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An object with a mass of #2 kg# is revolving around a point at a distance of #5 m#. If the object is making revolutions at a frequency of #1 Hz#, what is the centripetal force acting on the object?

Answer 1

Charlotte Aaron

#40pi^2#

Centripetal force is given by #F=(m*v^2)/r#

where ,

m is the mass of the particle , v is the velocity of the particle, and r is the radius of the circle of revolution.

Here frequency ( #nu#) is given as #1HZ#

#v=romega# #omega# is the angular velocity of the particle and is equal to #2pinu#

Therefore #omega=2pi*1# #v=r*2pi# #v=10pi#

#F=(2*(10pi)^2)/5#

#F=(200*pi^2)/5# #F=40pi^2# is the centripetal force acting on the particle

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Answer 2

Axel Acevedo

The centripetal force acting on the object is 20 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.