Question

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?

Answer 1

`40pi^2`

Explanation:

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