An object with a mass of #6 kg# is revolving around a point at a distance of #8 m#. If the object is making revolutions at a frequency of #2 Hz#, what is the centripetal force acting on the object?

Answer 1

I found: #7580N#

Centripetal Force is: #F_c=mv^2/r# where: #m=#mass; #v=# linear velocity; #r=# radius.

The frequency tells us the number of complete revolutions in one seconds (here 2 revolutions). We can ask ourselves what will be the Angular Velocity #omega# of our object, i.e., a kind of "curved" velocity involving not linear distance but angle described in time!

So we get:

#omega="angle"/"time"=(2pi)/T#

Where #T# will be the Period of time for a complete revolution (time to describe #2pi# radians).

The good thing is that the period is connected to frequency as #"frequency"=nu=1/T# and also the angular velocity can be changed into linear velocity simply considering at what distance from the center you are travelling....(basically, you include the radius)!!!! so: #v=omega*r#

in our case we get (collecting all our stuff):

#F_c=m(omega*r)^2/r=momega^2r=m(2pinu)^2r=6*(2*3.14*2)^2*8~~7580N#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Parker Abbott

The centripetal force acting on the object is 192 N.

By signing up, you agree to our Terms of Service and Privacy Policy

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.