# An object with a mass of 3 kg is revolving around a point at a distance of 4 m. If the object is making revolutions at a frequency of 5 Hz, what is the centripetal force acting on the object?

Jan 29, 2016

${F}_{c} = 1200 \setminus {\pi}^{2} N$

#### Explanation:

An object of mass $m = 3 k g$ and at a distance of $r = 4 m$ from the center of it's rotation revolves with a linear frequency of $f = 5 h z$ implying that angular frequency is $\setminus \omega = 10 \setminus {\pi}^{c} r a \mathrm{ds}$

Now, to find the centripetal force acting on the body, we'll have to substitute the above values into the equation ${F}_{c} = m {v}^{2} / r$

But wait, they didn't give us velocity $v$ with which the object is revolving. No need. We ourselves of course know that $v = \setminus \omega r$ so that means ${v}^{2} = \setminus {\omega}^{2} {r}^{2}$

Dividing the value for ${v}^{2}$ in the above equation by $r$ and multiplying by $m$, we get know that ${F}_{c} = m \setminus {\omega}^{2} r$

Now you see why there's an oddly present $\setminus {\pi}^{2}$ term there.