background-color: initial\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1807 size-full\" width=\"721\" height=\"314\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14.png 721w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14-300x131.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14-65x28.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14-225x98.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-14-350x152.png 350w\" sizes=\"auto, (max-width: 721px) 100vw, 721px\" \/><\/strong><\/div>\n<div><strong>3. Knowns and Unknowns<\/strong><\/div>\n<div><\/div>\n<div>Knowns:<\/div>\n<ul>\n<li>m<sub>A<\/sub> = 5 kg<\/li>\n<li>m<sub>B = <\/sub>8 kg<\/li>\n<li>h<sub>A\u00a0<\/sub>= 1 m<\/li>\n<li>h<sub>B<\/sub> = 2 m<\/li>\n<li>w<sub>A =\u00a0<\/sub>2m<\/li>\n<li>w<sub>B <\/sub>= 3 m<\/li>\n<li>d =\u00a0 2 m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>[latex]\\bar{x_A}[\/latex]<\/li>\n<li>[latex]\\bar{x_B}[\/latex]<\/li>\n<li>[latex]\\bar{y_a}[\/latex]<\/li>\n<li>[latex]\\bar{y_b}[\/latex]<\/li>\n<li>[latex]\\bar{x_s}[\/latex]<\/li>\n<li>[latex]\\bar{y_s}[\/latex]<\/li>\n<li>[latex]\\bar{r_s}[\/latex]<\/li>\n<\/ul>\n<div><\/div>\n<div><strong>4. Approach\u00a0<\/strong><\/div>\n<div>Use equations for the center of mass, the center of gravity, and the centroid. utilizing the advantage of symmetry whenever possible.<\/div>\n<div><\/div>\n<div><strong>5. Analysis\u00a0<\/strong><\/div>\n<div><\/div>\n<div><span style=\"text-decoration: underline\">FINDING CENTER OF MASS\u00a0<\/span><\/div>\n<div>Calculate the centre of mass of box A. Because it is symmetrical, you divide the height and width by 2 and add the distance from O.<\/div>\n<div>[latex]\\bar{x_A}=\\frac{w_A}{2}+1 = 2 m[\/latex]<\/div>\n<div>[latex]\\bar{y_A}=\\frac{h_A}{2} =\u00a0 0.5 m[\/latex]<\/div>\n<div><\/div>\n<div>Thus, in vector form center of mass of container A is [latex]{cm_A}=\\begin{bmatrix} 2 m\\\\ 0.5 m\\end{bmatrix}[\/latex]<\/div>\n<div><span style=\"background-color: initial;font-size: 1em\">Calculate the centre of mass of box B. Because it is symmetrical, you divide the height and width by 2, and add the distance from O.<\/span><\/div>\n<div>\n<div>[latex]\\bar{x_B}=\\frac{-w_B}{2}-1 \\\\=\\frac{-3}{2}-1 =-2.5 m[\/latex]<\/div>\n<div>[latex]\\bar{y_B}=\\frac{h_B}{2} = 1 m[\/latex]<\/div>\n<div>In vector form, [latex]{cm_B}=\\begin{bmatrix} -2.5 m\\\\ 1 m\\end{bmatrix}[\/latex]<\/div>\n<\/div>\n<div>Calculate the system centre of mass in x:<\/div>\n<div>[latex]cm_{sx} = \\frac {{m_A} \\cdot \\bar{x}_{ax} +m_B\\cdot \\bar{x}_{bx}} {m_A+m_B}\\\\ = \\frac{5kg \\cdot2m +8kg\\cdot(-2.5)m }{5 kg +8kg}\\\\ = -0.769 m[\/latex]<br \/>\nCalculate the System centre of mass in y:<br \/>\n[latex]cm_{sy} = \\frac {{m_A} \\cdot \\bar{x}_{ay} +m_B\\cdot \\bar{x}_{by}} {m_A+m_B}\\\\ = \\frac{5kg \\cdot 0.5m +8kg\\cdot 1m }{5 kg +8kg}\\\\ = 0.808m[\/latex]<\/div>\n<div><\/div>\n<div>In vector format center of mass of the system:<\/div>\n<div>[latex]{cm_S}=\\begin{bmatrix}-0.769 m \\\\0.808 m \\end{bmatrix}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><span style=\"text-decoration: underline\">FINDING CENTER OF GRAVITY<\/span><\/div>\n<div>The center of gravity of container A and container B is the same as the center of mass due to the uniform rectangular shape and orientation.<\/div>\n<div>Thus<\/div>\n<div>[latex]cg_{A} =\\begin{bmatrix}2 m \\\\ 0.5 m\\end{bmatrix}[\/latex]<\/div>\n<div>similarly, [latex]cg_{B} =\\begin{bmatrix}-2.5 m\u00a0 \\\\ 1 m\\end{bmatrix}[\/latex]<\/div>\n<div><\/div>\n<div>Since the system has a uniform gravitational field and constant density, the system&#8217;s centre of gravity is equal to that of the centre of mass found above.<\/div>\n<div>Thus<\/div>\n<div style=\"text-align: center\">[latex]{cg_S}=\\begin{bmatrix}-0.769\\\\0.808 m \\end{bmatrix}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><span style=\"text-decoration: underline\">FINDING CENTROID<\/span><\/div>\n<div>Similar to the center of gravity, the centroid of individual containers is also the same as the center of mass of the individual containers due to the uniform shape of the containers.<\/div>\n<div>In this specific problem, where the containers are of simple uniform rectangular shape, we can consider the symmetry and the dimensions of the system to find the centroid of the system.<\/div>\n<div><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16.png\" alt=\"Diagram shown to find the centroid.\" class=\"alignnone wp-image-1809 size-full\" width=\"688\" height=\"358\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16.png 688w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16-300x156.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16-65x34.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16-225x117.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Screenshot-16-350x182.png 350w\" sizes=\"auto, (max-width: 688px) 100vw, 688px\" \/><\/div>\n<div>Since the system has a rectangular shape, the centroid would be the geometrical center. The vector [latex]\\overrightarrow{r_S}[\/latex](blue vector line in the above diagram) can be found from the following equation derived from the above image.<\/div>\n<div>[latex]\\overrightarrow{r_{Sx}} = \\frac{t_s}{2}- w_B\\\\= \\frac{7m}{2}-3 m\\\\= 0.5 m[\/latex]; here t<sub>S<\/sub> is the total width of the system.<\/div>\n<div>[latex]\\overrightarrow{r_{Sy}} =\\frac{h_B}{2}\\\\ = 1m[\/latex]<\/div>\n<div>In vector form, [latex]\\overrightarrow{r_S}=\\begin{bmatrix}0.5m\\\\1 m \\end{bmatrix}[\/latex]<\/div>\n<div><\/div>\n<div><strong>6. Review<\/strong><\/div>\n<div>The centre of gravity being the same as the centre of mass is due to only looking at this problem in 2D. The requirements for the centre of mass and the centre of gravity being equal have been fulfilled.<\/div>\n<\/div>\n<div>\n<h1>Example 7.6.4: All of Ch 7, Submitted by Deanna Malone<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p style=\"text-align: left\">A woman doing yoga weighs 65 kg, and her legs are 85 cm long. The combined mass of both her legs is 11.7 kg, one of which is lifted 90 cm above the ground (as in the image below). Her torso contributes to 50% of her mass and is 60 cm long. Her arms are 4.5 kg each, and they are 90 cm from the fingertips to the shoulder. What is her center of gravity in the warrior 3 pose from the origin?<\/p>\n<figure id=\"attachment_2197\" aria-describedby=\"caption-attachment-2197\" style=\"width: 591px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back.jpg\" alt=\"A person in the warrior 3 pose.\" width=\"591\" height=\"331\" class=\"wp-image-2197\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back.jpg 1200w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back-300x168.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back-1024x573.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back-768x430.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back-65x36.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back-225x126.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Virabhadrasana_III_from_back-350x196.jpg 350w\" sizes=\"auto, (max-width: 591px) 100vw, 591px\" \/><figcaption id=\"caption-attachment-2197\" class=\"wp-caption-text\">Source:https:\/\/commons.wikimedia.org\/wiki\/File:Virabhadrasana_III_from_back.jpg<\/figcaption><\/figure>\n<\/div>\n<p style=\"text-align: justify\"><strong>2. Draw<\/strong><\/p>\n<figure id=\"attachment_2174\" aria-describedby=\"caption-attachment-2174\" style=\"width: 459px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554-300x173.jpg\" alt=\"Warrior pose 3 silhouette, and a FBD of the problem.\" width=\"459\" height=\"265\" class=\"wp-image-2174\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554-300x173.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554-65x38.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554-225x130.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554-350x202.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/fixed-deanna-malone-warrior-3-e1652283171554.jpg 660w\" sizes=\"auto, (max-width: 459px) 100vw, 459px\" \/><figcaption id=\"caption-attachment-2174\" class=\"wp-caption-text\">Source:https:\/\/www.vhv.rs\/viewpic\/ihxioJo_warrior-3-yoga-pose-silhouette-hd-png-download\/)<\/figcaption><\/figure>\n<p style=\"text-align: justify\">3. <strong>Knowns and Unknowns<\/strong><\/p>\n<p style=\"text-align: justify\">Knowns:<\/p>\n<ul style=\"text-align: justify\">\n<li>m= 65 kg<\/li>\n<li>m<sub>L<\/sub> = 11.7 kg<\/li>\n<li>m<sub>A<\/sub> = 4.5 kg<\/li>\n<li>m<sub>T<\/sub> = 50% of 65 kg<\/li>\n<\/ul>\n<p style=\"text-align: justify\">Unknowns:<\/p>\n<ul style=\"text-align: justify\">\n<li>[latex]\\overrightarrow{r}[\/latex], vector from origin to center of gravity<\/li>\n<\/ul>\n<p style=\"text-align: justify\"><span style=\"text-decoration: underline\">Assumptions<\/span><\/p>\n<p style=\"text-align: justify\">g= 9.81 m\/s<sup>2<\/sup><\/p>\n<p style=\"text-align: justify\">4.\u00a0 <strong>Approach<\/strong><\/p>\n<p style=\"text-align: justify\">Calculate mass and distance from origin, keeping in mind that the mass distribution is different for legs, torso, and arms, but assuming that the mass is evenly distributed within each of these parts. Use the center of gravity equation.<\/p>\n<p style=\"text-align: justify\">5.<strong> Analysis<\/strong><\/p>\n<p style=\"text-align: justify\">Point of mass for leg in air \u2013 11.7 kg at 90 cm up and 42.5 cm left from the origin.<\/p>\n<p style=\"text-align: justify\">Point mass for leg on ground \u2013 11.7 kg at 42.5 cm up and 90 cm left.<\/p>\n<p style=\"text-align: justify\">Point mass for torso \u2013 32.5 kg at 90 cm up and 115 cm left.<\/p>\n<p style=\"text-align: justify\">Point mass for both arms \u2013 9 kg at 90 cm up and 190 cm left.<\/p>\n<p style=\"text-align: justify\">[latex]\\bar{x_m}=\\frac{(m_L\\cdot0.425m)+(m_L\\cdot0.90m)+(m_T\\cdot1.15)+(2\\cdot m_A\\cdot1.90m)}{m}[\/latex]<\/p>\n<p style=\"text-align: justify\">[latex]\\bar{x_m} = \\frac{(11.7 kg \\cdot 0.425 m ) + (11.7 kg \\cdot 0.90 m )+(32.5 kg \\cdot 1.15 m )+(2 \\cdot 4.5 kg \\cdot 1.90 m) }{65 kg}[\/latex]<\/p>\n<p style=\"text-align: justify\">\u00a0[latex]\\bar{x_m }=1.08 m[\/latex]<\/p>\n<p>[latex]\\bar{y_m} = \\frac{(m_L\\cdot0. 9m)+(m_L\\cdot0.425m)+(m_T\\cdot0.9)+(2\\cdot m_A\\cdot0.90m)}{m}[\/latex]<\/p>\n<p>[latex]\\bar{y_m} = \\frac{(11.7 kg \\cdot 0.90 m ) + (11.7 kg \\cdot 0.425 m )+(32.5 kg \\cdot 0.90 m )+(2 \\cdot4.5 kg \\cdot 0.90 m) }{65 kg}[\/latex]<\/p>\n<p style=\"text-align: justify\">[latex]\\bar {y_m} = 0.81 m[\/latex]<\/p>\n<p>in vector form, [latex]\\overrightarrow{r}=\\begin{bmatrix}1.08m\\\\0.81m \\end{bmatrix}[\/latex]<\/p>\n<p style=\"text-align: justify\"><span style=\"background-color: initial;font-size: 1em\">6.\u00a0<\/span><strong style=\"background-color: initial;font-size: 1em\">Review<\/strong><\/p>\n<p style=\"text-align: justify\">The answer makes sense because it is close to where the torso is, and it has the most mass. It is a little lower than the torso, which makes sense as this pose would lower the center of gravity.<\/p>\n<\/div>\n<h1>Example 7.6.5: All of Ch 7, Submitted by Liam Murdock<\/h1>\n<div class=\"textbox\">\n<div style=\"text-align: justify\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>You are the engineer for a tech company, and your role is to design a new extended mount for their new phones. Currently, you are trying to figure out what distance the mount can be from the hand \u201cd\u201d.<\/p>\n<p>You can only use the equation for the midpoint of the rod Izz = (1\/12)ml2at your disposal, and the pole weighs 15 N. If a hand is held up the pole at the bottom, it can have a mass moment of inertia of 5 kgm2 before breaking (Iz\u2019z\u2019). Figure out what size the extended mount can be (d).<\/p>\n<figure id=\"attachment_1732\" aria-describedby=\"caption-attachment-1732\" style=\"width: 447px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun_Teotihuacan.jpg\" alt=\"A person taking a selfie with a selfie stick.\" class=\"wp-image-1732\" width=\"447\" height=\"398\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun_Teotihuacan.jpg 281w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun_Teotihuacan-65x58.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun_Teotihuacan-225x200.jpg 225w\" sizes=\"auto, (max-width: 447px) 100vw, 447px\" \/><figcaption id=\"caption-attachment-1732\" class=\"wp-caption-text\">Source:https:\/\/commons.wikimedia.org\/wiki\/File:Tourist_taking_selfie_with_stick_atop_Pyramid_of_the_Sun,_Teotihuac%C3%A1n.jpg<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7.jpg\" alt=\"A FBD of the problem.\" width=\"492\" height=\"393\" class=\"alignnone wp-image-2200 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7.jpg 492w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7-300x240.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7-65x52.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7-225x180.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/chapter-7-350x280.jpg 350w\" sizes=\"auto, (max-width: 492px) 100vw, 492px\" \/><\/p>\n<p><strong>3.<\/strong> <strong>Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>W<sub>p<\/sub> =15 N<\/li>\n<li>g = 9.81 m\/s<sup>2<\/sup><\/li>\n<li>Izz = (1\/12)ml<sup>2<\/sup><\/li>\n<li>Max Iz\u2019z\u2019 =5 kgm<sup>2<\/sup><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>L, length of stick<\/li>\n<\/ul>\n<p><strong>4.<\/strong> <strong>Approach\u00a0<\/strong><\/p>\n<p>Newton\u2019s Second Law, Parallel Axis Theorem, Understanding of Mass Moment of Inertia<\/p>\n<p><strong>5.<\/strong> <strong>Analysis\u00a0<\/strong><\/p>\n<p>[latex]m = \\frac{w_p}{g}[\/latex]<\/p>\n<p>[latex]m = \\frac{15N} {9.81 \\: m\/s^2}[\/latex]<\/p>\n<p>[latex]m = 1.529\\: kg[\/latex]<\/p>\n<p>[latex]I_g =\\frac{1}{12}\\:md^2[\/latex]<\/p>\n<p>[latex]Iz'z' = I_g + ml^2[\/latex]<\/p>\n<p>[latex]Iz'z' = \\frac{1}{12}\\:md^2+m\\times \\left(\\frac{1}{2}d\\right) ^2[\/latex]<\/p>\n<p>thus,<\/p>\n<p>[latex]5kgm^2 = \\left (\\frac{1}{12}\\times (1.529kg)\\times d^2 \\right)+\\left(1.529kg\\: \\times \\frac{1}{4}d^2\\right)[\/latex]<\/p>\n<p>[latex]d =3.13 m[\/latex]<\/p>\n<p>In conclusion, the maximum size the pole can be is 3.13 m without the device breaking.<\/p>\n<p><strong>6.<\/strong> <strong>Review\u00a0<\/strong><\/p>\n<p>Applying the derived value of d in<\/p>\n<p>[latex]Iz'z' = \\frac{1}{3}md^2[\/latex]<\/p>\n<p>[latex]Iz'z' =\\frac{1}{3}\\times (1.529kg)\\times(3.13m)^2[\/latex]<\/p>\n<p>[latex]Iz'z' = 4.99999kgm^2 =5.0kgm^2[\/latex]<\/p>\n<p>According to the equation, the max distance of 3.13 m does make sense<\/p>\n<\/div>\n<h1>Example 7.6.6: Center of Mass, Submitted by Dhruvil Kanani<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">A thin uniform metal disc with a diameter of 18cm and an initial mass of 50.87kg has a circular hole cut out of it, turning it into a crescent shape, shown in the figure below. The radius of the hole is 6cm. Assuming the mass of the disc is evenly distributed, calculate the mass of the crescent formed and its center of mass.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2.png\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1931\" width=\"258\" height=\"258\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2.png 354w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2-300x300.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2-150x150.png 150w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2-65x65.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2-225x225.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru2-350x350.png 350w\" sizes=\"auto, (max-width: 258px) 100vw, 258px\" \/><\/div>\n<div><strong>2. Draw<\/strong><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1930\" width=\"360\" height=\"359\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1.png 503w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1-300x300.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1-150x150.png 150w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1-65x65.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1-225x224.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru1-350x349.png 350w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Mass of Initial disc: m<sub>1<\/sub> = 50.87kg<\/li>\n<li>Radius of larger disc: r<sub style=\"text-align: initial;background-color: initial\">1<\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\"> = 18cm\/2 = 9cm<\/span><\/li>\n<li>Radius of cut: r<sub style=\"text-align: initial;background-color: initial\">2<\/sub><span style=\"text-align: initial;background-color: initial;font-size: 1em\"> = 6cm<\/span><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Mass of crescent: m<sub>2<\/sub><\/li>\n<li>Center of mass of crescent: CM<sub style=\"text-align: initial;background-color: initial\">c<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><br \/>\nUse the mass and radius of the larger disc to find the density of the material. Calculate the area of the crescent<br \/>\nUse the area and density to find the mass of the crescent.<br \/>\nFind the center of mass of the smaller circle.<br \/>\nUse the center of mass of each circle and the mass of each circle to determine the center of mass of the crescent.<\/p>\n<p><strong>5. Analysis<\/strong><br \/>\nDensity of materials:<br \/>\n[latex]\\rho = \\frac{m_1}{2\\pi \\cdot r_1 ^2} \\\\ \\rho = \\frac{50.87kg}{2\\pi \\cdot 9^2} \\\\ \\rho = 0.1kg\/cm^2[\/latex]<\/p>\n<p>Area of crescent A<sub>c<\/sub> :<br \/>\n[latex]A_c = 2\\pi \\cdot (r_1 ^2 - r_2 ^2) cm^2 \\\\ A_c = 2\\pi \\cdot (9^2 - 6^2) cm^2 \\\\ A_c = 90\\pi cm^2[\/latex]<\/p>\n<p>Mass of crescent:<br \/>\n[latex]m_2 = \\rho \\cdot A_c \\\\ m_2 = 0.1kg\/cm^2 \\cdot 90\\pi cm^2 \\\\ m_2 = 9\\pi kg \\approx 28.27kg[\/latex]<\/p>\n<p>Center of mass of the cut circle:<\/p>\n<p>Since the edge of the cut circle lines incident to the larger circle, the radius of the cut can be subtracted from the radius of the full disc to find the center of mass of the cut, CM<sub>cut<\/sub><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3.png\" alt=\"A FBD of the diagram to find its Center of mass cut.\" class=\"alignnone wp-image-1933\" width=\"292\" height=\"291\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3.png 503w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3-300x300.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3-150x150.png 150w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3-65x65.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3-225x224.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru3-350x349.png 350w\" sizes=\"auto, (max-width: 292px) 100vw, 292px\" \/><\/p>\n<p>[latex]CM_{cut x} = R_1 - R_2 \\\\ CM_{cut x} = 9cm - 6cm \\\\ CM_{cut x} = 3cm[\/latex]<\/p>\n<p>CM<sub>cut x<\/sub> is 3cm in the negative x direction.<\/p>\n<p>CM<sub>cut y<\/sub> is equal to 0 as the center of mass of both the disc and the cut is zero in the y-direction.<\/p>\n<p>The center of mass of the full disc is at [0cm, 0cm] as it&#8217;s centred at the origin.<\/p>\n<p>Center of mass of the crescent:<\/p>\n<p>The center of mass of the full circle (CM<sub>full<\/sub>) is [0cm, 0cm] and its mass m<sub>1<\/sub> = 50.87kg.<br \/>\nThe center of mass of the cut (CM<sub>cut<\/sub>) is [-3cm, 0cm], and its mass (m<sub>cut<\/sub>) is the mass of the crescent minus the mass of the full circle:\u00a0 28.27kg &#8211; 50.87kg = -22.6kg.<\/p>\n<p>The center of mass of the crescent CM<sub>c<\/sub> is:<\/p>\n<p>[latex]CM_{cx}\u00a0 = \\frac{(CM_{full x} \\cdot m_1) + (CM_{cut x} \\cdot m_{cut})}{m_2}[\/latex]<\/p>\n<p>[latex]CM_{cx}\u00a0 = \\frac{(0cm \\cdot 50.87kg) + (-3cm \\cdot -22.6kg)}{28.27kg}[\/latex]<\/p>\n<p>[latex]CM_{cx} = 2.4cm[\/latex]<\/p>\n<p>Both the center of mass of the full circle and the cut in the y-direction are 0cm, giving the center of mass of the crescent in the y-direction also 0cm.<\/p>\n<p>The complete center of mass of the crescent is therefore:<br \/>\n<strong>CM<sub>c<\/sub> = [2.4cm, 0cm]<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4.png\" alt=\"Diagram showing the complete center of mass.\" class=\"alignleft wp-image-1941\" width=\"373\" height=\"371\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4.png 503w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4-150x150.png 150w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4-65x65.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4-225x224.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/dhru4-350x349.png 350w\" sizes=\"auto, (max-width: 373px) 100vw, 373px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><strong>6. Review:<\/strong><\/p>\n<p>The center of mass of the crescent leans to the side with more mass, as one would expect. The mass and center of mass found are both in the expected units with a reasonable magnitude.<\/p>\n<\/div>\n<h1>Example 7.6.7: Mass moment of inertia, Submitted by <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Michael Oppong-Ampomah&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:156,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:1}\">Michael Oppong-Ampomah<\/span><\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>An empty tissue paper roll rolls down an angled floor. If the tissue paper roll weighs 2 lbs, has and inner radius 0.9 inches, total diameter 2 inches, and a length of 12 inches, then calculate the mass moment of inertia of the roll.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><strong>2. Draw<\/strong><\/p>\n<p><strong style=\"font-size: 1em\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908.png\" alt=\"A FBD of the problem.\" width=\"380\" height=\"318\" class=\"alignnone wp-image-2049 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908.png 380w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908-300x251.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908-65x54.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908-225x188.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/12\/moi-of-hollow-cylinder-e1674763444908-350x293.png 350w\" sizes=\"auto, (max-width: 380px) 100vw, 380px\" \/><\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>3. K<\/strong><strong style=\"text-align: initial;font-size: 1em\">nowns and Unknown<\/strong><strong style=\"text-align: initial;font-size: 1em\">s\u00a0<\/strong><\/p>\n<p>Knowns:<\/p>\n<p>w = 2 lbs<\/p>\n<p>Inner radius r<sub>1<\/sub> = 0.9 in<\/p>\n<p>Outer radius r<sub>2<\/sub> = 1 in<\/p>\n<p>h=12 in<\/p>\n<p>&nbsp;<\/p>\n<p>Unknowns:<\/p>\n<p>Mass moment of inertia<\/p>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Finding the mass of the roll, identifying the axis of rotation, and applying the corresponding mass moment of inertia of the cylinder.<\/p>\n<p>[latex]I_{x}= \\frac{1}{2} m(r_{1}^2 + r_{2}^2)[\/latex]<\/p>\n<p>[latex]I_{y}= I_{z}= \\frac{1}{12} m[3(r_{1}^2 + r_{2}^2)+h^2][\/latex]<\/p>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p>weight of the roll = 2 lb<\/p>\n<p>mass of\u00a0 the roll, [latex]m = \\frac{2 lb}{32.2 ft\/s^2} = 0.062 lb[\/latex]<\/p>\n<p>Since we used constant of gravity with a unit of [latex]ft\/ s^2[\/latex] it is important to convert other values from inches to ft.<\/p>\n<p>[latex]0.9 in = \\frac{0.9}{12}=0.075 ft[\/latex]<\/p>\n<p>[latex]1 in =\\frac{1}{12} = 0.083 ft[\/latex]<\/p>\n<p>[latex]12 in = 1 ft[\/latex]<\/p>\n<p>now, [latex]I_{x}= \\frac{1}{2} m(r_{1}^2 + r_{2}^2)[\/latex]<\/p>\n<p>[latex]I_{x} = \\frac{1}{2} 0.062(0.075^2 + 0.083^2) = 0.00039 \\quad lb ft^2[\/latex]<\/p>\n<p>[latex]I_{y}= I_{z}= \\frac{1}{12} m[3(r_{1}^2 + r_{2}^2)+h^2][\/latex]<\/p>\n<p>[latex]I_{y}= I_{z}= \\frac{1}{12} 0.062[3(0.075^2 + 0.083^2)+1^2] =0.0054 \\quad lb ft^2[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Although the weight of an empty paper roll is exaggerated for easier calculation, the answers make sense. The value of I<sub>y<\/sub> is higher than I<sub>x<\/sub><sub>,<\/sub> which means more force will be needed for rotation in that axis. From this, it is important to note that the roll will rotate along the x-axis. Also note that weight is converted to mass, and units must be the same.<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":60,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"7.7 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-577","chapter","type-chapter","status-publish","hentry"],"part":64,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/577","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/users\/60"}],"version-history":[{"count":34,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/577\/revisions"}],"predecessor-version":[{"id":2874,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/577\/revisions\/2874"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/64"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/577\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=577"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=577"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=577"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=577"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}