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A model train with a mass of #1 kg# is moving along a track at #4 (cm)/s#. If the curvature of the track changes from a radius of #144 cm# to #60 cm#, by how much must the centripetal force applied by the tracks change?

Answer 1

Adrian Ayala

The change in centripetal force is #=0.0016N#

Centripetal force is what

#F=(mv^2)/r#

The mass, #m=(1)kg#

The speed, #v=(0.04)ms^-1#

The radius, #=(r) m#

The centripetal force fluctuation is

#DeltaF=F_2-F_1#

#F_1=mv^2/r_1=1*0.04^2/1.44=0.0011N#

#F_2=mv^2/r_2=1*0.04^2/0.60=0.0027N#

#DeltaF=0.0027-0.0011=0.0016N#

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Answer 2

Charles Aeschliman

The centripetal force applied by the tracks must change by a factor of:

F = m * v^2 / r

Where: F = centripetal force m = mass of the train v = velocity of the train r = radius of curvature

Initially, the radius of curvature is 144 cm:

F_initial = (1 kg) * (4 cm/s)^2 / 144 cm = 0.111 N

After the change, the radius of curvature becomes 60 cm:

F_final = (1 kg) * (4 cm/s)^2 / 60 cm = 0.267 N

Therefore, the centripetal force applied by the tracks must increase by 0.267 N - 0.111 N = 0.156 N.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.