A model train with a mass of #6 kg# is moving along a track at #8 (cm)/s#. If the curvature of the track changes from a radius of #3 cm# to #15 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #6  kg# is moving along a track at #8 (cm)/s#.  If the curvature of the track changes from a radius of #3 cm# to #15 cm#, by how much must the centripetal force applied by the tracks change?

Sebastian Acosta · Accepted Answer

The change in centripetal force is #=1.024N#

Centripetal force is what

#F=(mv^2)/r#

The mass, #m=(6)kg#

The speed, #v=(0.08)ms^-1#

The radius, #=(r) m#

#r_1=0.03m#

#r_2=0.15m#

The centripetal force fluctuation is

#DeltaF=F_2-F_1#

#F_1=mv^2/r_1=6*0.08^2/0.03=1.28N#

#F_2=mv^2/r_2=6*0.08^2/0.15=0.256N#

#DeltaF=F_1-F_2=1.28-0.256=1.024N#

Axel Acevedo · Answer

undefined To calculate the change in centripetal force, we first need to find the initial centripetal force and the final centripetal force.

Initial centripetal force:
F_initial = m * v^2 / r_initial

Final centripetal force:
F_final = m * v^2 / r_final

Change in centripetal force:
ΔF = F_final - F_initial

Substituting the given values:
F_initial = (6 kg) * (0.08 m/s)^2 / (0.03 m)
F_final = (6 kg) * (0.08 m/s)^2 / (0.15 m)

Calculating:
F_initial ≈ 12.8 N
F_final ≈ 1.7067 N

ΔF ≈ 1.7067 N - 12.8 N
ΔF ≈ -11.0933 N

Therefore, the centripetal force applied by the tracks must decrease by approximately 11.0933 Newtons.