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The equation,

$$ a_c=\frac{4\pi^2r}{T^2} $$

Can be expressed as:

$$ F_c=\frac{m4\pi^2r}{T^2} $$

I am confused as to how to arrive at this second equation, and the relationship between these two equations.

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2 Answers 2

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If you have some object moving in a circle with a velocity $v$, then as any Physics textbook will tell you the acceleration towards the centre is:

$$a = \frac{v^2}{r}$$

To get the velocity we note that the circumference of the circle is $2\pi r$, so if the object takes a time $T$ to go round the circle the velocity is just distance divided by time:

$$ v = \frac{2\pi r}{T}$$


$$ v^2 = \frac{4\pi^2 r^2}{T^2}$$

and if you put this expression for $v^2$ in the first equation it gives:

$$ a = \frac{4\pi^2 r}{T^2}$$

The way to get the force is to note that Newton's first law tells us:

$$ F = ma $$

where $m$ is the mass of the moving object. Put our expression for $a$ into this equation and we get;

$$ F = m\frac{4\pi^2 r}{T^2}$$

or as you have written it:

$$ F = \frac{m4\pi^2 r}{T^2}$$

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In this context, a is the accelaration and F is the force. The connection between these two is simply Newton's second law, namely F = m*a, where m stands for the mass of the object. This equation is one of the most basic principles in classical mechanics.

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