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: Change of variable in an integral

  1. Nov 28, 2017 #1
    1. The problem statement, all variables and given/known data
    A massless string of length 2l connects two hockey pucks that lie on frictionless ice. Aconstant horizontal force F is applied to the midpoint of the string, perpendicular to it (see right figure). How much kinetic energy is lost when the pucks collide, assuming they stick together?


    3. The attempt at a solution
    Putting a y-axis along the string when it is straight and a x-axis along the direction of the force, the force acting on the puck "at the bottom" along the t-axis is ##\frac{F}{2}\cdot tan \theta##.
    So, by calculating the work for this puck I obtain ##W=\int_{-l}^{0} \frac{F\cdot tan \theta}{2}dy=\int_{-l}{0} \frac{F\cdot tan \theta}{2} d(l\cdot sin\theta)= \int_{\frac{-\pi}{2}}^{0} \frac{Fl}{2}\cdot sin\theta d\theta##.
    Can you explaine me how the last passage works?
     

    Attached Files:

  2. jcsd
  3. Nov 28, 2017 #2

    phyzguy

    User Avatar
    Science Advisor

    It's a change of variables [itex] y = l \sin(\theta)[/itex].

    Since l is constant: [itex]d(l \sin(\theta)) = l d(\sin(\theta)) = l \cos(\theta) d \theta[/itex].

    So: [itex]\frac{F \tan(\theta)}{2}d(l \sin(\theta)) = \frac{F l \tan(\theta) \cos(\theta)}{2}d \theta = \frac{F l \sin(\theta)}{2}d \theta[/itex].

    Then since [itex] y = l \sin(\theta)[/itex], you need to change the integration limit from [itex] y = -l [/itex] to [itex] \theta = \frac{-\pi}{2}[/itex].

    Does this answer your question?
     
  4. Nov 28, 2017 #3
    Perfectly, thank you!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted