Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 15, 2016 22:36:14 GMT
Vasco,
On p. 243 of Needham, we find
Is this just the integral of force over distance, with the force being proportional to unit mass and the square of the distance?
$\hspace{5em}\int r dr = r^2/2$
Gary
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Post by Admin on Apr 16, 2016 8:37:26 GMT
Vasco, On p. 243 of Needham, we find Is this just the integral of force over distance, with the force being proportional to unit mass and the square of the distance? $\hspace{5em}\int r dr = r^2/2$ Gary Gary No, it's the integral of Force(F) x dr = work done(dW) moving the mass a distance dr against the force F. In the case being considered in subsection 2 on page 243 the force F is proportional to r, as stated at the beginning of the subsection on page 241. In the last paragraph of subsection 1 on page 241 Needham says we are assuming unit mass and so $F = mr = r$, and it follows that $dW = F \times dr = r \times dr$ and then $W=\int_0^r r\times dr=\big[r^2/2\big]_0^r=r^2/2$ In subsection 3 Needham deals with gravity, i.e. when $F \propto 1/r^2$. Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 20, 2016 23:25:47 GMT
Vasco,
Thank you. I have it now. When I read that, I had forgotten that the context was linear rather than quadratic.
Gary
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Post by mondo on Aug 19, 2023 5:42:28 GMT
Vasco, On p. 243 of Needham, we find Is this just the integral of force over distance, with the force being proportional to unit mass and the square of the distance? $\hspace{5em}\int r dr = r^2/2$ Gary Gary No, it's the integral of Force(F) x dr = work done(dW) moving the mass a distance dr against the force F. In the case being considered in subsection 2 on page 243 the force F is proportional to r, as stated at the beginning of the subsection on page 241. In the last paragraph of subsection 1 on page 241 Needham says we are assuming unit mass and so $F = mr = r$, and it follows that $dW = F \times dr = r \times dr$ and then $W=\int_0^r r\times dr=\big[r^2/2\big]_0^r=r^2/2$ In subsection 3 Needham deals with gravity, i.e. when $F \propto 1/r^2$. Vasco There is something missing, the relation between work and potential energy should be $W = -E_p$ while in your calculation it looks like they are equal. I think the missing puzzle is the direction of force $r$ - it is positive when pointing into the direction of $o$ but when lift the particle, hence gaining it's potential energy we go into the opposite direction. So I think the integral should be $\int_{0}^{r} -r dr = \frac{-r^2}{2} = -E_p$
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Post by Admin on Aug 19, 2023 15:50:18 GMT
Mondo
Ok. To be precise then, the integral in your case is the work done by the central force, whereas in my case it's the potential energy.
Vasco
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