{"id":87027,"date":"2025-06-01T04:26:30","date_gmt":"2025-06-01T04:26:30","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87027"},"modified":"2025-06-01T04:26:30","modified_gmt":"2025-06-01T04:26:30","slug":"the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v\/","title":{"rendered":"The angular acceleration of a simple pendulum at an angle \u03b1 from the v"},"content":{"rendered":"

The angular acceleration of a simple pendulum at an angle \u03b1 from the vertical is proportional to<\/p>\n

[amp_mcq option1=”tan \u03b1” option2=”sin\u00b2 \u03b1” option3=”sin \u03b1” option4=”sin 2\u03b1” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nFor a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about the point of suspension is given by \u03c4 = -mgl sin \u03b1, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and \u03b1 is the angular displacement. According to Newton’s second law for rotation, the torque is also equal to I\u03b2, where I is the moment of inertia and \u03b2 is the angular acceleration. For a simple pendulum, I = ml\u00b2. Thus, ml\u00b2\u03b2 = -mgl sin \u03b1. This gives \u03b2 = -(g\/l) sin \u03b1. The magnitude of the angular acceleration is |\u03b2| = (g\/l) sin \u03b1. Since g and l are constants, the angular acceleration is proportional to sin \u03b1.
\n<\/section>\n
\n– Restoring torque \u03c4 = -mgl sin \u03b1.
\n– Torque \u03c4 = I\u03b2.
\n– Moment of inertia of a simple pendulum I = ml\u00b2.
\n– Angular acceleration \u03b2 = \u03c4\/I = -(g\/l) sin \u03b1.
\n– Magnitude of angular acceleration is proportional to sin \u03b1.
\n<\/section>\n
\nFor small angles (\u03b1 << 1 radian), sin \u03b1 \u2248 \u03b1, and the equation of motion becomes \u03b2 \u2248 -(g\/l)\u03b1. In this small angle approximation, the motion is Simple Harmonic Motion (SHM), and the angular acceleration is proportional to the angular displacement \u03b1. However, for any angle \u03b1, the angular acceleration is strictly proportional to sin \u03b1.\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The angular acceleration of a simple pendulum at an angle \u03b1 from the vertical is proportional to [amp_mcq option1=”tan \u03b1” option2=”sin\u00b2 \u03b1” option3=”sin \u03b1” option4=”sin 2\u03b1” correct=”option3″] This question was previously asked in UPSC Geoscientist – 2023 Download PDFAttempt Online For a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about … <\/p>\n

Detailed SolutionThe angular acceleration of a simple pendulum at an angle \u03b1 from the v<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1091],"tags":[1105,1129,1128],"class_list":["post-87027","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe angular acceleration of a simple pendulum at an angle \u03b1 from the v<\/title>\n<meta name=\"description\" content=\"For a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about the point of suspension is given by \u03c4 = -mgl sin \u03b1, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and \u03b1 is the angular displacement. According to Newton's second law for rotation, the torque is also equal to I\u03b2, where I is the moment of inertia and \u03b2 is the angular acceleration. For a simple pendulum, I = ml\u00b2. Thus, ml\u00b2\u03b2 = -mgl sin \u03b1. This gives \u03b2 = -(g\/l) sin \u03b1. The magnitude of the angular acceleration is |\u03b2| = (g\/l) sin \u03b1. Since g and l are constants, the angular acceleration is proportional to sin \u03b1. - Restoring torque \u03c4 = -mgl sin \u03b1. - Torque \u03c4 = I\u03b2. - Moment of inertia of a simple pendulum I = ml\u00b2. - Angular acceleration \u03b2 = \u03c4\/I = -(g\/l) sin \u03b1. - Magnitude of angular acceleration is proportional to sin \u03b1.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-\u03b1-from-the-v\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The angular acceleration of a simple pendulum at an angle \u03b1 from the v\" \/>\n<meta property=\"og:description\" content=\"For a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about the point of suspension is given by \u03c4 = -mgl sin \u03b1, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and \u03b1 is the angular displacement. According to Newton's second law for rotation, the torque is also equal to I\u03b2, where I is the moment of inertia and \u03b2 is the angular acceleration. For a simple pendulum, I = ml\u00b2. Thus, ml\u00b2\u03b2 = -mgl sin \u03b1. This gives \u03b2 = -(g\/l) sin \u03b1. The magnitude of the angular acceleration is |\u03b2| = (g\/l) sin \u03b1. Since g and l are constants, the angular acceleration is proportional to sin \u03b1. - Restoring torque \u03c4 = -mgl sin \u03b1. - Torque \u03c4 = I\u03b2. - Moment of inertia of a simple pendulum I = ml\u00b2. - Angular acceleration \u03b2 = \u03c4\/I = -(g\/l) sin \u03b1. - Magnitude of angular acceleration is proportional to sin \u03b1.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-\u03b1-from-the-v\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:26:30+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"The angular acceleration of a simple pendulum at an angle \u03b1 from the v","description":"For a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about the point of suspension is given by \u03c4 = -mgl sin \u03b1, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and \u03b1 is the angular displacement. According to Newton's second law for rotation, the torque is also equal to I\u03b2, where I is the moment of inertia and \u03b2 is the angular acceleration. For a simple pendulum, I = ml\u00b2. Thus, ml\u00b2\u03b2 = -mgl sin \u03b1. This gives \u03b2 = -(g\/l) sin \u03b1. The magnitude of the angular acceleration is |\u03b2| = (g\/l) sin \u03b1. Since g and l are constants, the angular acceleration is proportional to sin \u03b1. - Restoring torque \u03c4 = -mgl sin \u03b1. - Torque \u03c4 = I\u03b2. - Moment of inertia of a simple pendulum I = ml\u00b2. - Angular acceleration \u03b2 = \u03c4\/I = -(g\/l) sin \u03b1. - Magnitude of angular acceleration is proportional to sin \u03b1.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-\u03b1-from-the-v\/","og_locale":"en_US","og_type":"article","og_title":"The angular acceleration of a simple pendulum at an angle \u03b1 from the v","og_description":"For a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about the point of suspension is given by \u03c4 = -mgl sin \u03b1, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and \u03b1 is the angular displacement. According to Newton's second law for rotation, the torque is also equal to I\u03b2, where I is the moment of inertia and \u03b2 is the angular acceleration. For a simple pendulum, I = ml\u00b2. Thus, ml\u00b2\u03b2 = -mgl sin \u03b1. This gives \u03b2 = -(g\/l) sin \u03b1. The magnitude of the angular acceleration is |\u03b2| = (g\/l) sin \u03b1. Since g and l are constants, the angular acceleration is proportional to sin \u03b1. - Restoring torque \u03c4 = -mgl sin \u03b1. - Torque \u03c4 = I\u03b2. - Moment of inertia of a simple pendulum I = ml\u00b2. - Angular acceleration \u03b2 = \u03c4\/I = -(g\/l) sin \u03b1. - Magnitude of angular acceleration is proportional to sin \u03b1.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-\u03b1-from-the-v\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:26:30+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v\/","name":"The angular acceleration of a simple pendulum at an angle \u03b1 from the v","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:26:30+00:00","dateModified":"2025-06-01T04:26:30+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"For a simple pendulum at an angle \u03b1 from the vertical, the restoring torque about the point of suspension is given by \u03c4 = -mgl sin \u03b1, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and \u03b1 is the angular displacement. According to Newton's second law for rotation, the torque is also equal to I\u03b2, where I is the moment of inertia and \u03b2 is the angular acceleration. For a simple pendulum, I = ml\u00b2. Thus, ml\u00b2\u03b2 = -mgl sin \u03b1. This gives \u03b2 = -(g\/l) sin \u03b1. The magnitude of the angular acceleration is |\u03b2| = (g\/l) sin \u03b1. Since g and l are constants, the angular acceleration is proportional to sin \u03b1. - Restoring torque \u03c4 = -mgl sin \u03b1. - Torque \u03c4 = I\u03b2. - Moment of inertia of a simple pendulum I = ml\u00b2. - Angular acceleration \u03b2 = \u03c4\/I = -(g\/l) sin \u03b1. - Magnitude of angular acceleration is proportional to sin \u03b1.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-angular-acceleration-of-a-simple-pendulum-at-an-angle-%ce%b1-from-the-v\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"The angular acceleration of a simple pendulum at an angle \u03b1 from the v"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87027","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87027"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87027\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87027"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87027"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87027"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}