A model train, with a mass of 8 kg, is moving on a circular track with a radius of 1 m. If the train's rate of revolution changes from 3/5 Hz to 5/6 Hz, by how much will the centripetal force applied by the tracks change by?

May 14, 2018

The change in centripetal force is $= 105.6 N$

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, $r = \left(1\right) m$

The frequencies are

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

${f}_{2} = \left(\frac{5}{6}\right) H z$

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

The variation in centripetal force is

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

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 8 \cdot 1 \cdot {\left(2 \pi \cdot \frac{3}{5}\right)}^{2} = 113.7 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 8 \cdot 1 \cdot {\left(2 \pi \cdot \frac{5}{6}\right)}^{2} = 219.3 N$

$\Delta F = | {F}_{2} - {F}_{1} | = 219.3 - 113.7 = 105.6 N$