# A model train, with a mass of 12 kg, is moving on a circular track with a radius of 9 m. If the train's rate of revolution changes from 6 Hz to 8 Hz, by how much will the centripetal force applied by the tracks change by?

Mar 11, 2016

The change in Centripetal Force will be $48 N \to 85.3 N$.

#### Explanation:

Here:

$m = 12$
${v}_{1} = 6$
${v}_{2} = 8$
$r = 9$

We know:

Centripetal Force $= \frac{m {v}^{2}}{r}$

For first rate of revolution. i.e. $6 H z$

$C {F}_{1} = \frac{12 \cdot {6}^{2}}{9}$

$C {F}_{1} = \frac{12 \cdot 36}{9}$

$C {F}_{1} = \frac{432}{9}$

$C {F}_{1} = 48 N$

For second rate of revolution. i.e. $8 H z$

$C {F}_{2} = \frac{12 \cdot {8}^{2}}{9}$

$C {F}_{2} = \frac{12 \cdot 64}{9}$

$C {F}_{2} = \frac{768}{9}$

$C {F}_{2} = 85.3 N$

The change in Centripetal Force will be $48 N \to 85.3 N$.