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

Mar 16, 2018

The change in centripetal force is $= 111.03 N$

#### Explanation:

The centripetal force is

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

The mass of the train, $m = \left(5\right) k g$

The radius of the track, $r = \left(3\right) m$

The frequencies are

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

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

The angular velocity is $\omega = 2 \pi f$

The variation in centripetal force is

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

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 5 \cdot 3 \cdot {\left(2 \pi \cdot \frac{1}{2}\right)}^{2} = 148.04 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 5 \cdot 3 \cdot {\left(2 \pi \cdot \frac{1}{4}\right)}^{2} = 37.01 N$

$\Delta F = {F}_{1} - {F}_{2} = 148.04 - 37.01 = 111.03 N$