# A model train, with a mass of 8 kg, is moving on a circular track with a radius of 1 m. If the train's rate of revolution changes from 5/8 Hz to 5/4 Hz, by how much will the centripetal force applied by the tracks change by?

Mar 4, 2017

$123.37 N$

#### Explanation:

The centripetal force of an object is given by

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

You can also find $v$ by

$v = 2 \pi r f$

where $f$ is the frequency of revolution.

Substitute this into the above equation,

$F = \frac{m {\left(2 \pi r f\right)}^{2}}{r} = \frac{4 m {\pi}^{2} {r}^{2} {f}^{2}}{r}$

$= 4 m r {\pi}^{2} {f}^{2}$

or, since we're looking at the change in centripetal force from the change in frequency, then

$\Delta F = 4 m r {\pi}^{2} {\left(\Delta f\right)}^{2}$

where $\Delta f$ is $\frac{5}{4} H z - \frac{5}{8} H z = \frac{5}{8} H z$

Therefore, using the values that we know,

$\Delta F = 4 \times 8 \times 1 \times {\pi}^{2} \times {\left(\frac{5}{8}\right)}^{2}$

$= 123.37 N$