1. Not finding help here? Sign up for a free 30min 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!

Velocity in a Cycloidal Pendulum

  1. Sep 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Cycloidal Pendulum, with x= RΘ+RsinΘ and y = -RcosΘ
    I need to find the Lagragian.


    2. Relevant equations

    L = T - V

    3. The attempt at a solution

    I just want to know how do I find the velocity so I can find T, which is 1/2 mv². I thought it would be dx/dΘ but it didn't work. Can anyone explain to me how can I find the velocity?
     
  2. jcsd
  3. Sep 16, 2014 #2

    ZetaOfThree

    User Avatar
    Gold Member

    Remember that velocity is the derivative of position with respect to time and is a vector. What are its x and y components? Could you use these components to find ##v^2##?
     
  4. Sep 17, 2014 #3
    Ok, I realized that I must consider x and y and derivative in relation with time.

    v = dx/dt + dy/dt

    then v² would be (dx/dt + dy/dt)².

    I did the derivatives:

    x'= RΘ' + Rcos(Θ)Θ' and y' = Rsin(Θ)Θ',

    so x+y = RΘ' + RΘ' (cosΘ + sinΘ)

    Also, v² = ( RΘ' + RΘ' (cosΘ + sinΘ))² = (4R²Θ'² + R²Θ'²cos²Θsin²Θ)
    But in my answer it is

    4R²Θ²cos²(Θ/2)

    What am I doing wrong?
     
    Last edited: Sep 17, 2014
  5. Sep 17, 2014 #4

    ZetaOfThree

    User Avatar
    Gold Member

    ##v=\frac{dx}{dt}+\frac{dy}{dt}## is wrong. Remember that ##\vec{v}## is a vector! What is the x-component of ##\vec{v}##? What is the y-component of ##\vec{v}##? After you have the components, you can find ##v^2## by squaring the components and adding them, right?
     
  6. Sep 17, 2014 #5
    Ohhh, you are right! It worked now, thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted