HW 6 Mathematics 503, Mathematical Modeling, CSUF , June 24, 2007

Nasser M. Abbasi

June 15, 2014

1 Problem 1 (section 3.5,#9, page 197)
2 Problem 1 (section 3.5,#9, page 197)

1 Problem 1 (section 3.5,#9, page 197)

problem:

Consider a simple plane pendulum with a bob of mass $m$ attached to a string of length $l$ . After the pendulum is set in motion the string is shortened by a constant rate $ddlt = − α$ . Formulate Hamilton's principle and determine the equation of motion. Compare the Hamiltonian to the total energy. Is energy conserved?

Solution:

Assume initial string length is $l$ , and assume $t(0 )= 0$ , then at time $t$ we have

r(t) = l− αt

K.E. First note that

and

Now

P.E.

Hence

( ) ( ) L t,𝜃 (t), _𝜃(t) = 1m α2+ r2 _𝜃2 − mg (l− rcos𝜃 ) 2

(1)

Hence the Euler-Lagrange equations are

L − d-(L_)= 0 𝜃 dt 𝜃

(2)

But

|-----------------| | | | L 𝜃 = − mgrsin𝜃 | ------------------

and

|------------| | L = mr2 _𝜃 | ---_𝜃----------

and

d- ( 2 ) dt (L_𝜃)= m 2r_r_𝜃 + r ¨𝜃

But $r_= − α$ , the above becomes

|------------------------| | ( ) | | ddt(L𝜃_) = m r2 ¨𝜃 − 2rα _𝜃 | --------------------------

Hence $L𝜃 − d(L_𝜃)= 0 dt$ becomes

Hence the ODE becomes, after dividing by common factor $r$

|----------------------| | | | r¨𝜃 − 2α _𝜃 + gsin𝜃 = 0| ------------------------

This is a second order nonlinear diﬀerential equation. Notice that when $l = αt$ it will mean that the string has been pulled all the way back to the pivot and $r(t)= 0$ . So when running the solution it needs to run from $t = 0$ up to $t = l α$ .

A small simulation was done for the above solution which can be run for diﬀerent parameters to see the eﬀect more easily. Here is a screen shot.

Now we need to determine the Hamiltonian of the system.

H = − L (t,𝜃,ϕ (t,𝜃,p))+ ϕ (t,𝜃,p)p

(3)

Where we deﬁne a new variable $p$ called the canonical momentum by

Hence

_𝜃 = -p-- mr2

This implies that

ϕ (t,𝜃 ,p )= -p-2 mr

Then from (1) and (3), we now calculate $H$

Hence the Hamiltonian is

1 p2 1 H = ----2 + mg (l− rcos𝜃)−--mα2 2mr 2

(5)

Now we are asked to compare $H$ to the total energy. The total instantaneous energy of the system is $(T + V)$ , hence we need to determine if $H = T + V$ or not.

1- ( 2 2 _2) T + V = 2 m α + r𝜃 + mg(l− rcos𝜃 )

(6)

To make comparing (5) and (6) easier, I need to either replace $p$ by $mr2 _𝜃$ in (5) or replace $𝜃_$ by $-p2 mr$ . Lets do the later, hence $(6)$ becomes

If $H$ is the total energy, then (7)-(6) should come out to be zero, lets ﬁnd out

Hence we see that

$|-----------------| | H − (T + V)⁄= 0 | -------------------$

Hence $H$ does not represents the total energy, and the energy of the system is not conserved. $1ReadingareferenceonNoether'stheorem,totalenergy is writtenas − (T + V)not(T + V),this would notmakeadiﬀerence in showing thatH ⁄= totalenergy,justdiﬀerentcalculationsresultsasshown below, butthesame conclusion (1 ( ) ) − (T + V)=− 2 m α2+r2_𝜃2+ mg(l− rcos𝜃) (6) Tomake comparing(5)and(6)easier,Ineed toeitherreplacep bymr2_𝜃 in(5)orreplace _𝜃 by mpr2,letdothelater,hence(6) becomes ( 1 ( ( p )2) ) − (T + V)= − 2m α2+r2 mr2 +mg (l− rcos𝜃) ( ( 2 ) ) = − 1m α2+ p22-+ mg(l− rcos𝜃) ( 2 mr2 ) = − 1mα2+ 1-p2 + mg(l− rcos𝜃) (7) 2 2mr If H isthetotalenergy,then (7)- (6)should come outtobezero,letsﬁndout ( ) ( ) H− (− (T + V))= 1p2-+ mg(l− rcos𝜃)− 1mα2 + 1mα2+ 1-p2+ mg(l− rcos𝜃) 2mr2 2 2 2mr2 = -p2+ 2mg(l− rcos𝜃) mr2 p2 Now we ask,cantheabovebezero? Since mr2 isalways≥ 0,andsince (l− rcos𝜃)representstheremaininglength of thestring, 1henceitisapositive quantity(untilsuchtime the stringhasbeenpulled allthewayin),ThereforeRHS above >0.HenceH doesnot representthetotalenergy ofthe system. Hencetheenergy isnotconserved.$

2 Problem 1 (section 3.5,#9, page 197)

problem: A bead of mass $m$ slides down the rim of a circular hoop of radius $R$ . The hoop stands vertically and rotates about its diameter with angular velocity $ω$ . Determine the equation of motion of the bead.

Answer:

Kinetic energy $T$ is made up of 2 components, one due to motion of the bead along the hoop itself with speed $R_𝜃$ , and another due to motion with angular speed $ω$ which has speed given by $R sin𝜃 ω$