A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #6 m#. If the train's rate of revolution changes from #5/3 Hz# to #1/4 Hz#, by how much will the centripetal force applied by the tracks change by?

1 Answer
May 1, 2018

Answer:

The change in centripetal force is #=5145.35N#

Explanation:

The centripetal force is

#F=(mv^2)/r=mromega^2N#

The mass of the train, #m=(8)kg#

The radius of the track, #r=(6)m#

The frequencies are

#f_1=(5/3)Hz#

#f_2=(1/4)Hz#

The angular velocity is #omega=2pif#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=mromega_1^2=mr*(2pif_1)^2=8*6*(2pi*5/3)^2=5263.79N#

#F_2=mromega_2^2=mr*(2pif_2)^2=8*6*(2pi*1/4)^2=118.44N#

#DeltaF=F_2-F_1=5263.79-118.44=5145.35N#