A model train, with a mass of #5 kg#, is moving on a circular track with a radius of #9 m#. If the train's rate of revolution changes from #4 Hz# to #7 Hz#, by how much will the centripetal force applied by the tracks change by?

Lets first start out by defining what we know:

#m=5kg#
#r=9m#
#f_1=4Hz#
#f_2=7Hz#

The formula for the centripetal force exerted on an object with mass #m# in circular motion travelling at a speed #v#, a distance #r# away from the center of rotation is given by

#F_c=mv^2/r#

We still lack velocity, however, so we are not ready for this step yet. We must first find how fast this train is actually moving.

If the train has a frequency of #4Hz#, then it completes 4 cycles every second. From here we just need to define how long a cycle is in meters. If the track is circular and the radius is given, we can calculate the length of one revolution by finding the circumference.

#d=2pir=2pi(9m)approx56.5m#

Now, we just substitute the distance per cycle into the original velocity relationship.

#(4(56.5m))/(1s)=226m/s#

Now that we have a velocity, we can revisit the centripetal force equation with real values:

#F_c=(5kg)(226m/s)^2/(9m)=28424N#

This is how much force is being applied by the train tracks to keep the train from proceeding in a tangential direction to the circular path.

A more condensed form of the equations used so far will be utilized for this final portion of the problem.

Now we just need to find the centripetal force exerted on the train when the frequency is #7Hz#.

#F_c=m(2pirf)^2/r#

#F_c=(5kg)(2pi(9m)(7Hz))^2/(9m)=87050N#

The change in centripetal force exerted by the tracks onto the train is #approx58674N#