A particle of mass m starts at rest from the top of a smooth hemisphere of radiu

Question

A particle of mass m starts at rest from the top of a smooth hemisphere of radius R and slides down the surface. Using the constraint f(r, theta) = r - a = 0, use Lagrange multipliers to find the force of constraint and determine where (ie. what theta) the mass leaves the surface of the sphere. A uniform disk of mass m and radius a starts from rest at the top of a large cylinder of radius R that is fixed to the ground and rolls without slipping down the surface. Use Lagrange multipliers to determine where the disk leaves the surface of the hemisphere.

Explanation / Answer

Let the radial position of the particle is given by

r=r (sin, cos),

The radial velocity will be given by

r=r(sin,cos) + r(cos,sin).

The Kinetic Energy of the particle

T=1/2(mr2 )=1/2m(r2+r22).

Its Potential Energy is

V=mg r cos,

where rcos = particle height

c If we were not interested in finding a constraint force, the Lagrangian of this is

L=TV=1/2m(r2+r22)-mgr cos.

But the constraints are

f(r)=ra=0.

The new potential energy is

V’=V+ f(r).

(This is possible only because the constraint is holonomic)

Therefore, the new Lagrangian will be:

L=TV=1/2m(r2+r22)mgrcos(ra).

The equations of motion are :

(d/dt )(L/r)(L/r)=( )f/r

(d/dt)(L/)L/=()f/

leading to the two equations of motion:

mr¨mr2+mgcos=

mr2¨=mgr sin.

We know r¨=0 (constraints)

. Therefore, the equations of motion simplify in:

ma2+mgcos=

¨=g/asin.

Lambda corresponds to the force of constraint. Now We shall determine the value of this force of constraint as a function angle

.

First of all, we know that

d/dt 2=2¨,

and from the second equation of motion, and integrating both sideand assuming that the initial velocity is zero we get

2=2gacos.

By putting this relation into first equation of motion involving

we provide a constraint force which is a function of solely

=(2ma)(g/a)cos+mgcos=mg(3cos2)

When the constraint force is no longer active and therefore equals zero. From this we get

final=cos1(2/3)=48.19.

final=squareroot(2g/3a)

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