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Central Force Umiatin,M.Si
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The aim : to evaluate characteristic of motion under central force field
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A. Introduction Central Force always directed along the line connecting the center of the two bodies Occurs in : motion of celestial bodies and nuclear interaction
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Central Force Motion as One Body Problem Suppose isolated system consist two bodies and separated a distance r = |r| with interaction between them described by a central force F(r), we need six quantities used to describe motion of those particle :
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Method 1 : To describe those, we need six quantities ( three component of r 1 and three component of r 2 ). The equation of motion of those particle are : If F(r) > 0: repusive, F(r) <0 : Attractive. Coupled by :
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Method 2 Describe a system using center mass (R) and relative position (r). R describes the motion of the center of mass and r describes the relative motion of one particle with respect to the other
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No external forces are acting on the system, so the motion of the center of mass is uniform translational motion. R** = 0.
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Where reduced mass define by : Two bodies problem has been simplified into one body problem.
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Solve the equation of motion : The center of mass moves with uniform velocity : By choosing the initial condition, vo, to, Ro = 0, the origin of coordinate coincides with center of mass R.
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So the position of m1 and m2 which measured from center of mass :
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If m2 >> m1, then reduced mass: The eq of motion : Become :
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Hence the problem can be treated as a one body problem. Thus, whenever we use mass m instead of µ, we are indicating that the other mass is very large, whereas the use of µ indicates that either the two masses are comparable.
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B. General properties of Central Force 1.Central Force is Confined to a Plane If p is the linear momentum of a particle of mass µ, the torque τ about an axis passing through the center of force is :
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If the angular momentum L of mass µ is constant, its magnitude and direction are fixed in space. Hence, by definition of the cross product, if the direction of L is fixed in space, vectors r and p must lie in a plane perpendicular to L. That is, the motion of particle of mass µ is confined to a plane that is perpendicular to L.
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As we the force acting at body is central force, three dimensional problem can be reduced into two dimensional. Using polar coordinate system :
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2. Angular Momentum and Energy are Constant The angular momentum of a particle of mass µ at a distance r from the force center is :
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Since there are no dissipative systems and central forces are conservative, the total energy is constant :
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3. Law of Equal Areas Consider a mass µ at a distance r( θ ) at time t from the force center O :
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Subtituting
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C. Equation of Motion From the previous description : If we know V(r), these equations can be solved for θ (t) and r(t). The set [ θ (t), r(t)] describes the orbit of the particle.
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Solve the equation, we find : We will get t(r) then inverse r(t). But we are interested in the equation of the path in term r and θ
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We may write : And subtitute : Then :
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Suppose the Force is F(r) = Kr n K = constant If n = 1 the solution is motion of harmonic oscillator If n = -2, eq : coulomb and gravitation force
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Other method : Use Lagrangian Lagrangian of the system : We find
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To simplify, use other variable, for example : u in which = 1/r
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Next find Therefore :
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We can transform into :
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Example 1. Find the force law for a central force field that allow a particle to move in logarithmic spiral orbit given by (k and α are constant) :
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Solution : First determine :
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Now determine :
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2. Find r(t) and θ(t) ! Solution :
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3. What is the total energy ? Solution :
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We know that
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D. Planetary Motion The equation for the path of a particle moving under the influence of a central force whose magnitude is inversely proportional to the distance between the particle can be obtain from :
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If we define the origin of θ so that the minimum value of r occurs at θ = 0, so
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