# 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 5/3 Hz to 1/4 Hz, by how much will the centripetal force applied by the tracks change by?

Jul 8, 2016

${F}_{c} = 137 N$

#### Explanation:

In order to get the centripetal force for each instance we need to substitute mass, velocity and radius into the following equation:

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

Hertz are revolutions per second therefore we can calculate the velocity of the train in both situations.

So first calculate the distance of the track:

$d = 2 \pi r = 2 \pi \left(1 m\right) = 2 \pi$

I will leave the result in terms of pi for accuracy.

So $v = \frac{d}{t}$

Therefore since we know that a hertz is revolutions per second so know the time and we can obtain the distance by multiplying the hertz by the distance of track. So the velocity of the first $\frac{5}{3} H z$ is:

$v = \frac{\left(\frac{5}{3}\right) \left(2 \pi\right)}{1} = 10.47 m {s}^{-} 1$ (4 sf)

For the $\frac{1}{4} H z$

$v = \frac{\left(\frac{1}{4}\right) \left(2 \pi\right)}{1} = 1.571 m {s}^{-} 1$ (4 sf)

Now since we have all of the variables for the force calculation substitute them into that equation:

${F}_{c} = \left(8 k g\right) \frac{{\left(10.47 m {s}^{-} 1\right)}^{2}}{2 \pi} = 140 N$ (3sf)

And

${F}_{c} = \left(8 k g\right) \frac{{\left(1.571 m {s}^{-} 1\right)}^{2}}{2 \pi} = 3.14 N$ (3sf)

Go ahead and calculate the difference:

${F}_{c} = 140 N - 3.14 N = 137 N$ (3sf)