A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #5 m#. If the train's rate of revolution changes from #1/9 Hz# to #1/4 Hz#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #5 m#.  If the train's rate of revolution changes from #1/9 Hz# to #1/4 Hz#, by how much will the centripetal force applied by the tracks change by?

Benjamin Asiedu · Accepted Answer

Change in centripetal force#=mw_2^2r-mw_1^2r#
#=4pi^2mr(n_2^2-n_1^2)=4xxpi^2xx4xx5xx(1/4^2-1/9^2)#
#~~39.6N#

Adrian Ayala · Answer

undefined The centripetal force applied by the tracks on the model train can be calculated using the formula:

$$ F = m \times r \times (2 \times \pi \times f)^2 $$

Where:
$ F $ = centripetal force
$ m $ = mass of the train (4 kg)
$ r $ = radius of the circular track (5 m)
$ f $ = frequency of revolution

Initial frequency, $ f_1 = \frac{1}{9} $ Hz
Final frequency, $ f_2 = \frac{1}{4} $ Hz

Initial centripetal force, $ F_1 = 4 \times 5 \times (2 \times \pi \times \frac{1}{9})^2 $

Final centripetal force, $ F_2 = 4 \times 5 \times (2 \times \pi \times \frac{1}{4})^2 $

Change in centripetal force, $ \Delta F = F_2 - F_1 $

Calculating the values:

$ F_1 = 4 \times 5 \times (2 \times \pi \times \frac{1}{9})^2 $
$ F_1 \approx 8.791 $ N

$ F_2 = 4 \times 5 \times (2 \times \pi \times \frac{1}{4})^2 $
$ F_2 \approx 78.539 $ N

$ \Delta F = F_2 - F_1 $
$ \Delta F \approx 69.748 $ N

Therefore, the centripetal force applied by the tracks changes by approximately 69.748 N.