# A model train with a mass of 2 kg is moving along a track at 9 (cm)/s. If the curvature of the track changes from a radius of 6 cm to 3 cm, by how much must the centripetal force applied by the tracks change?

Jun 18, 2017

The centripetal force changes by $= 0.27 N$

#### Explanation:

The centripetal force is

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

mass, $m = 2 k g$

speed, $v = 0.09 m {s}^{-} 1$

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

The variation in centripetal force is

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

${F}_{1} = m {v}^{2} / {r}_{1} = 2 \cdot {0.09}^{2} / 0.06 = 0.27 N$

${F}_{2} = m {v}^{2} / {r}_{2} = 2 \cdot {0.09}^{2} / 0.03 = 0.54 N$

$\Delta F = 0.54 - 0.27 = 0.27 N$
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