# An object with a mass of 6 kg is revolving around a point at a distance of 8 m. If the object is making revolutions at a frequency of 2 Hz, what is the centripetal force acting on the object?

Feb 17, 2016

I found: $7580 N$

#### Explanation:

Centripetal Force is:
${F}_{c} = m {v}^{2} / r$
where:
$m =$mass;
$v =$ linear velocity;
$r =$ radius.

The frequency tells us the number of complete revolutions in one seconds (here 2 revolutions).
We can ask ourselves what will be the Angular Velocity $\omega$ of our object, i.e., a kind of "curved" velocity involving not linear distance but angle described in time!

So we get:

$\omega = \text{angle"/"time} = \frac{2 \pi}{T}$

Where $T$ will be the Period of time for a complete revolution (time to describe $2 \pi$ radians).

The good thing is that the period is connected to frequency as $\text{frequency} = \nu = \frac{1}{T}$
and also the angular velocity can be changed into linear velocity simply considering at what distance from the center you are travelling....(basically, you include the radius)!!!!
so:
$v = \omega \cdot r$

in our case we get (collecting all our stuff):

${F}_{c} = m {\left(\omega \cdot r\right)}^{2} / r = m {\omega}^{2} r = m {\left(2 \pi \nu\right)}^{2} r = 6 \cdot {\left(2 \cdot 3.14 \cdot 2\right)}^{2} \cdot 8 \approx 7580 N$