A model train, with a mass of #3 kg#, is moving on a circular track with a radius of #8 m#. If the train's rate of revolution changes from #5/4 Hz# to #3/8 Hz#, by how much will the centripetal force applied by the tracks change by?
1 Answer
The the centripetal force is decreased by
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 (
#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=(F_c)_f(F_c)_i#
#=mr(2pif_f)^2mr(2pif_i)^2#
We can simplify this equation:
#=>color(purple)(DeltaF_c=4mrpi^2(f_f^2f_i^2))#
We are provided with the following information:

#>"m=3"kg"# 
#>"r"=8"m"# 
#>f_i=5/4"s"^1# 
#>f_f=3/8"s"^1#
Substituting these values into the equation we derived above:
#DeltaF_c=4(3"kg")(8"m")pi^2[(3/8"s"^1)^2(5/4"s"^1)^2]#
#~~color(red)(1347"N")#
Therefore, the centripetal force is decreased by