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?

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Answer 1 · 2017-10-01T15:58:40

$40pi^2$

Centripetal force is given by $F=(m*v^2)/r$

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 $1HZ$

$v=romega$
$omega$ is the angular velocity of the particle and is equal to $2pinu$

Therefore
$omega=2pi*1$
$v=r*2pi$
$v=10pi$

$F=(2*(10pi)^2)/5$

$F=(200*pi^2)/5$
$F=40pi^2$ is the centripetal force acting on the particle

An object with a mass of 2 kg2 kg is revolving around a point at a distance of 5 m5 m. If the object is making revolutions at a frequency of 1 Hz1 Hz, what is the centripetal force acting on the object?