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 `3/8 Hz` to `5/4 Hz`, by how much will the centripetal force applied by the tracks change by?

Answer 1

We know that angular frequency `omega` and natural frequency `nu` are related as, `omega =2 pinu`

So,if the angular frequency changed from `omega_1` to `omega_2`

we can say, `omega_2 =2pinu_2=2pi(5/4)=7.854 rads^-1` and `omega_1=2pinu_1=2pi(3/8)=2.355 rads^-1`

Now,centripetal force is expressed as `momega^2r`

So,change in centripetal force is `mr(omega_2^2 - omega_1^2)=8*1*(7.854^2-2.355^2)=449.113 N`

Answer 2

The change in centripetal force is `=449.1N`

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=(3/8)Hz`

`f_2=(5/4)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*1*(2pi*3/8)^2=44.4N`

`F_2=mromega_2^2=mr*(2pif_2)^2=8*1*(2pi*5/4)^2=493.5N`

`DeltaF=F_1-F_2=493.5-44.4=449.1N`