Question

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?

Answer 1

`F_c = 137N`

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(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 = 2pir = 2pi(1m) = 2pi`

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

So `v = 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 `5/3 Hz` is:

`v = ((5/3)(2pi))/1 = 10.47 ms^-1` (4 sf)

For the `1/4 Hz`

`v = ((1 / 4)(2pi))/1 = 1.571 ms^-1` (4 sf)

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

`F_c = (8kg)((10.47 ms^-1)^2)/(2pi) = 140N` (3sf)

And

`F_c = (8kg)((1.571 ms^-1)^2)/(2pi) = 3.14N` (3sf)

Go ahead and calculate the difference:

`F_c = 140N - 3.14N = 137N` (3sf)