{"id":101,"date":"2016-03-01T16:05:48","date_gmt":"2016-03-01T16:05:48","guid":{"rendered":"http:\/\/physicscatalyst.com\/graduation\/?p=101"},"modified":"2020-02-25T06:20:53","modified_gmt":"2020-02-25T06:20:53","slug":"central-forces-2","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/graduation\/central-forces-2\/","title":{"rendered":"Motion Under Central Forces (part 2)"},"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 topics like Law of conservation of energy, Equation of motion and Form of motion under the effect of central forces.<\/p>\n

Law of conservation of energy<\/h2>\n

From equation 4 we consider
\n\\[f(r) = – \\frac{{du(r)}}{{dr}}\\]
\nthen this shows that \\(f(r)\\) is not only a central force but a conservative force. In this condition this equation<\/p>\n

also express the law of conservation of energy.
\nPutting equation 8 in 4 we get
\n\\[m\\frac{{{d^2}r}}{{d{t^2}}} – \\frac{{r{J^2}}}{{m{r^4}}} = – \\frac{{du}}{{dr}}\\]
\nor,
\n\\[m\\frac{{{d^2}r}}{{d{t^2}}} = – \\frac{{du}}{{dr}} + \\frac{{r{J^2}}}{{m{r^4}}}\\]
\nSince we know that
\n\\[\\frac{d}{{dr}}\\left( {\\frac{1}{{{r^2}}}} \\right) = – \\frac{2}{{{r^3}}}\\]
\nTherefore
\n\\[m\\frac{{{d^2}r}}{{d{t^2}}} = – \\frac{d}{{dr}}\\left( {u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right)\\tag{10}\\]
\n\\[m\\left( {\\frac{{dr}}{{dt}}} \\right)\\left( {\\frac{{{d^2}r}}{{d{t^2}}}} \\right) = – \\frac{{dr}}{{dt}}\\frac{d}<\/p>\n

{{dr}}\\left( {u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right)\\]
\nby which
\n\\[\\frac{d}{{dt}}\\left[ {\\frac{1}{2}m{{\\left( {\\frac{{dr}}{{dt}}} \\right)}^2}} \\right] = – \\frac{d}{{dt}}\\left( {u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right)\\]
\nor,<\/p>\n

\\[\\frac{d}{{dt}}\\left[ {\\frac{1}{2}m{{\\left( {\\frac{{dr}}{{dt}}} \\right)}^2} + u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right] = 0\\tag{11}\\]
\nor,
\n\\[\\left[ {\\frac{1}{2}m{{\\left( {\\frac{{dr}}{{dt}}} \\right)}^2} + u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right] = {\\rm{constant}} = E(let)\\tag{12}\\]
\nEquation 12 shows the conservation of total energy. In equation 12 term $\\frac{1}{2}m{\\left( {\\frac{{dr}}{{dt}}} \\right)^2}$ represents transversal kinetic energy<\/b><\/em> and term \\({\\frac{{{J^2}}}{{2m{r^2}}}}\\) is rotational <\/b><\/em><\/p>\n

kinetic energy and \\(u\\) is the potential energy<\/em>.
\nHere,
\n\\[\\frac{{{J^2}}}{{2m{r^2}}} = {\\left( {m{r^2}\\frac{{d\\theta }}{{dt}}} \\right)^2}\\frac{1}{{2m{r^2}}} = \\frac{1}{2}m{r^2}{\\left( {\\frac{{d\\theta }}{{dt}}} \\right)^2} = \\frac{1}{2}I{\\omega ^2}\\tag{13}\\]<\/p>\n

Equation of motion (equation of path of moving particle)<\/h2>\n

Equation 12 can be written as follows,
\n\\[\\frac{{dr}}{{dt}} = {\\left[ {\\frac{2}{m}\\left( {E – u – \\frac{{{J^2}}}{{2m{r^2}}}} \\right)} \\right]^ {{\\raise0.5ex\\hbox{$\\scriptstyle 1$}
\n\\kern-0.1em\/\\kern-0.15em
\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}\\]
\n\\[dt = {\\left[ {\\frac{2}{m}\\left( {E – u – \\frac{{{J^2}}}{{2m{r^2}}}} \\right)} \\right]^{ – {\\raise0.5ex\\hbox {$\\scriptstyle 1$}
\n\\kern-0.1em\/\\kern-0.15em
\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}dr\\tag{14}\\]
\nOn integrating left hand side from \\(\\theta \\) to \\(t\\) and right hand side from \\({{r_0}}\\) to \\({{r}}\\) we get
\n\\(\\int\\limits_0^t {dt} = \\int\\limits_{{r_0}}^r {{{\\left[ {\\frac{2}{m}\\left( {E – u – \\frac{{{J^2}}}{{2m{r^2}}}} \\right)} \\right]}^{ – {\\raise0.5ex\\hbox{$\\scriptstyle 1$}
\n\\kern-0.1em\/\\kern-0.15em
\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}dr} \\tag{15}\\)
\nWe also know that angular momentum
\n\\[J = m{r^2}\\left( {\\frac{{d\\theta }}{{dt}}} \\right)\\]
\n\\[d\\theta = \\frac{J}{{m{r^2}}}dt\\]
\n\\[\\theta = \\int\\limits_0^t {\\frac{J}{{m{r^2}}}dt} \\tag{16}\\]
\nHere \\({{r}}\\) is dependent on \\({{t}}\\).
\nEquations 15 and 16 are theoretically used to explain the relation between \\({{r}}\\) and \\({{t}}\\) and \\(\\theta \\) and \\({{t}}\\) if we know the nature of central forces from equation 14. Now if we put \\(dt\\) in terms of \\({{r}}\\) in equation 16 and solve it , then its solution shows the progress of motion. In other words we can say that polar co-ordinates \\({{r}}\\) and \\(\\theta \\) of a particle can be known with respect to time \\({{t}}\\) otherwise path of moving particle under central forces can be known by eleminating \\({{t}}\\) from equation 4.
\nWe know that
\n\\[J = m{r^2}\\left( {\\frac{{d\\theta }}{{dt}}} \\right)\\]
\n\\[\\frac{{d\\theta }}{{dt}} = \\frac{J}{{m{r^2}}}\\frac{{d(\\theta )}}{{d(\\theta )}}\\]
\nTherefore in the form of operator
\n\\[\\frac{d}{{dt}} = \\frac{J}{{m{r^2}}}\\frac{d}{{d\\theta }}\\]
\nand
\n\\[\\frac{d}{{dt}}\\left( {\\frac{d}{{dt}}} \\right) = \\frac{J}{{m{r^2}}}\\frac{d}{{d\\theta }}\\left( {\\frac{J}{{m{r^2}}} \\frac{d}{{d\\theta }}} \\right)\\]
\nthis implies that
\n\\[\\frac{{{d^2}}}{{d{t^2}}} = \\frac{{{J^2}}}{{{m^2}{r^4}}}\\frac{{{d^2}}}{{d{\\theta ^2}}}\\tag{17}\\]
\nNow from equation 4
\n\\[m\\left[ {\\frac{{{d^2}r}}{{d{t^2}}} – r{{\\left( {\\frac{{d\\theta }}{{dt}}} \\right)}^2}} \\right] = f(r)\\]
\n\\[m\\left[ {\\frac{{{J^2}}}{{{m^2}{r^4}}}\\frac{{{d^2}r}}{{d{\\theta ^2}}} – r\\frac{{{J^2}}}{{{m^2}{r^4}}}} \\right] = f(r)\\]
\n\\[\\frac{{{J^2}}}{{m{r^2}}}\\frac{d}{{d\\theta }}\\left( {\\frac{1}{{{r^2}}}\\frac{{dr}}{{d\\theta }}} \\right) – \\frac {{{J^2}}}{{m{r^3}}} = f(r)\\]
\nbut
\n\\[\\frac{1}{{{r^2}}}\\frac{{dr}}{{d\\theta }} = – \\frac{d}{{d\\theta }}\\left( {\\frac{1}{r}} \\right)\\]
\ntherefore
\n\\[\\frac{{{J^2}}}{{m{r^2}}}\\left[ {\\frac{{{d^2}}}{{d{\\theta ^2}}}\\left( {\\frac{1}{r}} \\right) + \\left( {\\frac{1}{r}} \\right)} \\right] = – f(r)\\tag{18}\\]
\nif \\(u = \\left( {\\frac{1}{r}} \\right)\\) , then
\n\\[\\frac{{{J^2}{u^2}}}{m}\\left[ {\\frac{{{d^2}u}}{{d{\\theta ^2}}} + u} \\right] = – F\\left( {\\frac{1}{u}} \\right) \\tag{19}\\]
\nThis equation is the equation of path of moving particle. This can be solved if nature and magnitude of force is known. If path of the moving particle is known then we can find about the nature of the force.<\/p>\n

Form of motion under the effect of central forces<\/h2>\n

According to equation 10
\n\\[m\\frac{{{d^2}r}}{{d{t^2}}} = – \\frac{d}{{dr}}\\left( {u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right) = – \\frac{{du’}}{{dr}} = f'(r)\\]
\nwhere,\\(u’ = \\left( {u + \\frac{{{J^2}}}{{2m{r^2}}}} \\right)\\)
\nthat is
\n\\[f'(r) = – \\frac{{du}}{{dr}} – \\frac{d}{{dr}}\\left( {\\frac{{{J^2}}}{{2m{r^2}}}} \\right)\\]
\n\\[f'(r) = f(r) + \\frac{{{J^2}}}{{m{r^3}}}\\tag{20}\\]
\nHere, \\({\\frac{{{J^2}}}{{2m{r^2}}}}\\) is potential energy. According to centrifugal forces
\n\\[\\frac{{{J^2}}}{{m{r^3}}} = m{\\omega ^2}r\\]
\nNow we can write the magnitude of total energy from equation 12
\n\\[E = \\frac{1}{2}m{\\left( {\\frac{{dr}}{{dt}}} \\right)^2} + u + \\frac{{{J^2}}}{{2m{r^2}}}\\]
\nor,
\n\\[E = \\frac{1}{2}m{\\left( {\\frac{{dr}}{{dt}}} \\right)^2} + u’\\tag{21}\\]
\nHence
\n\\[u’ = u + \\frac{{{J^2}}}{{2m{r^2}}}\\]
\nInverse square central forces curves for \\(u\\) , \\(u’\\) and \\(\\frac{{{J^2}}}{{2m{r^2}}}\\) are shown below in the figure.
\n\"central_forces\"
\nHere magnitude of kinetic energy is \\(\\frac{1}{2}m{\\left( {\\frac{{dr}}{{dt}}} \\right)^2}\\) and is always positive. Also centrifugal energy \\(\\frac{{{J^2}}}{{2m{r^2}}}\\) is positive whereas potential energy \\(u\\) can be positive as well as negative. \\(u’ = u + \\frac{{{J^2}}}{{2m{r^2}}}\\) that is effective potential energy can be positive or negative.
\nPositive magnitude of \\(u’\\) means
\n\\[\\left| {\\frac{{{J^2}}}{{2m{r^2}}}} \\right| > \\left| u \\right|\\]
\nNegative magnitude of \\(u’\\) means
\n\\[\\left| {\\frac{{{J^2}}}{{2m{r^2}}}} \\right| < \\left| u \\right|\\]
\nTherefore the positive and negative values of \\(u’\\) will depend on the relative magnitude of \\(u\\) and \\(\\frac{{{J^2}}}{{2m{r^2}}}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

This article covers topics like Law of conservation of energy, Equation of motion and Form of motion under the effect of central forces.<\/p>\n

Law of conservation of energy<\/h2>\n

From equation 4 we consider \\[f(r) = – \\frac{{du(r)}}{{dr}}\\] then this shows that \\(f(r)\\) is not only a central force but a conservative force. In this condition this equation<\/p>\n

<\/p>\n","protected":false},"author":1,"featured_media":0,"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-101","post","type-post","status-publish","format-standard","hentry","category-newtonian-mechanics"],"yoast_head":"\nMotion Under Central Forces (part 2) - Learn about education and B.Sc. 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