# A model train with a mass of 4 kg is moving along a track at 18 (cm)/s. If the curvature of the track changes from a radius of 25 cm to 42 cm, by how much must the centripetal force applied by the tracks change?

Oct 20, 2016

The centripetal force changes in a factor of $\frac{25}{42}$, i.e. approximately $0.6$ times greater.

#### Explanation:

The centripetal force acting on a moving mass $m$ traveling a circular path with radius $r$ at a constant speed $v$ is given by the formula:

${F}_{c} = m {v}^{2} / r$

If the path's radius is modified from a ${r}_{1}$ value to a ${r}_{2}$ one, the initial centripetal force ${F}_{c 1}$ changes to a new value ${F}_{c 2}$ which can be compared using the above formula:

$\frac{{F}_{c 2}}{{F}_{c 1}} = \frac{m {v}^{2} / {r}_{2}}{m {v}^{2} / {r}_{1}} = \frac{\frac{1}{r} _ 2}{\frac{1}{r} _ 1} = {r}_{1} / {r}_{2}$

Thus:

$\frac{{F}_{c 2}}{{F}_{c 1}} = {r}_{1} / {r}_{2} = \frac{25 c m}{42 c m} \approx 0.595 \ldots \Rightarrow {F}_{c 2} \approx 0.595 {F}_{c 1}$