[PDF] Classical Mechanics LECTURE 17: EFFECTIVE POTENTIAL & SIMPLE

Ueff(r) =J22mr2GmMr

IUeff(r) I

Etot<0 : Bound

(closed) orbit with r

1 I

Etothas minimum

energy atr=r0: dU effdr =0 , circular motion with _r=0 I

Etot>0 : Unbound

(open) orbit with r>r34

17.2 Examples

17.2.1 Example 1 : 2-D harmonic oscillatorI

F=kr(ignore the natural length of the spring)

I

Energy equation :

E=12 m_r2+Ueff(r) I

Ueff(r) =J22mr2+12

kr2 I

For circular motion_r=0 :

E minwhen@Ueff@rjr0=0 J2mr

30+kr0=0

whereJ=mv0r0 I

Leads tomv20r

0=k r0

as expectedIncluding the natural length: I

F=kr!F=k(ra)

I U=12 kr2!U=12 k(ra)2

Leads to

mv20r

0=k(r0a)5

Example continued

I

Ueff(r) =J22mr2+12

kr2 I

For general motion :

I

F=kr!mx=kx

!my=ky I

Solution for B.C"s att=0:

x=r2;y=0;_x=0 !x=r2cos!t !y=r1sin!t where!2=km

IEllipse:(xr

2)2+ (yr

1)2=16

17.2.2 Example 2 : Rotating ball on table

Two particles of massmare connected by a light inextensible string of length`. The particle on the table starts att=0 at a distance`=2 from the hole at a speedv0perpendicular to the string. Find the speed at which the particle below the table falls.I

Energy equation :

E=12 m_r2+J22mr2+U(r) Iquotesdbs_dbs2.pdfusesText_3