# A model train with a mass of 8 kg is moving along a track at 12 cms^-1. If the curvature of the track changes from a radius of 45 cm to 240 cm, by how much must the centripetal force applied by the tracks change?

May 21, 2017

The change in magnitude of the centripetal force from before the change of radius to after is $0.256 - 0.048 = 0.208$ $N$.

#### Explanation:

I'm going to work in metres, the SI unit of distance, rather than cm, because that will yield forces in newton (N) the SI unit of force. So, restating the question:

A model train with a mass of 8 kg is moving along a track at $0.12$  ms^−1. If the curvature of the track changes from a radius of $0.45$ $m$ to $2.40$ $m$, by how much must the centripetal force applied by the tracks change?

Centripetal force is given by:

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

Before the change of radius, the centripetal force required is:

$F = \frac{m {v}^{2}}{r} = \frac{8 \times {0.12}^{2}}{0.45} = 0.256$ $N$

After the change of radius, it is:

$F = \frac{m {v}^{2}}{r} = \frac{8 \times {0.12}^{2}}{2.4} = 0.0 .048$ $N$

The change in magnitude from before to after is $0.256 - 0.048 = 0.208$ $N$.