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

Aug 19, 2017

${F}_{c} \approx 5053 N$ radially inward.

#### Explanation:

The centripetal force is given in accordance with Newton's second law as:

${F}_{c} = m {a}_{c}$

where $m$ is the mass of the object and ${a}_{c}$ is the centripetal acceleration experienced by the object

The centripetal acceleration is given by:

${a}_{c} = {v}^{2} / r$

which is equivalent to $r {\omega}^{2}$. Therefore, we can write:

${F}_{c} = m r {\omega}^{2}$

where $r$ is the radius and $\omega$ is the angular velocity of the object

The angular velocity can also be expressed as:

$\omega = 2 \pi f$

where $f$ is the frequency of the revolution

And so our final expression becomes:

color(blue)(F_c=mr(2pif)^2

We are given:

• $\mapsto \text{m"=2"kg}$

• $\mapsto \text{r"=4"m}$

• $\mapsto f = 4 {\text{s}}^{-} 1$

Substituting these values into the equation we derived above:

F_c=(2"kg")(4"m")(2pi(4"s"^-1))^2

$= 5053.237 \text{N}$

$\approx 5053 N$ radially inward

This may also be expressed in scientific notation as $5 \times {10}^{3} \text{N}$ where significant figures are concerned.