# 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 36 cm to 35 cm, by how much must the centripetal force applied by the tracks change?

Jan 24, 2018

The change in centripetal force is $= 0.023 N$

#### Explanation:

The centripetal force is

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

The mass is of the train $m = 9 k g$

The velocity of the train is $v = 0.18 m {s}^{-} 1$

${r}_{1} = 0.36 m$

and

${r}_{2} = 0.35 m$

The variation in the centripetal force is

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

The centripetal forces are

${F}_{1} = 9 \cdot {0.18}^{2} / 0.36 = 0.833 N$

${F}_{2} = 9 \cdot {0.18}^{2} / 0.35 = 0.81 N$

$\Delta F = {F}_{1} - {F}_{2} = 0.833 - 0.81 = 0.023 N$