# A model train, with a mass of 2 kg, is moving on a circular track with a radius of 8 m. If the train's rate of revolution changes from 1/2 Hz to 2/5 Hz, by how much will the centripetal force applied by the tracks change by?

Feb 25, 2017

$\Delta {F}_{c} = 3 {\pi}^{2}$ Newtons

#### Explanation:

Here, the centripetal force is from the tracks keeping the train in a circle.

Finding the velocity from the frequency
$\frac{1}{2}$ $H z$

$= \left(1 \text{rotation")/(2"seconds}\right)$

$= \left(1 \text{rotation")/(2"seconds") * (2pir)/(1"rotation}\right)$

$= 8 \pi$ meters per second

Similarly, $\frac{2}{5}$ $H z$ = $7.2 \pi$ meters per second

Newtons second law
$F = m a$
${F}_{c} = m \cdot {a}_{c}$

Using the equation for centripetal acceleration
${F}_{c} = m \cdot \left({v}^{2} / r\right)$

Finding the difference in force

F_"(at 0.5Hz)" - F_"(at 0.4Hz)" = m*(v_"at0.5Hz"^2/r) - m*(v_"at0.4Hz"^2/r)

Substituting in values

$\Delta {F}_{c} = 2 \cdot \left({\left(8 \pi\right)}^{2} / 8\right) - 2 \cdot \left({\left(7.2 \pi\right)}^{2} / 8\right)$

$\Delta {F}_{c} = 16 {\pi}^{2} - 13 {\pi}^{2}$

$\Delta {F}_{c} = 3 {\pi}^{2}$ Newtons

Which is about $30$ Newtons