A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #7 m#. If the train's rate of revolution changes from #1/6 Hz# to #1/8 Hz#, by how much will the centripetal force applied by the tracks change by?

1 Answer
Aug 9, 2017

Answer:

The centripetal force will decrease by #~~13"N"#

Explanation:

The centripetal force is given in accordance with Newton's second law as:

#F_c=ma_c#

where #m# is the mass of the object and #a_c# is the centripetal acceleration experienced by the object

The centripetal acceleration can be expressed in terms of the angular velocity (#omega#) as:

#a_c=romega^2#

Therefore, we can state:

#F_c=mromega^2#

The angular velocity can also be expressed in terms of the frequency of the motion as:

#omega=2pif#

Putting this all together, we have:

#color(blue)(F_c=mr(2pif)^2#

To find the change in centripetal force as the frequency changes, we're being asked for #DeltaF_c#, where:

#DeltaF_c=(F_c)_f-(F_c)_i#

#=mr(2pif_f)^2-mr(2pif_i)^2#

We can simplify this equation:

#=>mr(4pi^2f_f^2)-mr(4pi^2f_i^2)#

#=>color(purple)(DeltaF_c=4mrpi^2(f_f^2-f_i^2))#

We are provided with the following information:

  • #->"m=4"kg"#

  • #->"r"=7"m"#

  • #->f_i=1/6"s"^-1#

  • #->f_f=1/8"s"^-1#

Substituting these values into the equation we derived above:

#DeltaF_c=4(4"kg")(7"m")pi^2[(1/8/"s"^-1)^2-(1/6"s"^-1)^2]#

#~~color(red)(-13.43"N")#

Therefore, the centripetal force is decreased by #13.43"N"#.