Motion on a paraboloid: Consider a particle of mass m that moves, without fricti

Question

Motion on a paraboloid: Consider a particle of mass m that moves, without friction, on the smooth paraboloid z = x^2 + y^2 while experiencing the constant force of gravity (so the gravitational potential is V = mgz). (a) Write down the Lagrangian for this system using the Cartesian (rectangular) coordinates x and y as your generalized coordinates. Determine the equations of motion and simplify, taking advantage of obvious first integrals. (b) Write down the Lagrangian for this system using the cylindrical coordinates r and theta as your generalized coordinates. Determine the equations of motion and simplify, taking advantage of obvious first integrals.

Explanation / Answer

(a)as given equation of constraint is

z = x2 + y2

now let us write time derivative of z as z.

now z. in terms of x and y is, z. = 2xx. + 2yy.

where x. and y. are time derivatives of x and y.

Now, kinetic energy of the system are

KE = 1/2 mx.2 + 1/2 my.2 + 1/2 mz.2

or KE = 1/2 mx.2 + 1/2 my.2 + 1/2 m(2xx. + 2yy.)2

and PE = mgz = mg(x2 + y2 )

hence langrangian of the system is

L = KE - PE

= 1/2 mx.2 + 1/2 my.2 + 1/2 m(2xx. + 2yy.)2 - mg(x2 + y2 )

hence eqn of motion for x xoordinate is given by

d/dt((d/dx.)partialL) - d/dx(L)partial = 0

which will give,

mx.. + m(2xx. + 2yy.)2x - 2mgx = 0

or mx.. + 4x2x. m+ 4yy.xm - 2mgx = 0

similarly for y coordinate:

my.. + 4xyx. m+ 4y2y.m - 2mgy = 0

These are the required equations of motion.

(b) in cylinderical coordinates

r2 = x2 + y2 = z

or, r = (z)1/2

z= r2

z. = 2rr.

now our coordinates are theta (t) and r

KE = 1/2mr.2 + 1/2mr2(th.)2 + 1/2 mz.2

PE = 1/2 mgz=1/2 mgr2

L= KE - PE

L = 1/2mr.2 + 1/2mr2(th.)2 + 2mr2r.2 - 1/2 mgr2

now equation of motion is given by:

in r coordinate:

d/dt((d/dr.)partialL) - d/dr(L)partial = 0

mr.. + 8mr(r.)2 + 4mr2r. + mgr= 0

in theta coordinate:

d/dt((d/d(th).)partialL) - d/d(th)(L)partial = 0

mr2(th).. = 0

These are the required equations of motion.

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