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?

1 Answer
Jan 29, 2016

Answer:

#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.