Particle constrained to move on a hemisphere

[Jeremy Wittkopp]

Homework Statement

A particle slides on the outer surface of an inverted hemisphere. Using Lagrangian multipliers, determine the reaction force on the particle. Where does the particle leave the hemispherical surface?

L - Lagrangian
q_i - Generalized ith coordinate
f(r) - Holonomic constraint
Q_i - Generalized force of constraint on the ith particle
λ_j - Lagrangian multiplier

Homework Equations

[tex]\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = \sum_{j=0}^k \lambda_j \frac{\partial f_j}{\partial q_i},\ i = 1, ..., n \ j = 1,... k\\ \\

x = r * \sin\theta * \cos\phi\\
y = r * \sin\theta * \sin\phi\\
z = r * \cos\theta\\

f(r) = r - R = 0\\

Q_i = \sum_{j=0}^k \lambda_j \frac{\partial f_j}{\partial q_i}[/tex]

The Attempt at a Solution

So I was able to derive the potential energy [tex]U = mgr \cos\theta[/tex] the kinetic energy [tex]T = \frac{1}{2}m(\dot r^2 + r^2 \dot\theta^2 + r^2 \dot\phi^2 \sin^2\theta)[/tex]I then inserted them into the Lagrangian and got the two equations
[tex]\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = \lambda_r * \frac{\partial f}{\partial r} \\

\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = \lambda_\theta * \frac{\partial f}{\partial \theta}[/tex]

Where [tex]\frac{\partial f}{\partial r} = 1[/tex] and [tex]\frac{\partial f}{\partial \theta} = 0[/tex]

But after inputting L into these two equations I get:

[tex]Q_r = m \ddot r - mg\sin\theta - mr \dot \theta^2 - mr \dot \phi^2 \sin^2\theta[/tex]

I think there is something wrong here, but it may just be me. Also, I already searched through other threads to find this answer.

 

[Orodruin]

The separate terms in your final expression look reasonable. However, the constraints require that ##\ddot r = 0##. What is left is centripetal force required to keep the particle on the hemisphere and the force required to cancel the radial component of the gravitational force. I have not checked the signs, but it feels like the gravitational term should have the opposite sign (depending on what you mean by "inverted hemisphere").