{"id":201140,"date":"2025-03-15T03:11:29","date_gmt":"2025-03-15T03:11:29","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201140"},"modified":"2025-03-15T03:11:32","modified_gmt":"2025-03-15T03:11:32","slug":"you-have-finished-watching-a-movie-on-blu-ray-and-the-disc-is-slowingto-a-sto","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/you-have-finished-watching-a-movie-on-blu-ray-and-the-disc-is-slowingto-a-sto\/","title":{"rendered":"You have finished watching a movie on Blu-ray and the disc is slowingto a sto"},"content":{"rendered":"\n

You have finished watching a movie on Blu-ray and the disc is slowing
to a stop. The disc\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2s angular velocity at t = 0 is 27.5 rad\/s, and its
angular acceleration is a constant -10.0 rad\/s2. A line PQ on the disc\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2s
surface lies along the +x-axis at t = 0 (Fig.). (a) What is the disc\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2s
angular velocity at t = 0.300 s? (b) What angle does the line PQ make
with the +x-axis at this time?<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Answer:<\/h3>\n\n\n\n

(a) The disc\u2019s angular velocity at t=0.300t = 0.300 s is: \u03c9=24.5 rad\/s\\omega = 24.5 \\text{ rad\/s}<\/p>\n\n\n\n

(b) The angle that the line PQ makes with the +x-axis at this time is: \u03b8=446.91\u2218\\theta = 446.91^\\circ<\/p>\n\n\n\n

\n\n\n\n

Explanation:<\/h3>\n\n\n\n

We are given the initial angular velocity (\u03c90\\omega_0), the angular acceleration (\u03b1\\alpha), and the time (tt), and we need to determine:<\/p>\n\n\n\n

    \n
  1. The angular velocity (\u03c9\\omega) at t=0.300t = 0.300 s.<\/li>\n\n\n\n
  2. The angle (\u03b8\\theta) the line PQ makes with the +x-axis at this time.<\/li>\n<\/ol>\n\n\n\n

    Step 1: Finding Angular Velocity<\/strong><\/h4>\n\n\n\n

    The angular velocity at any time tt can be found using the kinematic equation: \u03c9=\u03c90+\u03b1t\\omega = \\omega_0 + \\alpha t<\/p>\n\n\n\n

    Substituting the given values: \u03c9=27.5+(\u221210.0)(0.300)\\omega = 27.5 + (-10.0)(0.300) \u03c9=27.5\u22123.0\\omega = 27.5 – 3.0 \u03c9=24.5 rad\/s\\omega = 24.5 \\text{ rad\/s}<\/p>\n\n\n\n

    Thus, at t=0.300t = 0.300 s, the disc\u2019s angular velocity is 24.5 rad\/s<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Finding Angular Displacement<\/strong><\/h4>\n\n\n\n

    The angular displacement (\u03b8\\theta) is given by: \u03b8=\u03c90t+12\u03b1t2\\theta = \\omega_0 t + \\frac{1}{2} \\alpha t^2<\/p>\n\n\n\n

    Substituting the values: \u03b8=(27.5)(0.300)+12(\u221210.0)(0.300)2\\theta = (27.5)(0.300) + \\frac{1}{2}(-10.0)(0.300)^2 \u03b8=8.25\u22120.45\\theta = 8.25 – 0.45 \u03b8=7.8 rad\\theta = 7.8 \\text{ rad}<\/p>\n\n\n\n

    Since 1 radian = 57.296\u221857.296^\\circ, converting to degrees: \u03b8=7.8\u00d757.296\\theta = 7.8 \\times 57.296 \u03b8=446.91\u2218\\theta = 446.91^\\circ<\/p>\n\n\n\n

    \n\n\n\n

    Interpretation<\/strong><\/h3>\n\n\n\n
      \n
    • The angle 446.91\u2218446.91^\\circ<\/strong> means the line PQ has rotated more than a full revolution (360\u00b0) and an extra 86.91\u00b0<\/strong>.<\/li>\n\n\n\n
    • The position of PQ relative to the +x-axis is 86.91\u221886.91^\\circ counterclockwise.<\/li>\n<\/ul>\n\n\n\n

      Now, I will generate an image illustrating the rotation of the disc.<\/p>\n\n\n\n

      The image illustrates the rotation of the PQ line on the disc:<\/p>\n\n\n\n

        \n
      • The blue arrow<\/strong> represents the initial position of the PQ line along the +x-axis.<\/li>\n\n\n\n
      • The red arrow<\/strong> represents the rotated position of PQ after 0.300 seconds.<\/li>\n\n\n\n
      • The disc has rotated 446.91\u00b0<\/strong>, which means it has completed one full revolution (360\u00b0) and an additional 86.91\u00b0 counterclockwise<\/strong>.<\/li>\n<\/ul>\n\n\n\n

        This confirms that the final position of PQ is at an angle of 86.91\u00b0 relative to the +x-axis<\/strong>.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        You have finished watching a movie on Blu-ray and the disc is slowingto a stop. The disc\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2s angular velocity at t = 0 is 27.5 rad\/s, and itsangular acceleration is a constant -10.0 rad\/s2. A line PQ on the disc\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2ssurface lies along the +x-axis at t = 0 (Fig.). (a) What is the disc\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2sangular velocity […]<\/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-201140","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201140","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=201140"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201140\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201140"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201140"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}