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Center of mass of astroid

  1. Jan 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the center of mass of Astroid ##x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} ## for ##x,y\geq 0## and ##a>0##.


    2. Relevant equations

    ##x_T=\frac{\int x\left | \dot{\vec{r}}(t) \right |}{\int \left | \dot{\vec{r}}(t) \right |}##

    3. The attempt at a solution

    ##x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}##

    ##(\frac{x}{a})^{\frac{2}{3}}+(\frac{y}{a})^{\frac{2}{3}}=1##

    Now is it ok to say that ##\frac{x}{a}=\cos^3t## and ##\frac{y}{a}=\sin^3t## for ##t\in \left [ 0,\frac{\pi }{2} \right ]##?

    Now ##\left | \dot{\vec{r}}(t) \right |=3a \sint \cost##.

    Than ##x_T=\frac{\int x\left | \dot{\vec{r}}(t) \right |}{\int \left | \dot{\vec{r}}(t) \right |}## can be written as

    ##x_T=\frac{\int_{0}^{\frac{\pi }{2}} 3a^2\cos^4t\sint}{\int_{0}^{\frac{\pi }{2}} 3a\cost\sint}=\frac{2}{5}a##

    or... is that parameterization wrong?
     
  2. jcsd
  3. Jan 6, 2014 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks fine to me.
     
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