{"id":8127,"date":"2023-04-30T12:08:05","date_gmt":"2023-04-30T12:08:05","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=8127"},"modified":"2024-05-25T06:53:17","modified_gmt":"2024-05-25T06:53:17","slug":"moment-of-inertia-equation","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/moment-of-inertia-equation.htm","title":{"rendered":"Moment of inertia equation","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">3 Problems and solutions about Moment of inertia equation<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">1. A solid cylinder has a radius of 8 cm and a mass of 2 kg. Meanwhile, a solid ball has a radius of 5 cm and a mass of 4 kg. If the two objects rotate with an axis through their center, determine the ratio of the moment of inertia of the cylinder and the ball.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Known:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Solid cylinder radius (r) = 8 cm = 0.08 m<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Solid cylinder mass (m) = 2 kg<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Solid ball radius (r) = 5 cm = 0.05 m<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The mass of the solid ball (m) = 4 kg<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> Comparison of the moment of inertia of a cylinder and a ball<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution<\/u><u>:<\/u><!--more--><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The formula for the moment of inertia of a solid cylinder:<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">I = \u00bd M R<sup>2<\/sup> = \u00bd (2)(0,08)<sup>2 <\/sup>= 0,0064 <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The formula for the moment of inertia of a solid ball:<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">I = 2\/5 M R<sup>2<\/sup> = 1\/5 (4)(0,05)<sup>2<\/sup> = (0,8)(0,0025) = 0,002 <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Comparison of the moment of inertia of the cylinder and the ball:<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">0,0064 : 0,002<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">3,2 : 1<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">2. The tin in which the biscuits were placed was used as a toy. The mass of the can is 200 grams and the radius is 15 cm. The can is rolled on a horizontal floor. If the lid and bottom of the can are neglected, determine the moment of inertia of the can.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Known:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The can resembles a hollow cylinder.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Mass (m) = 200 grams = 0.2 kg<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Radius (r) = 15 cm = 0.15 m<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The formula for the moment of inertia of a hollow cylinder through the axis:<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">I = M R<sup>2<\/sup> = (0,2)(0,15)<sup>2<\/sup> = (0,2)(0,0225) = 0,0045 kg m<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">3. A disc which is free to rotate about a vertical axis is capable of rotating at a speed of 80 revolutions per minute. If a small object with a mass of 4 x 10-2 kg is attached to a disc 5 cm from the axis, it turns out that the rotation is 60 revolutions per minute, then determine the moment of inertia of the disc.<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Known:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Initial angular velocity (\u03c9<sub>o<\/sub>) = 80 put \/ 60 s = 4\/3 put\/s<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Initial moment of inertia (I<sub>o<\/sub>) = I<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Final angular velocity (\u03c9<sub>t<\/sub>) = 60 put \/ 60 s = 1 put\/s<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Final moment of inertia (I<sub>t<\/sub>) = I + m r<sup>2 <\/sup>= I + (0.04)(0.05)<sup>2<\/sup> = I + (0.04)(0.0025) = I + 0.0425<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> moment of inertia of the Disc <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution:<\/u><\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The law of conservation of angular momentum<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">I \u03c9<sub>o <\/sub>= I \u03c9<sub>t<\/sub> <\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">I (4\/3) = I + (0,0425 (1))<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">4 I \/ 3 = I + 0,0425<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">4 I = 3 I + 0,1275<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">4 I &#8211; 3 I = 0,1275<\/span><\/p>\n<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">I = 0,1275 kg m<sup>2<\/sup><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>3 Problems and solutions about Moment of inertia equation 1. A solid cylinder has a radius of 8 cm and a mass of 2 kg. Meanwhile, a solid ball has a radius of 5 cm and a mass of 4 kg. If the two objects rotate with an axis through their center, determine the ratio &#8230; <a title=\"Moment of inertia equation\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/moment-of-inertia-equation.htm\" aria-label=\"Read more about Moment of inertia equation\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"3","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Moment of inertia equation","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-8127","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8127","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=8127"}],"version-history":[{"count":6,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8127\/revisions"}],"predecessor-version":[{"id":9855,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8127\/revisions\/9855"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=8127"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=8127"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=8127"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}