An object with a mass of 6 kg is revolving around a point at a distance of 8 m. If the object is making revolutions at a frequency of 5 Hz, what is the centripetal force acting on the object?

Jan 31, 2016

The centripetal force is given by $F = m {\omega}^{2} r$ and is $4800 {\pi}^{2}$ $N$ or $47 , 374$ $N$

Explanation:

The centripetal acceleration is given by:

$a = {\omega}^{2} r$

Where:

$r$ is the radius $\left(m\right)$
$\omega$ is the rotational speed $\left(r a {\mathrm{ds}}^{-} 1\right)$

we were given the rotational frequency in $H z$ (cycles per second), and there are $2 \pi$ radians in a cycle, so to find the rotational speed multiply $5$ $H z$ by $2 \pi$ to give $10 \pi$ $r a {\mathrm{ds}}^{-} 1$.

Using Newton's Second Law , the centripetal force will be $m$ times the centripetal acceleration.

$F = m a = m {\omega}^{2} r = 6 \cdot {\left(10 \pi\right)}^{2} \cdot 8 = 4800 {\pi}^{2} N = 47 , 374 N$