An object with a mass of #6 kg# is revolving around a point at a distance of #2 m#. If the object is making revolutions at a frequency of #9 Hz#, what is the centripetal force acting on the object?

Answer 1

I found #~~38kN#

We have that centripetal force is: #F_c=ma_c=mv^2/r# Where: #v=("distance")/("time")=("circumference")/("period")=(2pir)/T# With: #T=1/("frequency")=1/9# So: #v=2*3.14*2*9=113.04m/s# So: #F_c=6*(113.04)^2/2=38,334N#

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Answer 2

Xavier Adame

The centripetal force acting on the object is ( F_c = m \times (2\pi f)^2 ), where ( m ) is the mass of the object (6 kg) and ( f ) is the frequency of revolutions (9 Hz). Plugging in the values:

[ F_c = 6 \times (2\pi \times 9)^2 ]

[ F_c \approx 1017.6 , \text{N} ]

Therefore, the centripetal force acting on the object is approximately 1017.6 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.