A model train with a mass of #5 kg# is moving along a track at #14 (cm)/s#. If the curvature of the track changes from a radius of #84 cm# to #28 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #5  kg# is moving along a track at #14 (cm)/s#.  If the curvature of the track changes from a radius of #84 cm# to #28 cm#, by how much must the centripetal force applied by the tracks change?

Julia Aguilar · Accepted Answer

The centripetal force changes by #=0.23N#

Centripetal force is what

#F=(mv^2)/r#

mass, #m=5kg#

speed, #v=0.14ms^-1#

radius, #=r#

The centripetal force fluctuation is

#DeltaF=F_2-F_1#

#F_1=5*0.14^2/0.84=0.12N#

#F_2=5*0.14^2/0.28=0.35N#

#DeltaF=0.35-0.12=0.23N#

Adrian Ayala · Answer

undefined The centripetal force applied by the tracks must change according to the difference in curvature of the track. The centripetal force is given by the formula:

$ F = \frac{mv^2}{r} $

where:
- $ F $ is the centripetal force,
- $ m $ is the mass of the train (5 kg),
- $ v $ is the velocity of the train (14 cm/s), and
- $ r $ is the radius of curvature of the track.

Initially, the radius of curvature of the track is 84 cm. Substituting the values into the formula:

$ F_1 = \frac{5 \times (14)^2}{84} $

$ F_1 = \frac{5 \times 196}{84} $

$ F_1 = \frac{980}{84} $

$ F_1 = 11.67 $ N

Now, the radius of curvature changes to 28 cm. Substituting the values into the formula:

$ F_2 = \frac{5 \times (14)^2}{28} $

$ F_2 = \frac{5 \times 196}{28} $

$ F_2 = \frac{980}{28} $

$ F_2 = 35 $ N

The change in centripetal force applied by the tracks is given by the difference between the forces:

$ \Delta F = F_2 - F_1 $

$ \Delta F = 35 - 11.67 $

$ \Delta F \approx 23.33 $ N

Therefore, the centripetal force applied by the tracks must increase by approximately 23.33 N when the curvature of the track changes from a radius of 84 cm to 28 cm.