{"id":2292,"date":"2018-04-30T06:52:02","date_gmt":"2018-04-29T22:52:02","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2292"},"modified":"2023-08-08T13:18:19","modified_gmt":"2023-08-08T13:18:19","slug":"rotation-of-rigid-bodies-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/rotation-of-rigid-bodies-problems-and-solutions.htm","title":{"rendered":"Rotation of rigid bodies \u2013 problems and solutions","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;\">Rotation of rigid bodies \u2013 problems and solutions<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><b>The m<\/b><b>oment of force <\/b><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">1. Three forces act on a beam with a length of 6 meters, as shown in the figure below. What is the <a href=\"https:\/\/gurumuda.net\/physics\/the-magnitude-of-net-torque-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">net torque<\/a> rotates the beam about the point O as the axis of rotation?<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-2293\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Rotation-of-rigid-bodies-\u2013-problems-and-solutions-1.png\" alt=\"Rotation of rigid bodies \u2013 problems and solutions 1\" width=\"250\" height=\"132\" \/><\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><i>The axis of rotation at point O. <\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 1 (F<sub>1<\/sub>) = F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The distance between the line of action of F<sub>1 <\/sub>with the axis of rotation (r<sub>1<\/sub>) = 3 meters <!--more--><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 2 (F<sub>2<\/sub>) = 2F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The distance between the line of action of F<sub>2<\/sub> with the axis of rotation (r<sub>2<\/sub>) = 2 meters <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 3 (F<sub>3<\/sub>) = 2F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The distance between the line of action of F<sub>3<\/sub> with the axis of rotation (r<sub>3<\/sub>) = 3 meters <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> The magnitude of the <a href=\"https:\/\/gurumuda.net\/physics\/moment-of-force-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">moment of force <\/a><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The moment of force 1 :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03c4<sub>1 <\/sub>= F<sub>1<\/sub> r<sub>1<\/sub> = (F)(3) = -3F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The <\/i><i>moment of force <\/i><i>1<\/i><i> rotates beam clockwise so we assign <\/i><i>a negative <\/i><i>sign.<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The moment of force 2 :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03c4<sub>2 <\/sub>= F<sub>2<\/sub> r<sub>2<\/sub> = (2F)(2) = 4F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><i>The moment of force 2 rotates beam counterclockwise so we assign positive sign.<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The moment of force 3 :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03c4<sub>3 <\/sub>= F<sub>3<\/sub> r<sub>3 <\/sub>sin 30<sup>o <\/sup>= (2F)(3)(0.5) = 3F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><i>The moment of force 2 rotates beam counterclockwise so we assign positive sign.<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The resultant of the moment of force :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03a3\u03c4 = \u03c4<sub>1 <\/sub>+ \u03c4<sub>2 <\/sub>+ \u03c4<sub>3<\/sub> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">\u03a3\u03c4 = -3F + 4F + 3F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">\u03a3\u03c4 = 4F<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The magnitude of the moment of force is 4F Newton-meter. The resultant of the moment of force rotates beam counterclockwise so we assign a positive sign.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">2. \u03b1 = 30<sup>o<\/sup>, length of AB = BC = 1 meter. What is the moment of force about the axis of rotation at point A?<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The axis of rotation at point <\/i><i>A. <img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-2294\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/04\/Rotation-of-rigid-bodies-\u2013-problems-and-solutions-2.png\" alt=\"Rotation of rigid bodies \u2013 problems and solutions 2\" width=\"237\" height=\"98\" \/><\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 1 (F<sub>1<\/sub>) = 10 N<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The distance between the line of action of F<sub>1 <\/sub>with the axis of rotation (r<sub>1<\/sub>) = 1 meter <\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 2 (F<sub>2<\/sub>) = 10 N<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The distance between the line of action of F<sub>2 <\/sub>with the axis of rotation (r<sub>2<\/sub>) = 1 meter <\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Force 3 (F<sub>3<\/sub>) = 20 N<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The distance between the line of action of F<sub>3 <\/sub>with the axis of rotation (r<sub>3<\/sub>) = 2 meters<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted:<\/u> The resultant of the moment of force<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The moment of force 1 :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03c4<sub>1 <\/sub>= F<sub>1<\/sub> r<sub>1<\/sub> sin 30<sup>o <\/sup>= (10)(1)(0.5) = 5 N m<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The <\/i><i>moment of force <\/i><i>1<\/i><i> rotates beam <\/i><i>counter<\/i><i>clockwise so we assig<\/i><i>n a positive <\/i><i>sign.<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The moment of force 2 :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03c4<sub>2 <\/sub>= F<sub>2<\/sub> r<sub>2 <\/sub>sin 30<sup>o <\/sup>= (10)(1)(0.5) = -5 Newton meter<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The <\/i><i>moment of force 2 <\/i><i>rotates bea<\/i><i>m <\/i><i>clockwise so we assig<\/i><i>n negative <\/i><i>sign.<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The moment of force 3 :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03c4<sub>3<\/sub> = F<sub>3<\/sub> r<sub>3 <\/sub>sin 60<sup>o <\/sup>= (20)(2)(0.5\u221a3) = -20\u221a3 Newton meter<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><i>The <\/i><i>moment of force 3 <\/i><i>rotates bea<\/i><i>m <\/i><i>clockwise so we assig<\/i><i>n a negative <\/i><i>sign.<\/i><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">The resultant of the moment of force :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03a3\u03c4 = \u03c4<sub>1<\/sub> + \u03c4<sub>2 <\/sub>+ \u03c4<sub>3<\/sub> <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03a3\u03c4 = 5 \u2013 5 &#8211; 20\u221a3 <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">\u03a3\u03c4 = &#8211; 20\u221a3 N m<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The magnitude of the moment of force is 20\u221a3 N m. The resultant of the moment of force rotates beam clockwise so we assign negative sign.<\/span><\/p>\n<div class=\"flex-1 overflow-hidden\">\n<div class=\"react-scroll-to-bottom--css-ezcfc-79elbk h-full dark:bg-gray-800\">\n<div class=\"react-scroll-to-bottom--css-ezcfc-1n7m0yu\">\n<div class=\"flex flex-col text-sm dark:bg-gray-800\">\n<div class=\"group w-full text-token-text-primary border-b border-black\/10 dark:border-gray-900\/50 bg-gray-50 dark:bg-[#444654]\">\n<div class=\"flex p-4 gap-4 text-base md:gap-6 md:max-w-2xl lg:max-w-[38rem] xl:max-w-3xl md:py-6 lg:px-0 m-auto\">\n<div class=\"relative flex w-[calc(100%-50px)] flex-col gap-1 md:gap-3 lg:w-[calc(100%-115px)]\">\n<div class=\"flex flex-grow flex-col gap-3\">\n<div class=\"min-h-[20px] flex flex-col items-start gap-3 overflow-x-auto whitespace-pre-wrap break-words\">\n<div class=\"markdown prose w-full break-words dark:prose-invert light\">\n<ol style=\"text-align: justify;\">\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is a rigid body, and how is it different from a non-rigid body?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> A rigid body is an idealized object in which the distance between any two given points within the body remains constant, regardless of external forces or torques. In contrast, a non-rigid body can deform, allowing the distance between points within the body to change.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How is the moment of inertia of a rigid body related to its mass distribution?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> The moment of inertia of a rigid body is a measure of its resistance to rotational motion about a given axis and depends on both the mass of the body and its distribution relative to the axis of rotation. It is calculated by the sum of the products of each mass element&#8217;s mass and the square of its distance from the axis of rotation.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the significance of the rotational kinetic energy of a rotating rigid body?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> Rotational kinetic energy is a measure of the energy due to the rotation of a rigid body. It depends on both the moment of inertia and the angular velocity of the body, given by 1\/2 <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mord mathnormal\">I <\/span><span class=\"mord\"><span class=\"mord mathnormal\">\u03c9<\/span><sup><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/sup><\/span><\/span><\/span><\/span><\/span>, where <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">I<\/span><\/span><\/span><\/span><\/span> is the moment of inertia, and <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">\u03c9<\/span><\/span><\/span><\/span><\/span> is the angular velocity.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What happens to the angular momentum of a system of particles if no external torques act on it?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> If no external torques act on a system of particles, the total angular momentum of the system is conserved. This is the principle of conservation of angular momentum.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the parallel-axis theorem help in finding the moment of inertia of a rigid body?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> The parallel-axis theorem allows one to calculate the moment of inertia of a rigid body about any axis parallel to and a distance <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">d<\/span><\/span><\/span><\/span><\/span> away from an axis through its center of mass. It states that <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">I<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">I<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathnormal mtight\">c<\/span><span class=\"mord mathnormal mtight\">m<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mbin\">+<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">m<\/span><span class=\"mord\"><span class=\"mord mathnormal\">d<\/span><sup><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/sup><\/span><\/span><\/span><\/span><\/span>, where <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">I<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathnormal mtight\">c<\/span><span class=\"mord mathnormal mtight\">m<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> is the moment of inertia about the center of mass, <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">m<\/span><\/span><\/span><\/span><\/span> is the total mass, and <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">d<\/span><\/span><\/span><\/span><\/span> is the distance between the two axes.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the difference between rolling without slipping and rolling with slipping?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> Rolling without slipping occurs when a rigid body rotates about a fixed axis while also translating, without any relative motion between the body and the surface. Rolling with slipping means there is relative motion or sliding between the body and the surface.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the radius of gyration relate to the moment of inertia?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> The radius of gyration is a measure that describes how the mass of a body is distributed about its axis of rotation. It is defined as the square root of the ratio of the moment of inertia to the mass and provides an equivalent distance from the axis where all the mass could be concentrated without changing the moment of inertia.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What effect does increasing the moment of inertia have on the angular acceleration of a rigid body for a given torque?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> For a given torque, increasing the moment of inertia will decrease the angular acceleration, as <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">\u03b1<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathnormal mtight\">\u03c4\/I<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, where <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">\u03b1<\/span><\/span><\/span><\/span><\/span> is angular acceleration, <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">\u03c4<\/span><\/span><\/span><\/span><\/span> is torque, and <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">I<\/span><\/span><\/span><\/span><\/span> is the moment of inertia.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Can a force applied to a rigid body cause both translational and rotational motion? Explain how.<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> Yes, a force applied to a rigid body can cause both translational and rotational motion. If the force is applied at a point that does not coincide with the center of mass, it can cause the body to translate (move linearly) and rotate. The translational motion is determined by the net force, while the rotational motion depends on the torque created by the force about the center of mass.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Why does a figure skater spin faster when they pull their arms close to their body?<\/strong><\/span><\/li>\n<\/ol>\n<ul style=\"text-align: justify;\">\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> By pulling their arms close to their body, a figure skater reduces their moment of inertia. According to the conservation of angular momentum, if the moment of inertia decreases and no external torque is applied, the angular velocity must increase. Thus, the skater spins faster.<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">These questions and answers provide an understanding of key concepts related to the rotation of rigid bodies.<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>Rotation of rigid bodies \u2013 problems and solutions The moment of force 1. Three forces act on a beam with a length of 6 meters, as shown in the figure below. What is the net torque rotates the beam about the point O as the axis of rotation? Known : The axis of rotation at &#8230; <a title=\"Rotation of rigid bodies \u2013 problems and solutions\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/rotation-of-rigid-bodies-problems-and-solutions.htm\" aria-label=\"Read more about Rotation of rigid bodies \u2013 problems and solutions\">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":"","_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":"Rotation of rigid bodies \u2013 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-2292","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\/2292","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=2292"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2292\/revisions"}],"predecessor-version":[{"id":8613,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2292\/revisions\/8613"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2292"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2292"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2292"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}