Question

A model train, with a mass of `8 kg`, is moving on a circular track with a radius of `6 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

The change in centripetal force is `=5145.35N`

Explanation:

The centripetal force is

`F=(mv^2)/r=mromega^2N`

The mass of the train, `m=(8)kg`

The radius of the track, `r=(6)m`

The frequencies are

`f_1=(5/3)Hz`

`f_2=(1/4)Hz`

The angular velocity is `omega=2pif`

The variation in centripetal force is

`DeltaF=F_2-F_1`

`F_1=mromega_1^2=mr*(2pif_1)^2=8*6*(2pi*5/3)^2=5263.79N`

`F_2=mromega_2^2=mr*(2pif_2)^2=8*6*(2pi*1/4)^2=118.44N`

`DeltaF=F_2-F_1=5263.79-118.44=5145.35N`