# 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 9 Hz, what is the centripetal force acting on the object?

Jan 15, 2016

$F = 1.535 \times {10}^{5} \textcolor{w h i t e}{x} \text{N}$

#### Explanation:

$f = 9 \text{Hz}$

This means the time period, i.e the time for 1 revolution is given by:

$T = \frac{1}{f} = \frac{1}{9} \text{s}$

Speed = distance travelled / time taken so:

$v = \frac{2 \pi r}{T} = \frac{2 \pi 8}{\frac{1}{9}} = 144 \pi \textcolor{w h i t e}{x} \text{m/s}$

The expression for centripetal force is given by:

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

$\therefore F = \frac{6 \times {\left(144 \pi\right)}^{2}}{8} = 1.535 \times {10}^{5} \textcolor{w h i t e}{x} \text{N}$