# A model train with a mass of 2 kg is moving along a track at 9 (cm)/s. If the curvature of the track changes from a radius of 6 cm to 25 cm, by how much must the centripetal force applied by the tracks change?

Apr 19, 2018

The change in centripetal force is $= 0.2052 N$

#### Explanation:

The centripetal force is

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

The mass is of the train $m = 2 k g$

The velocity of the train is $v = 0.09 m {s}^{-} 1$

The radii of the tracks are

${r}_{1} = 0.06 m$

and

${r}_{2} = 0.25 m$

The variation in the centripetal force is

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

The centripetal forces are

$| | {F}_{1} | | = 2 \cdot {0.09}^{2} / 0.06 = 0.27 N$

$| | {F}_{2} | | = 2 \cdot {0.09}^{2} / 0.25 = 0.0648 N$

$\Delta F = {F}_{1} - {F}_{2} = 0.27 - 0.0648 = 0.2052 N$

Apr 19, 2018

$0.378 \text{ N}$

#### Explanation:

Consider a diagram:

We know that centripetal acceleration, towards the centre, is $a = {v}^{2} / r$

$v = 9 {\text{ cm s}}^{-} 1$
$= 0.09 {\text{ m s}}^{-} 1$

Now, use the formula to find the acceleration in each case for each radius.

Let ${a}_{1}$ be the acceleration for $6 \text{cm}$ and ${a}_{2}$ be the acceleration for $25 \text{cm}$

${a}_{1} = {0.09}^{2} / 0.06$
$= 0.135 {\text{ m s}}^{-} 2$

${a}_{2} = {0.09}^{2} / 0.25$
$= 0.324 {\text{ m s}}^{-} 2$

Now, since the acceleration will be towards the centre of the circle, resolve forces towards the centre using Newton's Second Law

${F}_{1} = 2 \times 0.135$
$= 0.27 \text{ N}$

${F}_{2} = 2 \times 0.324$
$= 0.648 \text{ N}$

So the increase in centripetal force is:

$\Delta F = 0.648 - 0.27$
$= 0.378 \text{ N}$