1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Particle constrained to move on a hemisphere

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data
    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
    qi - Generalized ith coordinate
    f(r) - Holonomic constraint
    Qi - Generalized force of constraint on the ith particle
    λj - Lagrangian multiplier

    2. Relevant 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]

    3. 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.
  2. jcsd
  3. Jan 8, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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").
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted