# 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?

Apr 30, 2018

Given,

$m = 5 \text{kg}$

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

${r}_{1} = 0.16 \text{m}$

${r}_{2} = 0.24 \text{m}$

Now, recall,

${F}_{\text{R}} = m \cdot {\nu}^{2} / r$

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

$\Delta {F}_{\text{R" = mnu^2(1/r_1 - 1/r_2) approx 0.15"N}}$

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

${F}_{\text{R}} \propto \frac{1}{r}$

Apr 30, 2018

The change in centripetal force is $= 0.15 N$

#### Explanation:

The centripetal force is

${\vec{F}}_{C} = \frac{m {v}^{2}}{r} \cdot \vec{r}$

The mass is of the train $m = 5 k g$

The velocity of the train is $v = 0.12 m {s}^{-} 1$

The radii of the tracks are

${r}_{1} = 0.16 m$

and

${r}_{2} = 0.24 m$

The variation in the centripetal force is

$\Delta F = {F}_{2} - {F}_{1}$

The centripetal forces are

$| | {F}_{1} | | = 5 \cdot {0.12}^{2} / 0.16 = 0.45 N$

$| | {F}_{2} | | = 5 \cdot {0.12}^{2} / 0.24 = 0.30 N$

$\Delta F = {F}_{1} - {F}_{2} = 0.45 - 0.30 = 0.15 N$