# A model train with a mass of 5 kg is moving along a track at 4 (cm)/s. If the curvature of the track changes from a radius of 16 cm to 2 cm, by how much must the centripetal force applied by the tracks change?

Apr 2, 2016

0.35N

#### Explanation:

We have that:

• $m = 5 k g$
• $v = 4 c m {s}^{-} 1 = 0.04 m {s}^{-} 1$
• ${r}_{1} = 16 c m = 0.16 m$
• ${r}_{2} = 2 c m = 0.02 m$

The centripetal force is given by:

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

Thus we can work out the change in centripetal force:

$\Delta F = {F}_{2} - {F}_{1} = \frac{m {v}^{2}}{r} _ 2 - \frac{m {v}^{2}}{r} _ 1$

$= \frac{\left(5\right) {\left(0.04\right)}^{2}}{0.02} - \frac{\left(5\right) {\left(0.04\right)}^{2}}{0.16} = 0.35 N$