# A model train with a mass of 4 kg is moving along a track at 8 (cm)/s. If the curvature of the track changes from a radius of 54 cm to 27 cm, by how much must the centripetal force applied by the tracks change?

Mar 4, 2017

$0.0474 N$

#### Explanation:

Firstly let's just put everything in SI units - change the $c m$ to $m$ to make everything easier to solve.

$8 c m {s}^{-} 1 = 0.08 m {s}^{-} 1$
$54 c m = 0.54 m$
$27 = 0.27 m$

Now, the equation for centripetal force is

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

where $F$ is force, $a$ is acceleration, $m$ is mass$,$v$i s v e l o c i t y$ and $r$ is radius.

In this case, the radius is changing, so we can say that

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

Putting in the values that we know,

$\Delta F = \frac{4 \cdot {0.08}^{2}}{0.27} - \frac{4 \cdot {0.08}^{2}}{0.54}$

$= 0.0474 N$