# A model train with a mass of 2 kg is moving along a track at 18 (cm)/s. If the curvature of the track changes from a radius of 9 cm to 12 cm, by how much must the centripetal force applied by the tracks change?

Jul 7, 2016

0.18N

#### Explanation:

Since $F = m a$ and ${A}_{c} = \frac{{V}^{2}}{r}$, ${F}_{c}$ must be ${F}_{c} = m \frac{{V}^{2}}{r}$, by substituting the acceleration in a circle into the equation for Force.

First we need to convert all of the values into standard units, since if we make the calculation in cm we will not get Newtons as the answer since Newtons is $K g m {s}^{-} 1$ instead we would get $K g c m {s}^{-} 1$

So we need to convert the $c m {s}^{-} 1$ and $c m$ into $m {s}^{-} 1$ and $m$ respectively.

$V = \frac{18 c m {s}^{-} 1}{100} = 0.18 m {s}^{-} 1$
${r}_{1} = \frac{9 c m}{100} = 0.09 m$
${r}_{2} = \frac{12 c m}{100} = 0.12 m$

Since we now have the mass, radius and velocity we simply substitute the values for the 0.09m radius into the equation as follows:

${F}_{c 1} = 2 k g \left(\frac{{\left(0.18 m {s}^{-} 1\right)}^{2}}{0.09 m}\right) = 0.72 N$

Then we repeat for the 0.12m radius

${F}_{c 2} = 2 k g \left(\frac{{\left(0.18 m {s}^{-} 1\right)}^{2}}{0.12 m}\right) = 0.54 N$

Now that we have both Centripetal Forces we can calculate the difference:

${F}_{c} = {F}_{c 1} - {F}_{c 2} = 0.72 N - 0.54 N = 0.18 N$