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Integral with initial conditions

  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap2.jpg Snap4.jpg

    2. Relevant equations
    $$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$
    $$F=ma,~a=\frac{dv}{dt}$$

    3. The attempt at a solution
    $$-\frac{mgR}{s^2}=mv\left( \frac{dv}{ds} \right)\rightarrow~v^2=\frac{2gR}{s}+C$$
    $$v_0(s=R)=\sqrt{2gR}\rightarrow~C=2g(R-1)$$
    $$v^2=\frac{2gR}{s}+2g(R-1)$$
    Not good
     
  2. jcsd
  3. Mar 11, 2017 #2

    haruspex

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    The link to the image for the start of the question does not work for me.
    Your very first equation looks dimensionally wrong. If s is a displacement then gR/s2 has dimension T-2.

    Edit. On reloading the page the first link works, and I see that the incorrect equation is provided. It should be mgR2/s2.
     
    Last edited: Mar 11, 2017
  4. Mar 11, 2017 #3
    $$v=\frac{ds}{dt}=v_0\sqrt{2gR}\rightarrow~s=\int v_0\sqrt{2gR} \,dt$$
    But that is a function of s, not t. nowhere there is t, nor in ##~a=v\frac{dv}{ds}## nor in ##~v=v_0\sqrt{2gR}##
     
  5. Mar 11, 2017 #4

    haruspex

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    I have no idea what you are doing there. v is not v0√(2gR). Where did that come from?
     
  6. Mar 11, 2017 #5
    $$\left\{ \begin{array}{l} v^2=\frac{2gR^2}{s}+C \\ v_0(s=R)=\sqrt{2gR} \end{array} \right. \rightarrow~C=0$$
    $$v^2=\frac{2gR^2}{s} \rightarrow~v=R\sqrt{\frac{2g}{s}}=v_0\sqrt{\frac{R}{s}}$$
     
  7. Mar 12, 2017 #6

    haruspex

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    Yes.
     
  8. Mar 12, 2017 #7
    So now i return to my question in post #3:
     
  9. Mar 12, 2017 #8

    ehild

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    v=ds/dt.
     
  10. Mar 12, 2017 #9
    $$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$
     
  11. Mar 12, 2017 #10

    ehild

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    It does not go anywhere. You have a separable differential equation, you can integral a function of s with respect to s. and a function of t with respect to t.
     
  12. Mar 12, 2017 #11
    $$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\sqrt{s}\,ds=v_0\sqrt{R}\,dt$$
    $$s\sqrt{s}=v_0\sqrt{R}\cdot t \rightarrow~\sqrt{s^3}=v_0\sqrt{R}\cdot t$$
    It isn't ##~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]##
     
  13. Mar 12, 2017 #12

    ehild

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    Did you forget how to integrate a power function? And you also omitted the integration constant.
     
  14. Mar 12, 2017 #13
    $$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\int \sqrt{s}\,ds=\int v_0\sqrt{R}\,dt$$
    $$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+C$$
    $$s_0(t=0)=R\rightarrow~C=\frac{2}{3}R^{3/2}$$
    $$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+\frac{2}{3}R^{3/2}$$
    $$\rightarrow~~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]$$
     
  15. Mar 12, 2017 #14

    ehild

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    Correct, at last.
     
  16. Mar 12, 2017 #15
    Thank you ehild and Haruspex
     
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