# A model train, with a mass of 6 kg, is moving on a circular track with a radius of 2 m. If the train's rate of revolution changes from 2 Hz to 6 Hz, by how much will the centripetal force applied by the tracks change by?

Oct 9, 2017

The force will change by $15150 N$

#### Explanation:

The equation for centripetal force is

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

where $v$ is linear velocity, $\omega$ is angular velocity, $r$ is radius and $m$ is mass.

We know that the mass is $6 k g$ and the radius is $2 m$. We will use angular velocity for these calculations, but we could equally do it with linear velocity.

We know that

$\omega = 2 \pi f$

where $f$ is the frequency, which we have.

For $f = 2 H z$

then

$\omega = 2 \pi \cdot 2 = 12.6 \text{rad"/"sec}$
${F}_{c} = 6 \cdot {12.6}^{2} \cdot 2 = 1905 N$

whereas with $f = 6 H z$

$\omega = 2 \pi f = 2 \pi \cdot 6 = 37.7 \text{rad"/"sec}$

so

${F}_{c} = 6 \cdot {37.7}^{2} \cdot 2 = 17055 N$

Therefore the centripetal force will change by

$\Delta {F}_{c} = 17055 N - 1905 N = 15150 N$