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?

1 Answer
Feb 17, 2016

Answer:

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#