{"id":17175,"date":"2023-06-19T06:20:48","date_gmt":"2023-06-19T06:20:48","guid":{"rendered":"https:\/\/www.greatassignmenthelp.com\/questions\/?p=17175"},"modified":"2023-12-21T07:15:06","modified_gmt":"2023-12-21T07:15:06","slug":"an-8-0-cm-diameter-400-g-sphere-is-released-from-rest-ta-the-tip-of-a-2-1-m-long-25-degree-incline","status":"publish","type":"post","link":"https:\/\/www.greatassignmenthelp.com\/questions\/an-8-0-cm-diameter-400-g-sphere-is-released-from-rest-ta-the-tip-of-a-2-1-m-long-25-degree-incline\/","title":{"rendered":"An 8.0 cm diameter, 400 g sphere is released from rest ta the tip of a 2.1 m long, 25 degree incline. It rolls, without slipping, to the bottom."},"content":{"rendered":"<p>a) What is the sphere&#8217;s angular velocity at the bottom of the incline?<\/p>\n<p>b) What fraction of its kinetic energy is rotational?<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>a) 88.1 rad\/s<\/p>\n<p>b) 0.286<\/p>\n<p><strong>Explanation:<\/strong><\/p>\n<p><strong>given information:<\/strong><\/p>\n<p>diameter, d = 8 cm = 0.08 m<\/p>\n<p>sphere&#8217;s mass, m = 400 g = 0.4 kg<\/p>\n<p>the distance from rest to the tip, h = 2.1 m<\/p>\n<p>incline angle, \u03b8 = 25\u00b0<\/p>\n<ol>\n<li>a) What is the sphere&#8217;s angular velocity at the bottom of the incline?<\/li>\n<\/ol>\n<p>mg (h sin\u03b8) = 1\/2 I\u03c9\u00b2 + 1\/2mv\u00b2<\/p>\n<p>I of solid sphere = 2\/5 mr\u00b2, therefore<\/p>\n<p>mg (h sin\u03b8) = 1\/2 (2\/5 mr\u00b2) \u03c9\u00b2 + 1\/2 mv\u00b2, now we can eliminate the mass<\/p>\n<p>g h sin \u03b8 = 1\/5 r\u00b2 \u03c9\u00b2 + 1\/2 v\u00b2<\/p>\n<p>\u03c9 = v\/r, v = \u03c9r<\/p>\n<p>consequently,<\/p>\n<p>g h sin \u03b8 = 1\/5 r\u00b2 \u03c9\u00b2 + 1\/2 (\u03c9r)\u00b2<\/p>\n<p>g h sin \u03b8 = (7\/10) r\u00b2 \u03c9\u00b2<\/p>\n<p>\u03c9\u00b2 = 10 g h sin \u03b8\/7 r\u00b2<\/p>\n<p>\u03c9 = \u221a10 g h sin \u03b8\/7 r\u00b2<\/p>\n<p>= \u221a10 (9.8) (2.1) sin 25\u00b0 \/ 7 (0.04)\u00b2<\/p>\n<p>= 88.1 rad\/s<\/p>\n<ol>\n<li>b) What fraction of its kinetic energy (KE) is rotational?<\/li>\n<\/ol>\n<p>fraction of its kinetic energy = rotational KE \/ total KE<\/p>\n<p>total KE = total potential energy<\/p>\n<p>= m g h sin \u03b8<\/p>\n<p>= 0.4 x 9.8 x 2.1 sin 25\u00b0<\/p>\n<p>= 3.48 J<\/p>\n<p>rotational KE = 1\/2 I\u03c9\u00b2<\/p>\n<p>= 1\/5 mr\u00b2\u03c9\u00b2<\/p>\n<p>= 1\/5 0.4 (0.04) \u00b2(88.1)\u00b2<\/p>\n<p>= 0.99<\/p>\n<p>fraction of its KE = 0.99\/3.48<\/p>\n<p>= 0.286<\/p>\n","protected":false},"excerpt":{"rendered":"<p>a) What is the sphere&#8217;s angular velocity at the bottom of the incline? b) What fraction of its kinetic energy is rotational? Answer: a) 88.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[27],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v23.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>An 8.0 cm diameter, 400 g sphere is released from rest ta the tip of a 2.1 m long, 25 degree incline. It rolls, without slipping, to the bottom.<\/title>\n<meta name=\"description\" content=\"An 8.0 cm diameter, 400 g sphere is released from rest ta the tip of a 2.1 m long, 25 degree incline. 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