# A model train, with a mass of 4 kg, is moving on a circular track with a radius of 8 m. If the train's rate of revolution changes from 1/6 Hz to 1/2 Hz, by how much will the centripetal force applied by the tracks change by?

Mar 28, 2016

The change in centripetal force will be $\Delta {F}_{c} = 256 \frac{\pi}{9}$N or $\Delta {F}_{c} = 280.73 N$

#### Explanation:

The problem describes a train traveling on a circular track that starts at an initial speed, but then speeds up to a final speed (that is, it increases its rate of revolution).

A greater centripetal force will be required at the final rate of revolution, ${w}_{f}$, compared to the initial rate of revolution (or angular velocity), ${w}_{i}$, in order to keep the train traveling on a circular path (and not fly off the tracks!!). They want us to find the difference between these two centripetal forces.

We can express this difference mathematically as

1) $\Delta {F}_{c} = {F}_{{c}_{f}} - {F}_{{c}_{i}}$

here ${F}_{{c}_{f}}$ and ${F}_{{c}_{i}}$ are the final and initial centripetal forces on the train respectively.

In general, the formula for centripetal force can be expressed as

2) ${F}_{c} = m \cdot {v}^{2} / r$

here $v$ represents the tangential velocity of an object (having a mass $m$) traveling around a circle of radius $r$. ${F}_{c}$ is of course the centripetal force.

Centripetal force, ${F}_{c}$, can also be expressed as

3) ${F}_{c} = m \cdot {w}^{2} \cdot r$ (why? because $v = w \cdot r$!)

where $w$ is the angular velocity of the object (and all the other variables, i.e. ${F}_{c} , m , \mathmr{and} r$, represent the same physical quantities as before). These formulas for centripetal force, ${F}_{c}$, are equivalent, but using the second form makes the solution a bit more straight forward.

Ok, keeping this more useful form for ${F}_{c}$ in mind, let’s express equation 1) in more detail...

4) $\Delta {F}_{c} = {F}_{{c}_{f}} - {F}_{{c}_{i}} = m \cdot {w}_{f}^{2} \cdot r - m \cdot {w}_{i}^{2} \cdot r$

here the final rate of revolution is ${w}_{f} = \frac{1}{2} H z$ and initial rate of revolution is ${w}_{i} = \frac{1}{6} H z$.

Simplifying further, with a little more algebra (i.e. taking advantage of the distributive property).

5) $\Delta {F}_{c} = {F}_{{c}_{f}} - {F}_{{c}_{i}} = m \cdot r \left({w}_{f}^{2} - {w}_{i}^{2}\right)$

It’s given that the mass of the train is $m = 4 k g$ and the radius of the circular track is $8 m$. Noting that 1 revolution equals $2 \pi$ radians, and substituting, we get

6)$\Delta {F}_{c} = m \cdot r \left({w}_{f}^{2} - {w}_{i}^{2}\right) = 4 k g \cdot 8 m \cdot \left({\left(\frac{2 \cdot \pi}{2}\right)}^{2} - {\left(\frac{2 \cdot \pi}{6}\right)}^{2}\right)$

$\Delta {F}_{c} = 280.73 N$