Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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:

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.