# A model train, with a mass of 3 kg, is moving on a circular track with a radius of 2 m. If the train's rate of revolution changes from 5/3 Hz to 3/4 Hz, by how much will the centripetal force applied by the tracks change by?

May 31, 2016

It is -524.62N.

#### Explanation:

The centripetal force is

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

We have both $m$ and $r$ we need to find $v$ and we can calculate it.

The frequency tell us how many turns the train does in a second. A turn is long $2 \pi r$, so the velocity is the frequency multiplied by the circumference. For the initial velocity we have

${v}_{i} = \frac{5}{3} \cdot 2 \pi r = \frac{5}{3} \cdot 2 \pi \cdot 2 \setminus \approx 20.94$ m/s.

The initial centripetal force is then

${F}_{i} = m {v}_{i}^{2} / r = 3 \cdot {20.94}^{2} / 2 = 657.72$ N.

The final velocity is

${v}_{f} = \frac{3}{4} \cdot 2 \pi r = \frac{3}{4} \cdot 2 \pi \cdot 2 \setminus \approx 9.42$ m/s

and the final force is

${F}_{f} = m {v}_{f}^{2} / r = 3 \cdot {9.42}^{2} / 2 \setminus \approx 133.1$ N

The difference in force is then

${F}_{f} - {F}_{i} = 133.1 - 657.72 = - 524.62$ N.

The negative sign is because the centripetal force decreases since the train is lowering the speed.

May 31, 2016

We have the following numbers at our disposal:

• $m = \text{3 kg}$, the mass of the train
• $r = \text{2 m}$, the radius of the train's path
• ${\omega}_{i} = \text{5/3 rev"cdot"s"^(-1) xx (2pi " rad")/"rev}$, the initial angular velocity
• ${\omega}_{f} = \text{3/4 rev"cdot"s"^(-1) xx (2pi " rad")/"rev}$, the final angular velocity

Recall that the sum of the centripetal "forces" is:

$\setminus m a t h b f \left(\sum {F}_{c} = \frac{m {v}_{T}^{2}}{r}\right)$

Hence, since we are looking for the change in the sum of the centripetal forces, $\Delta \left(\sum {F}_{c}\right)$, we are looking for $\frac{m \Delta {v}_{T}^{2}}{r}$, where

$\setminus m a t h b f \left({v}_{T} = r \omega\right)$

is the tangential velocity in $\text{m/s}$, while the mass and radius are constant.

The rates of revolution were given in the question, but they are saying:

$\text{5/3 Hz}$ $=$ $\text{5/3 of a}$ $\setminus m a t h b f \left(\text{full revolution}\right)$ $\text{per second}$

So, this angular velocity is actually

omega = (5/3 cancel("rev"))/"s" xx (2pi "rad")/cancel("rev") = (10pi)/3 "rad/s".

Therefore, the change in the sum of the centripetal forces is:

$\textcolor{b l u e}{\Delta \left(\sum {F}_{c}\right)}$

$= \frac{m \Delta {v}_{T}^{2}}{r}$

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

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

= (("3 kg")[(("2 m")(3/4*2pi " rad/s"))^2 - (("2 m")(5/3*2pi " rad/s"))^2])/("2 m")

= (("3 kg")(9pi^2 "m"^2"/s"^2 - (400pi^2)/9 "m"^2"/s"^2))/("2 m")

$=$ $\textcolor{b l u e}{- \text{524.7 N}}$

Thus, the sum of the centripetal forces has decreased by $\text{524.7 N}$.