{"id":184024,"date":"2025-01-18T02:36:37","date_gmt":"2025-01-18T02:36:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=184024"},"modified":"2025-01-18T02:36:39","modified_gmt":"2025-01-18T02:36:39","slug":"gligible-friction-in-the-groove-are-two-metal-blocks","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/18\/gligible-friction-in-the-groove-are-two-metal-blocks\/","title":{"rendered":"gligible friction in the groove are two metal blocks"},"content":{"rendered":"\n<p>gligible friction in the groove are two metal blocks, each with a mass of 0.064 kg, and they are connected to each other by a spring.<\/p>\n\n\n\n<p>At a particular moment in time (the \u201cinitial state\u201d) the blocks are 0.24 m apart, with zero radial velocity (that is, they are not moving toward or away from each other). At this moment the angular speed of the disk is ?i is 17 radians\/s. The spring pulls the blocks toward each other, and at a later time (the \u201cfinal state\u201d) the blocks are 0.08 m apart. Now what is the angular speed ?f? Approximate the metal blocks as point particles.<\/p>\n\n\n\n<p>?f= radians\/s<\/p>\n\n\n\n<p>B. A hollow lightweight grooved disk whose moment of inertia is 4.0 \u00d7 10-3 kg\u00b7m2 rotates with negligible friction around a vertical axis. Free to slide with negligible friction in the groove are two metal blocks, each with a mass of 0.060 kg, and they are connected to each other by a spring.<\/p>\n\n\n\n<p>At a particular moment in time (the \u201cinitial state\u201d) the blocks are 0.48 m apart, with zero radial velocity (that is, they are not moving toward or away from each other). At this moment the angular speed of the disk is ?i is 15 radians\/s. The spring pulls the blocks toward each other, and at a later time (the \u201cfinal state\u201d) the blocks are 0.16 m apart.<\/p>\n\n\n\n<p>Now what is the angular speed ?f? Approximate the metal blocks as point particles.<\/p>\n\n\n\n<p>?f= radians\/s<\/p>\n\n\n\n<p>What is the change in the potential energy of the spring, including sign?<\/p>\n\n\n\n<p>??U= J<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve the problem, we use the principle of conservation of angular momentum and the concept of change in the spring&#8217;s potential energy.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conservation of Angular Momentum:<\/h3>\n\n\n\n<p>The total angular momentum of the system (disk + blocks) must remain constant because no external torque acts on it.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Initial Angular Momentum<\/strong>: The angular momentum is the sum of the moment of inertia of the disk and the blocks: Li=Idisk\u03c9i+2mri2\u03c9iL_i = I_{\\text{disk}} \\omega_i + 2 m r_i^2 \\omega_i where:<ul><li>Idisk=4.0\u00d710\u22123\u2009kg\\cdotpm2I_{\\text{disk}} = 4.0 \\times 10^{-3} \\, \\text{kg\u00b7m}^2 (moment of inertia of the disk)<\/li><li>m=0.060\u2009kgm = 0.060 \\, \\text{kg} (mass of each block)<\/li><li>ri=0.48\u2009mr_i = 0.48 \\, \\text{m} (initial distance of each block from the center)<\/li><li>\u03c9i=15\u2009rad\/s\\omega_i = 15 \\, \\text{rad\/s} (initial angular speed of the disk)<\/li><\/ul>Substituting: Li=(4.0\u00d710\u22123)(15)+2(0.060)(0.48)2(15)L_i = (4.0 \\times 10^{-3}) (15) + 2 (0.060) (0.48)^2 (15)<\/li>\n\n\n\n<li><strong>Final Angular Momentum<\/strong>: Similarly: Lf=Idisk\u03c9f+2mrf2\u03c9fL_f = I_{\\text{disk}} \\omega_f + 2 m r_f^2 \\omega_f where:<ul><li>rf=0.16\u2009mr_f = 0.16 \\, \\text{m} (final distance of each block from the center)<\/li><li>\u03c9f\\omega_f (final angular speed to be found)<\/li><\/ul>Substituting: Lf=(4.0\u00d710\u22123)\u03c9f+2(0.060)(0.16)2\u03c9fL_f = (4.0 \\times 10^{-3}) \\omega_f + 2 (0.060) (0.16)^2 \\omega_f<\/li>\n\n\n\n<li><strong>Angular Momentum Conservation<\/strong>: Li=LfL_i = L_f Equating the two expressions: (4.0\u00d710\u22123)(15)+2(0.060)(0.48)2(15)=(4.0\u00d710\u22123)\u03c9f+2(0.060)(0.16)2\u03c9f(4.0 \\times 10^{-3}) (15) + 2 (0.060) (0.48)^2 (15) = (4.0 \\times 10^{-3}) \\omega_f + 2 (0.060) (0.16)^2 \\omega_f Simplify and solve for \u03c9f\\omega_f.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Change in Potential Energy of the Spring:<\/h3>\n\n\n\n<p>The change in potential energy (\u0394U\\Delta U) is related to the work done by the spring in pulling the blocks closer together. Using the formula for elastic potential energy: U=12kx2U = \\frac{1}{2} k x^2<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>kk is the spring constant<\/li>\n\n\n\n<li>xx is the separation between the blocks.<\/li>\n<\/ul>\n\n\n\n<p>The initial and final potential energies are: Ui=12k(0.48)2,Uf=12k(0.16)2U_i = \\frac{1}{2} k (0.48)^2, \\quad U_f = \\frac{1}{2} k (0.16)^2<\/p>\n\n\n\n<p>The change in potential energy: \u0394U=Uf\u2212Ui=12k[(0.16)2\u2212(0.48)2]\\Delta U = U_f &#8211; U_i = \\frac{1}{2} k \\left[ (0.16)^2 &#8211; (0.48)^2 \\right]<\/p>\n\n\n\n<p>Since kk is not given, you must express \u0394U\\Delta U in terms of kk. Alternatively, if kk is provided or can be deduced from other context, substitute and compute \u0394U\\Delta U.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Let me compute the angular speed \u03c9f\\omega_f and \u0394U\\Delta U numerically.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Results:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Final Angular Speed (\u03c9f\\omega_f):<\/strong> \u03c9f=67.13\u2009rad\/s\\omega_f = 67.13 \\, \\text{rad\/s}<\/li>\n\n\n\n<li><strong>Change in Spring Potential Energy (\u0394U\\Delta U):<\/strong> The change in spring potential energy is: \u0394U=12k(\u22120.1024)\\Delta U = \\frac{1}{2} k (-0.1024) The factor \u22120.1024-0.1024 represents the proportional change in potential energy, which depends on the spring constant kk. The negative sign indicates a decrease in the spring&#8217;s potential energy as the blocks move closer together.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>If you know the spring constant kk, substitute its value to compute the exact change in potential energy. Let me know if further clarification or calculations are needed!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>gligible friction in the groove are two metal blocks, each with a mass of 0.064 kg, and they are connected to each other by a spring. At a particular moment in time (the \u201cinitial state\u201d) the blocks are 0.24 m apart, with zero radial velocity (that is, they are not moving toward or away from [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-184024","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184024","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=184024"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184024\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184024"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184024"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184024"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}