# 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 1/4 Hz to 1/7 Hz, by how much will the centripetal force applied by the tracks change by?

Jul 23, 2018

The change in centripetal force is $= 39.88 N$

#### 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, $r = \left(8\right) m$

The frequencies are

${f}_{1} = \left(\frac{1}{4}\right) H z$

${f}_{2} = \left(\frac{1}{7}\right) H z$

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

The variation in centripetal force is

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

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 3 \cdot 8 \cdot {\left(2 \pi \cdot \frac{1}{4}\right)}^{2} = 59.22 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 3 \cdot 8 \cdot {\left(2 \pi \cdot \frac{1}{7}\right)}^{2} = 19.34 N$

$\Delta F = | {F}_{2} - {F}_{1} | = 59.22 - 19.34 = 39.88 N$