An object with a mass of #6 kg# is revolving around a point at a distance of #4 m#. If the object is making revolutions at a frequency of #1 Hz#, what is the centripetal force acting on the object?

Answer 1

#F_c=48*pi#

#"The centripetal force is given by formula :" F_c=m*v^2/r# #"we can write: " v=omega*r # #F_c=m*omega^2*r^2/r# #F_c=m*omega^2*r# #omega=2*pi*f# #F_c=6*2*pi*1*4# #F_c=48*pi#

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Answer 2

Xavier Adame

The centripetal force acting on an object in circular motion is given by the formula ( F = m \cdot r \cdot \omega^2 ), where ( F ) is the centripetal force, ( m ) is the mass of the object, ( r ) is the radius of the circular path, and ( \omega ) is the angular velocity in radians per second.

Given that the mass ( m = 6 ) kg, the radius ( r = 4 ) m, and the frequency ( f = 1 ) Hz, we can calculate the angular velocity using the formula ( \omega = 2\pi f ). Substituting the values, we get ( \omega = 2\pi \cdot 1 = 2\pi ) rad/s.

Now, we can calculate the centripetal force:

[ F = m \cdot r \cdot \omega^2 = 6 \cdot 4 \cdot (2\pi)^2 \approx 150.8 \text{ N} ]

Therefore, the centripetal force acting on the object is approximately 150.8 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.