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 `5/8 Hz` to `5/4 Hz`, by how much will the centripetal force applied by the tracks change by?

Answer 1

`123.37N`

Explanation:

The centripetal force of an object is given by

`F = (mv^2)/r`

You can also find `v` by

`v = 2pirf`

where `f` is the frequency of revolution.

Substitute this into the above equation,

`F = (m(2pirf)^2)/r = (4mpi^2r^2f^2)/r`

`=4mrpi^2f^2`

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

`DeltaF = 4mrpi^2(Deltaf)^2`

where `Deltaf` is `5/4Hz - 5/8Hz = 5/8Hz`

Therefore, using the values that we know,

`DeltaF = 4xx8xx1xxpi^2xx(5/8)^2`

`= 123.37N`