A model train with a mass of #5 kg# is moving along a track at #12 (cm)/s#. If the curvature of the track changes from a radius of #32 cm# to #24 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 #12 (cm)/s#.  If the curvature of the track changes from a radius of #32 cm# to #24 cm#, by how much must the centripetal force applied by the tracks change?

Elena Albarran · Accepted Answer

The change in centripetal force is #=0.075N#

Centripetal force is what

#F=(mv^2)/r#

The mass, #m=(5)kg#

The speed, #v=(0.12)ms^-1#

The radius, #=(r) m#

The centripetal force fluctuation is

#DeltaF=F_2-F_1#

#F_1=mv^2/r_1=5*0.12^2/0.32=0.225N#

#F_2=mv^2/r_2=5*0.12^2/0.24=0.3N#

#DeltaF=0.3-0.225=0.075N#

Emily Abbate · Answer

undefined The centripetal force applied by the tracks must change according to the change in curvature of the track.

Using the formula for centripetal force:

F = (m * v^2) / r

Where:
F = centripetal force
m = mass of the train (5 kg)
v = velocity of the train (12 cm/s)
r = radius of curvature of the track

Initial centripetal force (F_initial) with a radius of 32 cm:
F_initial = (5 kg * (12 cm/s)^2) / 32 cm

Final centripetal force (F_final) with a radius of 24 cm:
F_final = (5 kg * (12 cm/s)^2) / 24 cm

Change in centripetal force:
ΔF = F_final - F_initial