A model train with a mass of $5 k g$ $5 kg$ 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 $4 c m$ $4 cm$ to $2 c m$ $2 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2018-02-03T07:52:01

The change in ventripetal force is $= 0.2 N$ $=0.2N$

The centripetal force is

${\to F}_{C} = \frac{m v^{2}}{r} \cdot \to r$ $vecF_C=(mv^2)/r*vecr$

The mass is of the train $m = 5 k g$ $m=5kg$

The velocity of the train is $v = 0.04 m s^{- 1}$ $v=0.04ms^-1$

The radii are

$r_{1} = 0.04 m$ $r_1=0.04m$

and

$r_{2} = 0.02 m$ $r_2=0.02m$

The variation in the centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$||F_1||=5*0.04^2/0.04=0.2N$

$||F_2||=5*0.04^2/0.02=0.4N$

$DeltaF=F_2-F_1=0.4-0.2=0.2N$

A model train with a mass of 5 kg5kg5 kg is moving along a track at 4 (cm)/s4cms4 (cm)/s. If the curvature of the track changes from a radius of 4 cm4cm4 cm to 2 cm2cm2 cm, by how much must the centripetal force applied by the tracks change?