Question

A model train with a mass of `5 kg` is moving along a track at `12 (cm)/s`. If the curvature of the track changes from a radius of `16 cm` to `24 cm`, by how much must the centripetal force applied by the tracks change?

Answer 1

Given,

`m = 5"kg"`

`nu = (0.12"m")/"s"`

`r_1 = 0.16"m"`

`r_2 = 0.24"m"`

Now, recall,

`F_"R" = m * nu^2/r`

Hence, the centripetal force applied by the tracks on the model train must change by,

`DeltaF_"R" = mnu^2(1/r_1 - 1/r_2) approx 0.15"N"`

Note that as the radius increases, the centripetal force will generally decrease, such that,

`F_"R" propto 1/r`

Answer 2

The change in centripetal force is `=0.15N`

Explanation:

The centripetal force is

`vecF_C=(mv^2)/r*vecr`

The mass is of the train `m=5kg`

The velocity of the train is `v=0.12ms^-1`

The radii of the tracks are

`r_1=0.16m`

and

`r_2=0.24m`

The variation in the centripetal force is

`DeltaF=F_2-F_1`

The centripetal forces are

`||F_1||=5*0.12^2/0.16=0.45N`

`||F_2||=5*0.12^2/0.24=0.30N`

`DeltaF=F_1-F_2=0.45-0.30=0.15N`