# A model train, with a mass of 4 kg, is moving on a circular track with a radius of 7 m. If the train's rate of revolution changes from 1/6 Hz to 1/8 Hz, by how much will the centripetal force applied by the tracks change by?

Aug 9, 2017

#### Answer:

The centripetal force will decrease by $\approx 13 \text{N}$

#### 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 the angular velocity ($\omega$) as:

${a}_{c} = r {\omega}^{2}$

Therefore, we can state:

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

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

$\omega = 2 \pi f$

Putting this all together, we have:

color(blue)(F_c=mr(2pif)^2

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

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

$= m r {\left(2 \pi {f}_{f}\right)}^{2} - m r {\left(2 \pi {f}_{i}\right)}^{2}$

We can simplify this equation:

$\implies m r \left(4 {\pi}^{2} {f}_{f}^{2}\right) - m r \left(4 {\pi}^{2} {f}_{i}^{2}\right)$

$\implies \textcolor{p u r p \le}{\Delta {F}_{c} = 4 m r {\pi}^{2} \left({f}_{f}^{2} - {f}_{i}^{2}\right)}$

We are provided with the following information:

• $\to \text{m=4"kg}$

• $\to \text{r"=7"m}$

• $\to {f}_{i} = \frac{1}{6} {\text{s}}^{-} 1$

• $\to {f}_{f} = \frac{1}{8} {\text{s}}^{-} 1$

Substituting these values into the equation we derived above:

DeltaF_c=4(4"kg")(7"m")pi^2[(1/8/"s"^-1)^2-(1/6"s"^-1)^2]

$\approx \textcolor{red}{- 13.43 \text{N}}$

Therefore, the centripetal force is decreased by $13.43 \text{N}$.