A model train with a mass of 8 kg is moving along a track at 12 (cm)/s. If the curvature of the track changes from a radius of 48 cm to 120 cm, by how much must the centripetal force applied by the tracks change?

Apr 7, 2017

The centripetal force changes by $= 0.144 N$

Explanation:

The centripetal force is

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

mass, $m = 8 k g$

speed, $v = 0.12 m {s}^{-} 1$

radius, $= r$

The variation in centripetal force is

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

${F}_{1} = 8 \cdot {0.12}^{2} / 0.48 = 0.24 N$

${F}_{2} = 8 \cdot {0.12}^{2} / 1.20 = 0.096 N$

$\Delta F = 0.24 - 0.096 = 0.144 N$