An object with a mass of 16 kg is revolving around a point at a distance of 3 m. If the object is making revolutions at a frequency of 13 Hz, what is the centripetal force acting on the object?

Jan 28, 2016

I found: $320 , 250 N$

Explanation:

Centripetal force is equal to mass time centripetal acceleration:
${F}_{c} = m \cdot {a}_{c} = m \cdot {v}^{2} / r$
where:
$m = 16 k g =$mass;
v=?=linear velocity;
$r = 3 m =$radius.

Now:
Angular Velocity $\omega$ (a kind of circular velocity!) will be equal to distance divided time or:
$\omega = \frac{2 \pi}{T}$

in this case the distance will be the circumference of $2 \pi$ radians divided by the time or period $T$ to desribe it.
But period is related to frequency $\nu$ as:
$\nu = \frac{1}{T}$ so we can write:
$\omega = 2 \pi \nu$

But we want a Linear velocity!
We simply introduce the radius into the $\omega$ to get:
$v = \omega r = 2 \pi \nu r$
The centripetal force will then become:
F_c=m(4pi^2)nu^2r^2)/r=m(4pi^2)(nu^2)r=16*(4pi^2)(13^2)*3=320,248.9N~~320,250N