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An object with a mass of #7 kg# is revolving around a point at a distance of #8 m#. If the object is making revolutions at a frequency of #4 Hz#, what is the centripetal force acting on the object?

Answer 1

Lily Adams

Data:-

Mass#=m=7kg# Distance#=r=8m# Frequency#=f=4Hz# Centripetal Force#=F=??#

Sol:-

We know that:

The centripetal acceleration #a# is given by #F=(mv^2)/r................(i)#

Where #F# is the centripetal force, #m# is the mass, #v# is the tangential or linear velocity and #r# is the distance from center.

Also we know that #v=romega#

Where #omega# is the angular velocity.

Put #v=romega# in #(i)#

#implies F=(m(romega)^2)/r implies F =mromega^2...........(ii)#

The relation between angular velocity and frequency is

#omega=2pif#

Put #omega=2pif# in #(ii)#

#implies F=mr(2pif)^2# #implies F=4pi^2rmf^2#

Now, we are given with all the values

#implies F=4(3.14)^2*8*7*(4)^2=4*9.8596*8*16=35336.8064#

#implies F=35336.8064N#

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Answer 2

Nathan Ashcroft

The centripetal force acting on the object is approximately 1256 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.