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Integral with initial conditions

  • Thread starter Karol
  • Start date
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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
 

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:
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$$v=\frac{ds}{dt}=v_0\sqrt{2gR}\rightarrow~s=\int v_0\sqrt{2gR} \,dt$$
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}##
 

haruspex

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

haruspex

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$$\left\{ \begin{array}{l} v^2=\frac{2gR^2}{s}+C \\ v_0(s=R)=\sqrt{2gR} \end{array} \right. \rightarrow~C=0$$
$$v^2=\frac{2gR^2}{s} \rightarrow~v=R\sqrt{\frac{2g}{s}}=v_0\sqrt{\frac{R}{s}}$$
Yes.
 
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So now i return to my question in post #3:
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{\frac{R}{s}}##
 
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$$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$
 

ehild

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$$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$
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.
 
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$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\sqrt{s}\,ds=v_0\sqrt{R}\,dt$$
$$s\sqrt{s}=v_0\sqrt{R}\cdot t \rightarrow~\sqrt{s^3}=v_0\sqrt{R}\cdot t$$
It isn't ##~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]##
 

ehild

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$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\sqrt{s}\,ds=v_0\sqrt{R}\,dt$$
$$s\sqrt{s}=v_0\sqrt{R}\cdot t \rightarrow~\sqrt{s^3}=v_0\sqrt{R}\cdot t$$
It isn't ##~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]##
Did you forget how to integrate a power function? And you also omitted the integration constant.
 
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$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\int \sqrt{s}\,ds=\int v_0\sqrt{R}\,dt$$
$$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+C$$
$$s_0(t=0)=R\rightarrow~C=\frac{2}{3}R^{3/2}$$
$$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+\frac{2}{3}R^{3/2}$$
$$\rightarrow~~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]$$
 

ehild

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Correct, at last.
 
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Thank you ehild and Haruspex
 

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