# 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 1/3 Hz to 2/9 Hz, by how much will the centripetal force applied by the tracks change by?

Mar 20, 2018

The change in centripetal force is $= 36.56 N$

#### 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

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

${f}_{2} = \left(\frac{2}{9}\right) H z$

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

The variation in centripetal force is

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

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 5 \cdot 3 \cdot {\left(2 \pi \cdot \frac{1}{3}\right)}^{2} = 65.80 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 5 \cdot 3 \cdot {\left(2 \pi \cdot \frac{2}{9}\right)}^{2} = 29.24 N$

$\Delta F = {F}_{1} - {F}_{2} = 65.80 - 29.24 = 36.56 N$