A model train with a mass of 2 kg is moving along a track at 36 (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?

Mar 5, 2017

$4.32 N$

Explanation:

Put the values we have into SI units:

$36 c m {s}^{-} 1 = 0.36 m {s}^{-} 1$
$6 c m = 0.06 m$
$3 c m = 0.03 m$

The equation for centripetal force is

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

so the change in centripetal force is

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

Plug in the values we are given,

$\Delta F = \frac{2 \times {0.36}^{2}}{0.03} - \frac{2 \times {0.36}^{2}}{0.06}$

$= 8.64 - 4.32 = 4.32 N$