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#

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.

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Answer 2

Axel Acevedo

The centripetal force acting on the object is 300 N.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.