# An object with a mass of 2 kg is revolving around a point at a distance of 4 m. If the object is making revolutions at a frequency of 3 Hz, what is the centripetal force acting on the object?

Oct 24, 2016

I found: $2842 N$

#### Explanation:

Centripetal force is given as:
${F}_{c} = m {v}^{2} / r$..............................(1)

We can evaluate the velocity knowing that the object moves of a (linear) distance equal to:
$d = 2 \pi r$
and that the frequency (representing the number of turns in one second) is $3 H z$ meaning that in one second it will go through $3$ complete turns:

So the total liner distance will be:

$d = 3 \cdot 2 \pi r = 3 \cdot 2 \pi \cdot 4 = 75.4 m$ in one second corresponding to a velocity (linear):

$v = \frac{d}{t} = \frac{75.4}{1} = 75.4 \frac{m}{s}$

Using this information into the expression for centripetal force (1) we get:

${F}_{c} = 2 \cdot {\left(75.4\right)}^{2} / 4 = 2842 N$