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

Question

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

Lily Adams · Accepted Answer

The centripetal force change by #=19.74N#

Centripetal force is what

#F=mr omega^2#

The centripetal force fluctuation is

#DeltaF=mr(Delta omega)^2#

#=2*4*((1/2-1/4)2pi)^2#

#=8*4/16*pi^2=2pi^2#

#=19.74N#

Carson Alexander · Answer

undefined To calculate the change in centripetal force, we can use the formula:

$$ F = m \cdot \omega^2 \cdot r $$

Where:
- $ F $ is the centripetal force,
- $ m $ is the mass of the train (2 kg),
- $ \omega $ is the angular velocity (given by $ 2\pi f $, where $ f $ is the frequency),
- $ r $ is the radius of the circular track (4 m).

Initially:
$$ f_1 = \frac{1}{2} \text{ Hz} \Rightarrow \omega_1 = 2\pi f_1 $$
$$ f_2 = \frac{2}{8} \text{ Hz} \Rightarrow \omega_2 = 2\pi f_2 $$

Then, we calculate the initial and final centripetal forces:

$$ F_1 = m \cdot \omega_1^2 \cdot r $$
$$ F_2 = m \cdot \omega_2^2 \cdot r $$

Finally, we find the change in centripetal force:
$$ \Delta F = F_2 - F_1 $$