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

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Answer 1 · 2017-10-02T06:00:18

The change in centripetal force is $= 0.0016 N$ $=0.0016N$

The centripetal force is

$F = \frac{m v^{2}}{r}$ $F=(mv^2)/r$

The mass, $m = (1) k g$ $m=(1)kg$

The speed, $v = (0.04) m s^{- 1}$ $v=(0.04)ms^-1$

The radius, $= (r) m$ $=(r) m$

The variation in centripetal force is

$DeltaF=F_2-F_1$

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

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

$DeltaF=0.0027-0.0011=0.0016N$

A model train with a mass of 1 kg1kg1 kg is moving along a track at 4 (cm)/s4cms4 (cm)/s. If the curvature of the track changes from a radius of 144 cm144cm144 cm to 60 cm60cm60 cm, by how much must the centripetal force applied by the tracks change?