# A model train with a mass of 4 kg is moving along a track at 21 (cm)/s. If the curvature of the track changes from a radius of 420 cm to 140 cm, by how much must the centripetal force applied by the tracks change?

Jul 20, 2017

The change in certipetal force is $= 0.084 N$

#### Explanation:

The centripetal force is

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

mass, $m = 4 k g$

speed, $v = 0.21 m {s}^{-} 1$

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

The variation in centripetal force is

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

${F}_{1} = m {v}^{2} / {r}_{1} = 4 \cdot {0.21}^{2} / 4.2 = 0.042 N$

${F}_{2} = m {v}^{2} / {r}_{2} = 4 \cdot {0.21}^{2} / 1.4 = 0.126 N$

$\Delta F = 0.126 - 0.042 = 0.084 N$