# An object with a mass of 2 kg is revolving around a point at a distance of 5 m. If the object is making revolutions at a frequency of 1 Hz, what is the centripetal force acting on the object?

Oct 1, 2017

$40 {\pi}^{2}$

#### Explanation:

Centripetal force is given by $F = \frac{m \cdot {v}^{2}}{r}$

where ,

m is the mass of the particle ,
v is the velocity of the particle,
and r is the radius of the circle of revolution.

Here frequency ( $\nu$) is given as $1 H Z$

$v = r \omega$
$\omega$ is the angular velocity of the particle and is equal to $2 \pi \nu$

Therefore
$\omega = 2 \pi \cdot 1$
$v = r \cdot 2 \pi$
$v = 10 \pi$

$F = \frac{2 \cdot {\left(10 \pi\right)}^{2}}{5}$

$F = \frac{200 \cdot {\pi}^{2}}{5}$
$F = 40 {\pi}^{2}$ is the centripetal force acting on the particle