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 #16 cm# to #24 cm#, by how much must the centripetal force applied by the tracks change?

2 Answers
Apr 30, 2018

Given,

#m = 5"kg"#

#nu = (0.12"m")/"s"#

#r_1 = 0.16"m"#

#r_2 = 0.24"m"#

Now, recall,

#F_"R" = m * nu^2/r#

Hence, the centripetal force applied by the tracks on the model train must change by,

#DeltaF_"R" = mnu^2(1/r_1 - 1/r_2) approx 0.15"N"#

Note that as the radius increases, the centripetal force will generally decrease, such that,

#F_"R" propto 1/r#

Apr 30, 2018

Answer:

The change in centripetal force is #=0.15N#

Explanation:

The centripetal force is

#vecF_C=(mv^2)/r*vecr#

The mass is of the train #m=5kg#

The velocity of the train is #v=0.12ms^-1#

The radii of the tracks are

#r_1=0.16m#

and

#r_2=0.24m#

The variation in the centripetal force is

#DeltaF=F_2-F_1#

The centripetal forces are

#||F_1||=5*0.12^2/0.16=0.45N#

#||F_2||=5*0.12^2/0.24=0.30N#

#DeltaF=F_1-F_2=0.45-0.30=0.15N#