# 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 3/8 Hz to 5/4 Hz, by how much will the centripetal force applied by the tracks change by?

Mar 10, 2018

We know that angular frequency $\omega$ and natural frequency $\nu$ are related as, $\omega = 2 \pi \nu$

So,if the angular frequency changed from ${\omega}_{1}$ to ${\omega}_{2}$

we can say, ${\omega}_{2} = 2 \pi {\nu}_{2} = 2 \pi \left(\frac{5}{4}\right) = 7.854 r a {\mathrm{ds}}^{-} 1$ and ${\omega}_{1} = 2 \pi {\nu}_{1} = 2 \pi \left(\frac{3}{8}\right) = 2.355 r a {\mathrm{ds}}^{-} 1$

Now,centripetal force is expressed as $m {\omega}^{2} r$

So,change in centripetal force is $m r \left({\omega}_{2}^{2} - {\omega}_{1}^{2}\right) = 8 \cdot 1 \cdot \left({7.854}^{2} - {2.355}^{2}\right) = 449.113 N$

Mar 10, 2018

The change in centripetal force is $= 449.1 N$

#### 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

${f}_{1} = \left(\frac{3}{8}\right) H z$

${f}_{2} = \left(\frac{5}{4}\right) H z$

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

The variation in centripetal force is

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

${F}_{1} = m r {\omega}_{1}^{2} = m r \cdot {\left(2 \pi {f}_{1}\right)}^{2} = 8 \cdot 1 \cdot {\left(2 \pi \cdot \frac{3}{8}\right)}^{2} = 44.4 N$

${F}_{2} = m r {\omega}_{2}^{2} = m r \cdot {\left(2 \pi {f}_{2}\right)}^{2} = 8 \cdot 1 \cdot {\left(2 \pi \cdot \frac{5}{4}\right)}^{2} = 493.5 N$

$\Delta F = {F}_{1} - {F}_{2} = 493.5 - 44.4 = 449.1 N$