# A model train with a mass of 9 kg is moving along a track at 18 (cm)/s. If the curvature of the track changes from a radius of 32 cm to 45 cm, by how much must the centripetal force applied by the tracks change?

Apr 17, 2016

$- 0.26325 N$

#### Explanation:

Centripetal (center-seeking) force is given by the equation

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

where $m$ is mass, $v$ is velocity and $r$ is radius.

At the beginning, when the radius is $0.32 m$ and velocity $0.18 m {s}^{-} 1$, the centripetal force would be

$F = \frac{9 k g \cdot 0.0324 {m}^{2} {s}^{-} 2}{0.32 m}$
$0.91125 N$

In the second part, when the radius has expanded to $0.45 m$, centripetal force is

$F = \frac{9 k g \cdot 0.0324 {m}^{2} {s}^{-} 2}{0.45 m}$
$= 0.648 N$

The change in centripetal force is then found by simple subtraction,

$\Delta F = 0.648 N - 0.91125 N$
$= - 0.26325 N$