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Deriving Equation

Ian

Posted 11 October 2005 - 07:03 PM

I would greatly appreciate it if someone could answer this question

Write an expression for the radial acceleration of a particle moving in a circle of radius r, with a speed of v

Hence show that the centrapertal force F acting on this particle of mass m is inversly proportional to the square of the period T

Thanks a lot

Ian

dfx

Posted 11 October 2005 - 08:58 PM

a = v power2.gif / r

F = ma = mv power2.gif / r

BUT v = rw and w = theta.gif/T = 2 pi.gif /T . So v = r 2 pi.gif / T

Then F = ( m 4 pi.gif power2.gif r power2.gif ) / T power2.gif r

F = ( m 4 pi.gif power2.gif r )/ T power2.gif

therefore.gif F is inversely proportional to T power2.gif

dfx

Posted 11 October 2005 - 09:01 PM

However, I have a question.

We know that F = ( mv power2.gif ) / r .

Say instead of v we substitute r.

So we know that V = ( 2 pi.gif r ) / T implies.gif r = vT/2 pi.gif

So F = ( m v power2.gif ) / r = ( m v power2.gif 2 pi.gif ) / vT = ( m v 2 pi.gif ) / T

So is this not contradictory to our previous conclusion? In this case F is inversely proportional to T only. huh.gif

Ian

Posted 11 October 2005 - 09:49 PM

Thanks a lot

werlop

Posted 12 October 2005 - 09:45 AM

I know this has been answered already, but I'll just put it in Tex for ease of reading.


\[
\begin{array}{l}
 a = \frac{{v^2 }}{r} \\ 
 F = \frac{{mv^2 }}{r}{\rm  } \\ 
 {\rm but}\;v = r\omega {\rm ,}\;v^2  = r^2 \omega ^2 {\rm   so }\;\underline {F = mr\omega ^2 }  \\ 
 \omega  = \frac{{2\pi }}{T}{\rm ,}\;\omega ^2  = \frac{{4\pi ^2 }}{{T^2 }} \\ 
 F = mr\omega ^2  = \frac{{4mr\pi ^2 }}{{T^2 }} \\ 
  \Rightarrow F = \frac{1}{{T^2 }} \times 4mr\pi ^2  \\ 
 \underline{\underline {\;F\;{\rm is}\;{\rm inversely}\;{\rm proportional}\;{\rm to}\;T^2 }}  \\ 
 \end{array}
\]

werlop

Posted 12 October 2005 - 09:52 AM

QUOTE(dfx @ Oct 11 2005, 10:01 PM)
However, I have a question.

We know that F = ( mv power2.gif ) / r .

Say instead of v we substitute r.

So we know that V = ( 2 pi.gif r ) / T implies.gif r = vT/2 pi.gif

So F = ( m v power2.gif ) / r = ( m v power2.gif 2 pi.gif ) / vT =  ( m v 2 pi.gif ) / T

So is this not contradictory to our previous conclusion? In this case F is inversely proportional to T only.  huh.gif

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I did this really quickly so may have made a mistake, but it still works even if you substitute r:


\[
\begin{array}{l}
 a = \frac{{v^2 }}{r} \\ 
 F = \frac{{mv^2 }}{r} \\ 
 r = \frac{v}{\omega } \\ 
 {\rm but}\;v = r\omega {\rm ,}\;v^2  = r^2 \omega ^2 {\rm   so }\;\underline {F = mr\omega ^2 }  \\ 
 F = mr\omega ^2  = r\omega (m\omega ) = \underline {mv\omega }  \\ 
 \omega  = \frac{{2\pi }}{T}{\rm ,}\;v = \frac{{2\pi r}}{T} \\ 
 F = mv\omega  = m \times \frac{{2\pi r}}{T} \times \frac{{2\pi }}{T} \\ 
 F = \frac{{4mr\pi ^2 }}{{T^2 }} \\ 
  \Rightarrow F = \frac{1}{{T^2 }} \times 4mr\pi ^2  \\ 
 \underline{\underline {\;F\;{\rm is}\;{\rm inversely}\;{\rm proportional}\;{\rm to}\;T^2 }}  \\ 
 \end{array}
\]

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