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

Dec 30, 2015

I found $\approx 38 k N$

#### Explanation:

We have that centripetal force is:
${F}_{c} = m {a}_{c} = m {v}^{2} / r$
Where:
$v = \left(\text{distance")/("time")=("circumference")/("period}\right) = \frac{2 \pi r}{T}$
With: $T = \frac{1}{\text{frequency}} = \frac{1}{9}$
So:
$v = 2 \cdot 3.14 \cdot 2 \cdot 9 = 113.04 \frac{m}{s}$
So:
${F}_{c} = 6 \cdot {\left(113.04\right)}^{2} / 2 = 38 , 334 N$