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 `3/5 Hz` to `5/6 Hz`, by how much will the centripetal force applied by the tracks change by?

Answer 1

The change in centripetal force is `=105.6N`

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=(1)m`

The frequencies are

`f_1=(3/5)Hz`

`f_2=(5/6)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*1*(2pi*3/5)^2=113.7N`

`F_2=mromega_2^2=mr*(2pif_2)^2=8*1*(2pi*5/6)^2=219.3N`

`DeltaF=|F_2-F_1|=219.3-113.7=105.6N`