Mechanics 1: Motion in a Central Force Field

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1 Mechanics : Motion in a Cental Foce Field We now stud the popeties of a paticle of (constant) ass oving in a paticula tpe of foce field, a cental foce field. Cental foces ae ve ipotant in phsics and engineeing. Fo eaple, the gavitional foce of attaction between two point asses is a cental foce. The Coulob foce of attaction and epulsion between chaged paticles is a cental foce. Because of thei ipotance the deseve special consideation. We begin b giving a pecise definition of cental foce, o cental foce field. Cental Foces: The Definition. Suppose that a foce acting on a paticle of ass has the popeties that: the foce is alwas diected fo towad, o awa, fo a fied point O, the agnitude of the foce onl depends on the distance fo O. Foces having these popeties ae called cental foces. The paticle is said to ove in a cental foce field. The point O is efeed to as the cente of foce. Matheaticall, F is a cental foce if and onl if: F = f() = f(), () whee = is a unit vecto in the diection of. If f() < 0 the foce is said to be attactive towads O. If f() > 0 the foce is said to be epulsive fo O. We give a geoetical illustation in Fig.. z F=f() O Figue : Geoetical illustation of a cental foce. Popeties of a Paticle Moving unde the Influence of a Cental Foce. If a paticle oves in a cental foce field then the following popeties hold:. The path of the paticle ust be a plane cuve, i.e., it ust lie in a plane.. The angula oentu of the paticle is conseved, i.e., it is constant in tie. 3. The paticle oves in such a wa that the position vecto (fo the point O) sweeps out equal aeas in equal ties. In othe wods, the tie ate of change in aea is constant. This is efeed to as the Law of Aeas. We will descibe this in oe detail, and pove it, shotl.

2 Equations of Motion fo a Paticle in a Cental Foce Field. Now we will deive the basic equations of otion fo a paticle oving in a cental foce field. Fo Popet above, the otion of the paticle ust occu in a plane, which we take as the plane, and the cente of foce is taken as the oigin. In Fig. we show the plane, as well as the pola coodinate sste in the plane. θ j O θ i cosθ sinθ Figue : Pola coodinate sste associated with a paticle oving in the plane. Since the vectoial natue of the cental foce is epessed in tes of a adial vecto fo the oigin it is ost natual (though not equied!) to wite the equations of otion in pola coodinates. In ealie lectues we deived the epession fo the acceleation of a paticle in pola coodinates: a = ( θ ) + ( θ + ṙ θ)θ. () Then, using Newton s second law, and the atheatical fo fo the cental foce given in (), we have: o ( θ ) + ( θ + ṙ θ)θ = f(), (3) ( θ ) = f(), (4) ( θ + ṙ θ) = 0. (5) These ae the basic equations of otion fo a paticle in a cental foce field. The will be the stating point fo an of ou investigations. Fo these equations we can deive a useful constant of the otion. This is done as follows. Fo (5) we have: o ( θ + ṙ θ) = ( θ + ṙ θ) = d dt ( θ) = 0, θ = constant = h. (6) This is an inteesting elation that, we will see, is elated to popeties and 3 above. Howeve, one use fo it should be appaent. If ou know the coponent of the otion it allows ou to copute the θ coponent b integation. This is anothe eaple of how constants of the otion allow us to integate the equations of otion. It also eplain wh constants of the otion ae often efeed to as integals of the otion. Now, let us etun to popet 3 above and deive the Law of Aeas.

3 Suppose that in tie t the position vecto oves fo to +. Then the aea swept out b the position vecto in this tie is appoiatel half the aea of a paalleloga with sides and. We give a poof of this: Aea of paalleloga = height, = sin θ, =, see Fig. 3. z Ȧ = aeal velocit = _ ( v ) + Δ Aea = ΔA Q θ Δ P Figue 3: Hence, A =. Dividing this epession b t, and letting t 0, gives: A li t 0 t = li t 0 t = v, o Ȧ = v. Now we need to evaluate v. Using =, we have: Theefoe we have: v = (ṙ + θθ ) = ṙ( ) + θ( θ ) = θk. The vecto: θ = Ȧ = constant. (7) Ȧ = Ȧk = θk, is called the aeal velocit. 3

4 Altenative Fos to the Basic Equations of Motion fo a Paticle in a Cental Foce Field. Recall the basic equations of otion as the will be ou stating point: we deived the following constant of the otion: ( θ ) = f(), (8) ( θ + ṙ θ) = 0. (9) θ = h = constant. (0) This constant of the otion will allow ou to deteine the θ coponent of otion, povided ou know the coponent of otion. Howeve, (8) and (9) ae coupled (nonlinea) equations fo the and θ coponents of the otion. How could ou solve the without solving fo both the and θ coponents? This is whee altenative fos of the equations of otion ae useful. Let us ewite (8) in the following fo (b dividing though b the ass ): θ = f(). () Now, using (0), () can be witten entiel in tes of : h 3 = f(). () We can use () to solve fo (t), and the use (0) to solve fo θ(t). Equation () is a nonlinea diffeential equation. Thee is a useful change of vaiables, which fo cetain ipotant cental foces, tuns the equation into a linea diffeential equation with constant coefficients, and these can alwas be solved analticall. Hee we descibe this coodinate tansfoation. Let = u. This is pat of the coodinate tansfoation. We will also use θ as a new tie vaiable. Coodinate tansfoation ae effected b the chain ule, since this allows us to epess deivatives of old coodinates in tes of the new coodinates. We have: and ṙ = d dt = d dθ dθ dt = h d dθ = h d du du dθ = hdu dθ, (3) = dṙ dt = d ( h du ) = d ( h du ) dθ dt dθ dθ dθ dt = h u d u dθ, (4) whee, in both epessions, we have used the elation θ = h at stategic points. Now θ = h 4 = h u 3. (5) Substituting this elation, along with (4) into (8), gives: o ( ) h u d u dθ h u 3 = f ( ), u d u dθ + u = f ( u) h u. (6) Now if f() = K, whee K is soe constant, (6) becoes a linea, constant coefficient equation. 4

5 Cental Foce Fields ae Consevative. Now we will show that cental foces ae consevative foces. We alead know that thee ae an ipotant iplications that will follow fo this fact, such as consevation of total eneg. If a cental foce is consevative then the wok done b the foce in oving a paticle between two points is independent of the path taken between the two points, i.e., it onl depends on the endpoints of the path. In this case we ust have: F d = dv whee V is a scala valued function (the potential). Evaluating the left-hand-side of this epession gives: F d = f() d = f()d, whee we have used the elation d = d. Theefoe, fo which it follows that: dv = f()d, V = f()d. (7) Hence, if we know the cental foce field, (7) tells us how to copute the potential. Consevation of Eneg fo a Paticle in a Cental Foce Field. Since cental foces ae consevative foces, we know that total eneg ust be conseved. Now we deive epessions fo the total eneg of a paticle of ass in a cental foce field. We will do this in two was. Fist Method. Fist we copute the kinetic eneg. The velocit is given b: and theefoe: The kinetic eneg is given b: Theefoe we have: v = ṙ + θθ, v v = v = ṙ + θ. v + V = E. ( ) ṙ + θ f()d = E. (8) Second Method. The second ethod deals diectl with the equations of otion and ealizes the epession fo the total eneg as an integal of the equations of otion. We ultipl (4) b ṙ, ultipl (5) b θ, and add the esulting two equations to obtain: o, (ṙ + θ θ + ṙ θ ) = f()ṙ = ṙ d d f()d = d d dt d d ( ) ṙ + θ = d dt dt f()d. Integating both sides of this equation with espect to tie gives: ( ) ṙ + θ f()d = E = constant. f()d = d dt f()d, You can deive this elation b noting that =, and then coputing the diffeential of this equalit. 5

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