# A model train with a mass of 2 kg is moving along a track at 6 (cm)/s. If the curvature of the track changes from a radius of 3 cm to 8 cm, by how much must the centripetal force applied by the tracks change?

Aug 12, 2017

The centripetal force changes by $= 0.15 N$

#### Explanation:

The centripetal force is

$F = \frac{m {v}^{2}}{r}$

The mass, $m = \left(2\right) k g$

The speed, $v = \left(0.06\right) m {s}^{-} 1$

The radius, $= \left(r\right) m$

The variation in centripetal force is

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

${F}_{1} = m {v}^{2} / {r}_{1} = 2 \cdot {0.06}^{2} / 0.03 = 0.24 N$

${F}_{2} = m {v}^{2} / {r}_{2} = 2 \cdot {0.06}^{2} / 0.08 = 0.09 N$

$\Delta F = 0.24 - 0.09 = 0.15 N$

Aug 12, 2017

The centripetal force is decreased by $0.15 \text{N}$.

#### Explanation:

The centripetal force is given in accordance with Newton's second law as:

${F}_{c} = m {a}_{c}$

where $m$ is the mass of the object and ${a}_{c}$ is the centripetal acceleration experienced by the object

The centripetal acceleration can be expressed in terms of velocity as:

${a}_{c} = \frac{{v}^{2}}{r}$

Therefore, we can state:

${F}_{c} = \frac{m {v}^{2}}{r}$

The angular velocity can also be expressed in terms of the frequency of the motion as:

To find the change in centripetal force as the radius changes, we're being asked for $\Delta {F}_{c}$, where:

$\Delta {F}_{c} = {\left({F}_{c}\right)}_{f} - {\left({F}_{c}\right)}_{i}$

$= \frac{m {v}^{2}}{r} _ f - \frac{m {v}^{2}}{r} _ i$

We can simplify this equation:

$\implies m {v}^{2} \left(\frac{1}{r} _ f - \frac{1}{r} _ i\right)$

$\implies \textcolor{p u r p \le}{\Delta {F}_{c} = m {v}^{2} \left(\frac{1}{r} _ f - \frac{1}{r} _ i\right)}$

We are provided with the following information:

• $\to \text{m=2"kg}$

• $\mapsto v = 0.06 \text{m"//"s}$

• $\to \text{r_i"=0.03"m}$

• $\to \text{r_f"=0.08"m}$

Substituting these values into the equation we derived above:

$\Delta {F}_{c} = {\left(2 \text{kg")(0.06"m"//"s}\right)}^{2} \left(\frac{1}{0.08} - \frac{1}{0.03}\right)$

$\implies \textcolor{c r i m s o n}{- 0.15 \text{N}}$

Therefore, the centripetal force is decreased by $0.15 \text{N}$.