Integral with initial conditions

[Karol]

Homework Statement

Homework Equations

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

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]

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.

 

[Karol]

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

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

 

[Karol]

$$\left\{ \begin{array}{l} v^2=\frac{2gR^2}{s}+C \\ v_0(s=R)=\sqrt{2gR} \end{array} \right. \rightarrow~C=0$$
$$v^2=\frac{2gR^2}{s} \rightarrow~v=R\sqrt{\frac{2g}{s}}=v_0\sqrt{\frac{R}{s}}$$

 

[haruspex]

$$\left\{ \begin{array}{l} v^2=\frac{2gR^2}{s}+C \\ v_0(s=R)=\sqrt{2gR} \end{array} \right. \rightarrow~C=0$$
$$v^2=\frac{2gR^2}{s} \rightarrow~v=R\sqrt{\frac{2g}{s}}=v_0\sqrt{\frac{R}{s}}$$

Yes.

 

[Karol]

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 $~v=v_0\sqrt{\frac{R}{s}}$

 

[ehild]

So now i return to my question in post #3:

v=ds/dt.

 

[Karol]

$$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$

 

[ehild]

$$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$

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.

 

[Karol]

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

 

[ehild]

$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\sqrt{s}\,ds=v_0\sqrt{R}\,dt$$
$$s\sqrt{s}=v_0\sqrt{R}\cdot t \rightarrow~\sqrt{s^3}=v_0\sqrt{R}\cdot t$$
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.

 

[Karol]

$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\int \sqrt{s}\,ds=\int v_0\sqrt{R}\,dt$$
$$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+C$$
$$s_0(t=0)=R\rightarrow~C=\frac{2}{3}R^{3/2}$$
$$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+\frac{2}{3}R^{3/2}$$
$$\rightarrow~~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]$$

 

[ehild]

Correct, at last.

 

[Karol]

Thank you ehild and Haruspex