# An object with a mass of 5 kg is revolving around a point at a distance of 3 m. If the object is making revolutions at a frequency of 19 Hz, what is the centripetal force acting on the object?

Jul 17, 2016

I found $213 k N$

#### Explanation:

You know that Centripetal Force is:
${F}_{C} = m {v}^{2} / r$
Where
$v$ is velocity;
$r$ is the radius.
We need the velocity!
We can calculate it considering the length of the circumference decribed, i.e., $2 \pi r$ and divide it by the time $T$ taken.
The time taken for a complete revolution, or period $T$, is given as:
$T = \frac{1}{f} = \frac{1}{19} s$ where $f$ is the frequency.
The velocity will then be:
$v = \frac{2 \pi r}{T} = \left(2 \cdot \pi \cdot 3\right) \cdot 19 = 358 \frac{m}{s}$
So finally we have:
${F}_{C} = m {v}^{2} / r = 5 {\left(358\right)}^{2} / 3 \approx 213 k N$