A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #1 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?

Question

A model train, with a mass of #8 kg#, is moving on a circular track with a radius of #1 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?

Elena Albarran · Accepted Answer

#F_c = 137N#

In order to get the centripetal force for each instance we need to substitute mass, velocity and radius into the following equation:

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

Hertz are revolutions per second therefore we can calculate the velocity of the train in both situations.

So first calculate the distance of the track:

#d = 2pir = 2pi(1m) = 2pi#

I will leave the result in terms of pi for accuracy.

So #v = d / t#

Therefore since we know that a hertz is revolutions per second so know the time and we can obtain the distance by multiplying the hertz by the distance of track. So the velocity of the first #5/3 Hz# is:

#v = ((5/3)(2pi))/1 = 10.47 ms^-1# (4 sf)

For the #1/4 Hz#

#v = ((1 / 4)(2pi))/1 = 1.571 ms^-1# (4 sf)

Now since we have all of the variables for the force calculation substitute them into that equation:

#F_c = (8kg)((10.47 ms^-1)^2)/(2pi) = 140N# (3sf)

And

#F_c = (8kg)((1.571 ms^-1)^2)/(2pi) = 3.14N# (3sf)

Go ahead and calculate the difference:

#F_c = 140N - 3.14N = 137N# (3sf)

Nora Arzate · Answer

undefined The centripetal force will decrease by approximately 243.84 Newtons.