# A model train, with a mass of 5 kg, is moving on a circular track with a radius of 9 m. If the train's rate of revolution changes from 4 Hz to 5 Hz, by how much will the centripetal force applied by the tracks change by?

Apr 5, 2018

See below:

#### Explanation:

I think the best way to do this is to figure out how the time period of rotation changes:

Period and frequency are each other's reciprocal:

$f = \frac{1}{T}$

So the time period of rotation of the train changes from 0.25 seconds to 0.2 seconds. When the frequency increases. (We have more rotations per second)

However, the train still has to cover the full distance of the circumference of the circular track.

Circumference of circle: $18 \pi$ meters

Speed=distance/time

$\frac{18 \pi}{0.25} = 226.19 m {s}^{-} 1$ when frequency is 4 Hz (time period=0.25 s)

$\frac{18 \pi}{0.2} = 282.74 m {s}^{-} 1$ when frequency is 5 Hz. (time period=0.2 s)

Then we can find the centripetal force in both scenarios:

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

So, when the frequency is 4 Hz:

$F = \frac{\left(8\right) \times {\left(226.19\right)}^{2}}{9}$

$F \approx 45.5 k N$

When frequency is 5Hz:
$F = \frac{\left(8\right) \times {\left(282.74\right)}^{2}}{9}$

$F \approx 71 k N$

Change in force:
$71 - 45.5 = 25.5 k N$

So the total force increases by about $25.5 k N$.