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!

Center of mass of an arc

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    In the 1968 Olympic Games, University of Oregon jumper Dick Fosbury introduced a new technique of high jumping called the "Fosbury flop." It contributed to raising the world record by about 30 cm and is presently used by nearly every world-class jumper. In this technique, the jumper goes over the bar face up while arching his back as much as possible, as shown below. This action places his center of mass outside his body, below his back. As his body goes over the bar, his center of mass passes below the bar. Because a given energy input implies a certain elevation for his center of mass, the action of arching his back means his body is higher than if his back were straight. As a model, consider the jumper as a thin, uniform rod of length L. When the rod is straight, its center of mass is at its center. Now bend the rod in a circular arc so that it subtends an angle of θ = 80.5° at the center of the arc, as shown in Figure (b) below. In this configuration, how far outside the rod is the center of mass? Report your answer as a multiple of the rod length L.

    http://www.webassign.net/pse/p9-44alt.gif


    2. Relevant equations

    L=rθ
    where L is arc (rod) length.

    [itex]r_{cm}=\int r \: dm[/itex]


    3. The attempt at a solution

    So the TA for my discussion section said it was far too advanced to expect us to do it on our own, and did a healthy amount of work for us by giving us this equation...

    [itex]y_{cm}=\frac{1}{M}\int ^{ \theta / 2 }_{ - \theta / 2 } r cos\left(\theta \right) \frac{M}{L} r \: d\theta[/itex]

    I've attached a picture showing what the different variables I use are. Sadly, since I think this might be where my problem is, I don't remember how he derived this, but I know it was using the equations in the "relevant equations" section, and possibly using linear momentum as well, although I'm not sure.

    So I took his equation, pulled out r, L, and M, and integrated.

    [itex]y_{cm}=\frac{r^{2}}{L} \left( sin\left(\frac{\theta}{2} \right) - sin\left(\frac{-\theta}{2} \right) \right) [/itex]

    Plugging in (L/θ) for r gives...

    [itex]y_{cm}=\frac{L}{\theta^{2}} \left( sin\left(\frac{\theta}{2} \right) - sin\left(\frac{-\theta}{2} \right) \right) [/itex]

    Lastly, (since I'm looking for Δy, not y_cm) I used

    Δy = r - y_cm
    and
    r=(L/θ)

    to give me

    [itex]\Delta y=\frac{L}{θ} - \frac{L}{\theta^{2}} \left( sin\left(\frac{\theta}{2} \right) - sin\left(\frac{-\theta}{2} \right) \right) [/itex]

    By factoring out L and computing with θ=(80.5π/180) i get .699L .... Which isn't right, which is why I'm here. I'm not sure if the equation given to me by my TA was wrong, or if I didn't integrate right, or what, but I'm stumped..

    Thanks in advance for any help.
     

    Attached Files:

  2. jcsd
  3. Nov 8, 2011 #2
    Oh man...

    As a word of advice for anyone who needs this advice.

    ALWAYS ALWAYS ALWAYS.... Check to make sure your calculator is in (degrees/radians) before you spend an hour staring at your work, an hour looking for help in real life, an hour looking for help on the internet, and an hour typing up your work in Latex on Physics Forums. Your head and your wall will thank you.

    http://aberrospecus.wordpress.com/2011/08/14/today-is-fail/

    Anyway, if anyone can walk me through how my TA got that equation, I would greatly appreciate it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook