# A model train with a mass of 9 kg is moving along a track at 18 (cm)/s. If the curvature of the track changes from a radius of 24 cm to 42 cm, by how much must the centripetal force applied by the tracks change?

Nov 21, 2016

The centripetal force applied by the tracks change is ${F}_{c} = 1.62 N$.

#### Explanation:

Centripetal force ${F}_{c} = m {v}^{2} / r$
As we know change in force '$\Delta {F}_{c}$' is inversely proportional to the change in the radius '$\Delta r$'.
So $\Delta r = 42 c m - 24 c m = 18 c m$
Conversion into metres for SI units $\therefore \Delta r = 18 \times {10}^{-} 2 m e t r e$.
Similarly to the velocity$\therefore v = 18 \times {10}^{-} 2 m / s$.
Substituting these values in above force equation we get,
${F}_{c} = \frac{\left(9 k g\right) {\left(18 \times {10}^{-} 2 m / s\right)}^{2}}{18 \times {10}^{-} 2 m} = 1.62 N$.