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m0nk3y
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Equilibrium - Particle in Bowl, looking for guidance!
A point particle of mass m is confines to the frictionless surface of a sphere cal bowl. There are 2 degrees of freedom. Prove that the equilibrium point is the bottom of the bowl. Near the bottom of the bowl, what is the most general form possible for the shape of the bowl in order to maintain the stability of the equilibrium point at the bottom?
spherical coordinates: rsin[tex]\thetacos\varphi + rsin\thetasin\varphi + rcos\theta[/tex]
Lagrange = Kinetic Energy(T)- Potential (V)
I started off by getting the Lagrange and got:
[tex]\stackrel{1}{2}[/tex]r^2 (theta)'^2 - mgrcos(theta)
Then I got the E.O.M
(mr^2 (theta)'')/2 - mgsin(theta)
I have to prove that it is at equilibrium at [tex]\theta[/tex]=0
But when I plug in 0 I am still left with (mr^2 (theta)'')/2
What am I doing wrong
[NOTATION::(theta)' is theta dot or velocity and (theta)'' is theta double dot or acceleration]
Any help is appreciated!
Thanks
Homework Statement
A point particle of mass m is confines to the frictionless surface of a sphere cal bowl. There are 2 degrees of freedom. Prove that the equilibrium point is the bottom of the bowl. Near the bottom of the bowl, what is the most general form possible for the shape of the bowl in order to maintain the stability of the equilibrium point at the bottom?
Homework Equations
spherical coordinates: rsin[tex]\thetacos\varphi + rsin\thetasin\varphi + rcos\theta[/tex]
Lagrange = Kinetic Energy(T)- Potential (V)
The Attempt at a Solution
I started off by getting the Lagrange and got:
[tex]\stackrel{1}{2}[/tex]r^2 (theta)'^2 - mgrcos(theta)
Then I got the E.O.M
(mr^2 (theta)'')/2 - mgsin(theta)
I have to prove that it is at equilibrium at [tex]\theta[/tex]=0
But when I plug in 0 I am still left with (mr^2 (theta)'')/2
What am I doing wrong
[NOTATION::(theta)' is theta dot or velocity and (theta)'' is theta double dot or acceleration]
Any help is appreciated!
Thanks