An object with a mass of #2 kg# is revolving around a point at a distance of #5 m#. If the object is revolving at a rate of #5 Hz#, what is the centripetal force on the object?

Answer 1

I got #9860N#

The rate of #5Hz# means that the object makes #5# revolutions in one second.

This means that it will travel a total linear distance of #d=5*2pir# (five times the circumference) during one second, or:

#d=10*pi*5=157m#

so, also its velocity will be: #d/t=157/1=157m/s#

The general expression for a centripetal force can now be used:

#F_c=ma_c=mv^2/r#

#F=2*(157^2)/5=9859.6~~9860N#

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Answer 2

Elena Albarran

Centripetal force = ( m \cdot r \cdot (\omega)^2 ) Where: ( m = 2 , \text{kg} ) ( r = 5 , \text{m} ) ( \omega = 2 \pi \cdot \text{frequency} )

Centripetal force = ( 2 \cdot 5 \cdot (2 \pi \cdot 5)^2 )

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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.