{"id":92,"date":"2016-03-01T16:00:20","date_gmt":"2016-03-01T16:00:20","guid":{"rendered":"http:\/\/physicscatalyst.com\/graduation\/?p=92"},"modified":"2020-02-25T06:20:59","modified_gmt":"2020-02-25T06:20:59","slug":"central-forces","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/graduation\/central-forces\/","title":{"rendered":"Motion Under Central Forces (part 1)"},"content":{"rendered":"
\n

Motion Under Central forces (series List)<\/strong><\/h4>\n

Introduction<\/a>
\nEquation of motion under central forces<\/a>
\n
\nLaw of conservation of angular momentum<\/a>
\nLaw of conservation of energy<\/a>
\nEquation of motion (equation of path of moving particle)<\/a>
\n
\nForm of motion under the effect of central forces
\n<\/a><\/p>\n<\/div>\n

This article covers an introduction to central forces , equation of motion under central forces and<\/p>\n

Introduction<\/h2>\n

If the force \\(\\mathop f\\limits^ \\to \\) acting on a body has following characteristics then it is a central force
\n(i) it depends on the distance between two particles
\n(ii) it is always directed towards or away from a fixed point.
\nGravitational force is an example of central forces. Mathematically if we
\nconsider central point as origin
\n\\[\\overrightarrow f (r) = f(r)( \\pm \\hat r)\\]
\nwhere + stands for repulsion forces and – for attractive forces , \\(f(r)\\) is
\nthe magnitude of the central forces and \\(\\hat r\\) is unit vector in the
\ndirection of central forces.<\/p>\n

Theorem:- <\/strong>Under the action of central forces particle
\nalways move in the same plane.
\nMultiplying equation (1) by \\(\\vec r\\) on both the sides we get<\/p>\n

\\[\\vec r \\times \\vec f(r) = f(r)\\left\\{ {\\vec r \\times \\hat r} \\right\\}\\]<\/p>\n

This gives
\n\\[f(r) = 0\\]
\nsince \\({\\vec r \\times \\hat r = 0}\\)
\nbut we know that
\n\\[\\vec f(r) = m\\frac{{{d^2}\\vec r}}{{d{t^2}}}\\]
\ntherefore \\[\\vec r \\times m\\frac{{{d^2}\\vec r}}{{d{t^2}}} = 0\\]
\nor, \\[\\vec r \\times \\frac{{{d^2}\\vec r}}{{d{t^2}}} = 0\\]
\nnow,<\/p>\n

\\[\\frac{d}{{dt}}\\left( {\\vec r \\times \\frac{{d\\vec r}}{{dt}}} \\right) = \\vec r \\times \\frac{{{d^2}\\vec r}}{{d{t^2}}} + 0\\]
\n\\[\\frac{d}{{dt}}\\left( {\\vec r \\times \\frac{{d\\vec r}}{{dt}}} \\right) = 0\\]
\n\\[\\left( {\\vec r \\times \\frac{{d\\vec r}}{{dt}}} \\right) = cons\\tan t(\\vec h)\\]
\nwhere \\({\\vec h}\\) is a vector which does not depend on time and is perpandicular to the plane formed by the position vector \\({\\vec r}\\) and velocity \\({\\frac{{d\\vec r}}{{dt}}}\\) . Thus the plane formed by \\({\\vec r}\\) and velocity \\({\\frac{{d\\vec r}}{{dt}}}\\) will also remains constant. Therefore the particle will always move in the same plane.
\nThe torque due to central forces can be found out by the above expression. Torque
\n\\[\\tau = \\vec r \\times \\vec f(r) = \\vec r \\times f(r)\\hat r = 0\\]
\nbut
\n\\[\\tau = \\frac{{d\\vec J}}{{dt}}\\]
\nThis implies thet \\({\\vec J}\\) is a constant quantity and it is known as angular momentum. Thus angular momentum is also conserved. But
\n\\[\\vec J = \\vec r \\times \\vec p = \\vec r \\times m\\frac{{d\\vec r}}{{dt}} = \\vec h\\]<\/p>\n

Equation of motion under central forces<\/h2>\n

The equation of particle moving in a plane under
\ncentral forces can be obtained in terms of polar co-ordinates. Let \\(m\\) be the
\nmass of the particle, \\((x,y)\\) be its cartesian co-ordinates and \\((r,\\theta
\n)\\) be the polar co-ordinates of the particle.
\ntherefore, \\(x = r\\cos \\theta \\) and \\(y = r\\sin \\theta \\)
\nDifferentiating it with respect to time we get
\n\\[\\frac{{dx}}{{dt}} = \\frac{{dr}}{{dt}}\\cos \\theta – r\\sin \\theta
\n\\frac{{d\\theta }}{{dt}}\\]
\n\\[\\frac{{dy}}{{dt}} = \\frac{{dr}}{{dt}}\\sin \\theta + r\\cos \\theta
\n\\frac{{d\\theta }}{{dt}}\\]
\nagain differentiating above two equations with respect to time we get
\n\\[\\frac{{{d^2}x}}{{d{t^2}}} = \\left[ {\\frac{{{d^2}r}}{{d{t^2}}} – r{{\\left(
\n{\\frac{{d\\theta }}{{dt}}} \\right)}^2}} \\right]\\cos \\theta – \\left[
\n{r\\frac{{{d^2}\\theta }}{{d{t^2}}} + 2\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}}
\n\\right]\\sin \\theta \\]
\nsimilarly,
\n\\[\\frac{{{d^2}y}}{{d{t^2}}} = \\left[ {\\frac{{{d^2}r}}{{d{t^2}}} – r{{\\left(
\n{\\frac{{d\\theta }}{{dt}}} \\right)}^2}} \\right]\\sin \\theta + \\left[
\n{r\\frac{{{d^2}\\theta }}{{d{t^2}}} + 2\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}}
\n\\right]\\cos \\theta \\]
\nnow acceleration
\n\\[\\vec a = \\left( {\\hat i\\frac{{{d^2}x}}{{d{t^2}}} + \\hat
\nj\\frac{{{d^2}y}}{{d{t^2}}}} \\right) = \\left[ {\\frac{{{d^2}r}}{{d{t^2}}} –
\nr{{\\left( {\\frac{{d\\theta }}{{dt}}} \\right)}^2}} \\right]\\left( {\\hat i\\cos
\n\\theta + \\hat j\\sin \\theta } \\right) – \\left[ {r\\frac{{{d^2}\\theta }}{{d{t^2}}}
\n+ 2\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}} \\right]\\left( { – \\hat i\\sin
\n\\theta + \\hat j\\cos \\theta } \\right)\\]
\nWe know that in radial direction unit vector is
\n\\(\\hat r = \\left( {\\hat i\\cos \\theta + \\hat j\\sin \\theta } \\right)\\)
\nand in transverse direction unit vector is
\n\\(\\hat \\theta = \\left( { – \\hat i\\sin \\theta + \\hat j\\cos \\theta } \\right)\\)<\/p>\n

this vector is perpandicular to the radial vector.
\nNow acceleration is \\(\\vec a = (\\hat r{a_r} + \\hat \\theta {a_\\theta })\\) and
\nthis implies that<\/p>\n

\\[\\vec a = \\left[ {\\frac{{{d^2}r}}{{d{t^2}}} – r{{\\left( {\\frac{{d\\theta
\n}}{{dt}}} \\right)}^2}} \\right]\\hat r – \\left[ {r\\frac{{{d^2}\\theta }}{{d{t^2}}}
\n+ 2\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}} \\right]\\hat \\theta \\]
\n\\[{a_r} = \\left[ {\\frac{{{d^2}r}}{{d{t^2}}} – r{{\\left( {\\frac{{d\\theta }}{{dt}}} \\right)}^2}} \\right]\\tag{2}\\]<\/p>\n

\\[{a_\\theta } = \\left[ {r\\frac{{{d^2}\\theta }}{{d{t^2}}} + 2\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}} \\right]\\tag{3}\\]
\nAccording to Newton’s second law of motion
\n\\[\\vec f(r) = f(r)\\hat r\\]
\n\\[f(r)\\hat r = m(\\hat r{a_r} + \\hat \\theta {a_\\theta })\\]
\nOn comparison
\n\\[m\\left[ {\\frac{{{d^2}r}}{{d{t^2}}} – r{{\\left( {\\frac{{d\\theta }}{{dt}}} \\right)}^2}} \\right] = f(r)\\tag{4}\\]
\nand
\n\\[m\\left[ {r\\frac{{{d^2}\\theta }}{{d{t^2}}} + 2\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}} \\right] = 0\\tag{5}\\]
\nEquations 4 and 5 are called the equation of motion of the particle moving under central forces.<\/p>\n

Law of conservation of angular momentum<\/h2>\n

Multiplying equation (5) of previous page by r on
\nboth the sides we get
\n\\[m\\left[ {{r^2}\\frac{{{d^2}\\theta }}{{d{t^2}}} + 2r\\frac{{dr}}{{dt}}\\frac{{d\\theta }}{{dt}}} \\right] = 0\\]
\nThis implies that
\n\\[\\frac{d}{{dt}}\\left( {m{r^2}\\frac{{d\\theta }}{{dt}}} \\right) = 0\\tag{6}\\]
\nor,
\n\\[\\frac{{dJ}}{{dt}} = 0\\tag{7}\\]
\nwhere,\\(J = \\left( {m{r^2}\\frac{{d\\theta }}{{dt}}} \\right) = constant\\tag{8}\\)<\/p>\n

Angular momentum of a particle moving with rotational motion is \\(\\left( {m{r^2}\\frac{{d\\theta }}{{dt}}} \\right)\\) therefore equation 7 shows law of conservation of angular momentum of moving particle under the influence of central force.
\n\\(J = \\left( {m{r^2}\\frac{{d\\theta }}{{dt}}} \\right)\\) is always conserved. Now if equation (6) is multiplied by the quantity \\(\\frac{1}{{2m}}\\) then we get another physical quantity which is AREAL VELOCITY. It also remains constant for the particle moving under central forces
\nThis implies that
\n\\[\\frac{d}{{dt}}\\left( {\\frac{1}{2}{r^2}\\frac{{d\\theta }}{{dt}}} \\right) = 0\\]
\nor,<\/p>\n

\\[\\frac{1}{2}{r^2}\\frac{{d\\theta }}{{dt}} = constant(invarient)\\tag{9}\\]
\nIt defines areal velocity as the area swept by position vector \\({\\vec r}\\) of particle in unit time. Therefore we come across to know that the law of conservation of angular momentum of central forces expresses the invariance of areal velocity. It is actually Kepler’s Second law of planetary motion which states that “Every planet moves with a constant areal velocity around the sun”<\/em><\/strong><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

This article covers an introduction to central forces , equation of motion under central forces and<\/p>\n

Introduction<\/h2>\n

If the force \\(\\mathop f\\limits^ \\to \\) acting on a\u00a0body has following characteristics then it is a central force (i) it depends on the distance between two particles (ii) it is always directed towards or away from a fixed point. Gravitational force is an example of central forces. Mathematically if we consider central point as origin <\/p>\n","protected":false},"author":1,"featured_media":122,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[10],"tags":[],"class_list":["post-92","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-newtonian-mechanics"],"yoast_head":"\nMotion Under Central Forces (part 1) - Learn about education and B.Sc. 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