# A model train with a mass of 3 kg is moving along a track at 16 (cm)/s. If the curvature of the track changes from a radius of 15 cm to 32 cm, by how much must the centripetal force applied by the tracks change?

Oct 30, 2017

The change in centripetal force is $= 0.27 N$

#### Explanation:

The centripetal force is

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

The mass, $m = \left(3\right) k g$

The speed, $v = \left(0.16\right) m {s}^{-} 1$

The radius, $= \left(r\right) m$

The variation in centripetal force is

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

${F}_{1} = m {v}^{2} / {r}_{1} = 3 \cdot {0.16}^{2} / 0.15 = 0.512 N$

${F}_{2} = m {v}^{2} / {r}_{2} = 3 \cdot {0.16}^{2} / 0.32 = 0.24 N$

$\Delta F = 0.512 - 0.24 = 0.27 N$