{"id":3036,"date":"2018-06-04T17:38:14","date_gmt":"2018-06-05T00:38:14","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3036"},"modified":"2018-06-04T17:38:14","modified_gmt":"2018-06-05T00:38:14","slug":"rolling-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/rolling-motion-problems-and-solutions.htm","title":{"rendered":"Rolling motion – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span>A hollow cylindrical object (I = m R<\/span>2<\/span><\/sup>) moves to roll without slipping up a rough inclined plane<\/a> with an initial velocity of 10 m\/s. The inclined plane has an elevation angle \u03b8 with tan \u03b8 = 0.75. If the gravitational acceleration g = 10 m.s<\/span>-2<\/span><\/sup>, the velocity<\/a> of the object is reduced to 5 m.s<\/span>-1<\/span><\/sup> then the distance on the inclined plane<\/a> of the object is\u2026<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Moment of inertia<\/a> of the hollow cylinder (I) = m R<\/span>2<\/span><\/sup><\/span><\/p>\n

Elevation angle = <\/span>\u03b8<\/span>, where tan <\/span>\u03b8<\/span> = 0.75 = 75\/100 = opp \/ adj<\/span><\/span><\/p>\n

Sin <\/span>\u03b8 <\/span>= opp \/ hyp = 75\/125 = 3\/5 = 0.6<\/span><\/span><\/p>\n

Acceleration due to gravity<\/a> (g) = 10 m\/s<\/span>2<\/span><\/sup><\/span><\/p>\n

Initial velocity (v<\/span>o<\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span>t<\/sub>) = 5 m\/s<\/span><\/span><\/p>\n

Wanted :<\/u><\/span> Distance \"Rolling<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Calculate the height reached by the cylinder using equation of the conservation of mechanical energy<\/a>.<\/span><\/span><\/p>\n

\"Rolling<\/p>\n

Initial height (h<\/span>o<\/sub>) = 0 meter <\/span><\/span><\/p>\n

Initial velocity (v<\/span>o<\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span>t<\/sub>) = 5 m\/s<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span>2<\/span><\/sup><\/span><\/p>\n

\"Rolling<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

The cylinder reaches a height of 7.5 meters.<\/span><\/span><\/p>\n

Distance traveled by cylinder :<\/span><\/span><\/p>\n

\"Rolling<\/p>\n

The distance traveled by the cylinder is <\/span>12.5 meters.<\/span><\/span><\/p>\n

The correct answer is A.<\/span><\/span><\/p>\n

Description of the equation :<\/span><\/span><\/p>\n

o<\/span><\/sub> = initial, <\/span>t<\/span><\/sub> = final, ME = mechanical energy<\/a>, PE<\/span> = potential energy, KE<\/span> = kinetic energy<\/a>, m = mass<\/a>, g = acceleration due to gravity, h = height, v = linear velocity, <\/span>\u03c9 = angular velocity<\/a>, I = moment of inertia, R = radius of cylinder<\/span><\/span><\/p>\n

2. <\/span>A solid cylinder (I = \u00bd m R<\/span>2<\/span><\/sup>) with a mass of 3 kg moves to roll without slipping up a rough inclined plane having an elevation angle \u03b8 with sin \u03b8 = 0.6. If the gravitational acceleration g = 10 m.s<\/span>-2 <\/span><\/sup>and the initial velocity of the object is 10 m\/s, then the length of the inclined plane traveled by an object is\u2026<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Moment of inertia of solid cylinder (I) = \u00bd m R<\/span>2<\/span><\/sup><\/span><\/p>\n

Mass of cylinder = 3 kg<\/span><\/span><\/p>\n

Initial velocity (v<\/span>o<\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span>t<\/sub>) = 0 m\/s (object stop)<\/span><\/span><\/p>\n

Elevation angle (<\/span>\u03b8<\/span>) = <\/span>\u03b8<\/span>, where sin <\/span>\u03b8 <\/span>= 0.6 = 6\/10 = opp\/hyp<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span>2\"Rolling<\/span><\/sup><\/span><\/p>\n

Wanted :<\/u><\/span> T<\/span>he length of the inclined plane traveled by object<\/span><\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Calculate the height reached by the cylinder using equation of the conservation of mechanical energy.<\/span><\/span><\/span><\/p>\n

\"Rolling<\/p>\n

\"Rolling<\/p>\n

Initial height (h<\/span>o<\/sub>) = 0 meter <\/span><\/span><\/p>\n

Initial velocity (v<\/span>o<\/sub>) = 10 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span>t<\/sub>) = 0 m\/s (object stop)<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span>2<\/span><\/sup><\/span><\/p>\n

\"Rolling<\/p>\n

The cylinder reaches a height of 7.5 meters.<\/span><\/span><\/p>\n

Distance traveled by cylinder :<\/span><\/span><\/p>\n

\"Rolling<\/p>\n

The distance traveled by the cylinder is 12.5 meters.<\/span><\/span><\/p>\n

3. <\/span>A hollow cylindrical object (I = m R<\/span>2<\/span><\/sup>) with radius R, moves to roll without slipping up a rough inclined plane with an angle of \u03b1 where sin \u03b1 = 0.8. If the gravitational acceleration g = 10 m.s<\/span>-2<\/span><\/sup> and the initial velocity is 8 m.s<\/span>-1<\/span><\/sup> then the length of the inclined plane reached by the object before it stops is\u2026<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Moment of inertia of solid cylinder (I) = m R<\/span>2<\/span><\/sup><\/span><\/p>\n

Initial velocity (v<\/span>o<\/sub>) = 8 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span>t<\/sub>) = 0 m\/s (object stop)<\/span><\/span><\/p>\n

Elevation angle of inclined plane (<\/span>\u03b8<\/span>) = <\/span>\u03b8<\/span>, where sin <\/span>\u03b8 <\/span>= 0.8 <\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span>2<\/span><\/sup><\/span><\/p>\n

Wanted :<\/u><\/span> Distance traveled by cylinder<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

Calculate the height reached by the cylinder using equation of the conservation of mechanical energy.<\/span><\/span><\/span><\/p>\n

\"Rolling<\/p>\n

\"Rolling<\/p>\n

Initial height (h<\/span>o<\/sub>) = 0 meter <\/span><\/span><\/p>\n

Initial velocity (v<\/span>o<\/sub>) = 8 m\/s<\/span><\/span><\/p>\n

Final velocity (v<\/span>t<\/span><\/sub>) = 0 m\/s<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span>2<\/span><\/sup><\/span><\/p>\n

\"Rolling<\/p>\n

The cylinder reaches a height of 6.4 meters.<\/span><\/span><\/p>\n

Distance traveled by cylinder :<\/span><\/span><\/p>\n

\"Rolling<\/p>\n

The distance traveled by the cylinder is 8 meters.<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. A hollow cylindrical object (I = m R2) moves to roll without slipping up a rough inclined plane with an initial velocity of 10 m\/s. The inclined plane has an elevation angle \u03b8 with tan \u03b8 = 0.75. If the gravitational acceleration g = 10 m.s-2, the velocity of the object is reduced to … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","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":"","_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":"Rolling motion - problems and solutions","_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-3036","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\/3036","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=3036"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3036\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3036"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3036"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3036"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}