Question

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 `2 Hz`, what is the centripetal force acting on the object?

Answer 1

I found: `7580N`

Explanation:

Centripetal Force is:
`F_c=mv^2/r`
where:
`m=`mass;
`v=` linear velocity;
`r=` radius.

The frequency tells us the number of complete revolutions in one seconds (here 2 revolutions).
We can ask ourselves what will be the Angular Velocity `omega` of our object, i.e., a kind of "curved" velocity involving not linear distance but angle described in time!

So we get:

`omega="angle"/"time"=(2pi)/T`

Where `T` will be the Period of time for a complete revolution (time to describe `2pi` radians).

The good thing is that the period is connected to frequency as `"frequency"=nu=1/T`
and also the angular velocity can be changed into linear velocity simply considering at what distance from the center you are travelling....(basically, you include the radius)!!!!
so:
`v=omega*r`

in our case we get (collecting all our stuff):

`F_c=m(omega*r)^2/r=momega^2r=m(2pinu)^2r=6*(2*3.14*2)^2*8~~7580N`