# A model train, with a mass of 8 kg, is moving on a circular track with a radius of 6 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?

May 1, 2018

The change in centripetal force is $= 5145.35 N$

#### Explanation:

The centripetal force is

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

The mass of the train, $m = \left(8\right) k g$

The radius of the track, $r = \left(6\right) m$

The frequencies are

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

${f}_{2} = \left(\frac{1}{4}\right) H z$

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

The variation in centripetal force is

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

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 8 \cdot 6 \cdot {\left(2 \pi \cdot \frac{5}{3}\right)}^{2} = 5263.79 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 8 \cdot 6 \cdot {\left(2 \pi \cdot \frac{1}{4}\right)}^{2} = 118.44 N$

$\Delta F = {F}_{2} - {F}_{1} = 5263.79 - 118.44 = 5145.35 N$