# An object with a mass of 6 kg is revolving around a point at a distance of 2 m. If the object is making revolutions at a frequency of 15 Hz, what is the centripetal force acting on the object?

Jan 22, 2016

$108 {\pi}^{2} \textcolor{w h i t e}{x} \text{N}$

#### Explanation:

Centripetal force is given by:

$F = \frac{m {v}^{2}}{r}$

To find $v :$

$f = 15 \text{Hz}$

So the time period $T$ is given by:

$T = \frac{1}{f} = \frac{1}{15} s$

$\therefore v = \frac{2 \pi r}{\frac{1}{15}} = 30 \pi r$

$\therefore F = \frac{6 \times {\left(3 \pi r\right)}^{2}}{r}$

$F = \frac{6 \times 9 {\pi}^{2} {r}^{\cancel{2}}}{\cancel{r}}$

$F = 54 {\pi}^{2} \times 2 = 108 {\pi}^{2} \textcolor{w h i t e}{x} \text{N}$