Question

An object with a mass of `3 kg` is revolving around a point at a distance of `4 m`. If the object is making revolutions at a frequency of `5 Hz`, what is the centripetal force acting on the object?

Answer 1

`F_c=1200\pi^2N`

Explanation:

An object of mass `m=3kg` and at a distance of `r=4m` from the center of it's rotation revolves with a linear frequency of `f=5hz` implying that angular frequency is `\omega=10\pi^crads`

Now, to find the centripetal force acting on the body, we'll have to substitute the above values into the equation `F_c=mv^2/r`

But wait, they didn't give us velocity `v` with which the object is revolving. No need. We ourselves of course know that `v=\omegar` so that means `v^2=\omega^2r^2`

Dividing the value for `v^2` in the above equation by `r` and multiplying by `m`, we get know that `F_c=m\omega^2r`

Now you see why there's an oddly present `\pi^2` term there.