# A model train, with a mass of 3 kg, is moving on a circular track with a radius of 2 m. If the train's rate of revolution changes from 5/3 Hz to 1/8 Hz, by how much will the centripetal force applied by the tracks change by?

Jan 23, 2018

The change in centripetal force is $= 654.3 N$

#### Explanation:

The centripetal force is

$F = \frac{m {v}^{2}}{r} = m r {\omega}^{2} N$

The mass of the train, $m = \left(3\right) k g$

The radius of the track, $r = \left(2\right) m$

The frequencies are

${f}_{1} = \left(\frac{5}{3}\right) H z$

${f}_{2} = \left(\frac{1}{8}\right) H z$

The variation in centripetal force is

$\Delta F = {F}_{2} - {F}_{1}$

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 3 \cdot 2 \cdot {\left(2 \pi \cdot \frac{5}{3}\right)}^{2} = 658.0 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 3 \cdot 2 \cdot {\left(2 \pi \cdot \frac{1}{8}\right)}^{2} = 3.7 N$

$\Delta F = {F}_{1} - {F}_{2} = 658 - 3.7 = 654.3 N$