# A model train with a mass of 8 kg is moving along a track at 9 (cm)/s. If the curvature of the track changes from a radius of 27 cm to 81 cm, by how much must the centripetal force applied by the tracks change?

Change in centripetal force $\Delta F = - 16000 \text{ }$Dynes

#### Explanation:

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

Solve for $\Delta F$

Replace $F$ with $F + \Delta F$ and $r$ with $r + \Delta r$

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

Subtract $F$ from both sides

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

$\cancel{F} + \Delta F - \cancel{F} = \frac{m {v}^{2}}{r + \Delta r} - \frac{m {v}^{2}}{r}$

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

$\Delta F = \left(m {v}^{2}\right) \left(\frac{1}{r + \Delta r} - \frac{1}{r}\right)$

But $\Delta r = 81 - 27 = 54$

$\Delta F = \left(8000 {\left(9\right)}^{2}\right) \left(\frac{1}{27 + 54} - \frac{1}{27}\right)$

$\Delta F = - 16000 \text{ }$Dynes

God bless....I hope the explanation is useful.