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 $24 c m$ $24 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-08-24T15:35:54

The change in centripetal force is $= 0.167 N$ $=0.167N$

The centripetal force is

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

The mass, $m = (5) k g$ $m=(5)kg$

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

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=5*0.04^2/0.04=0.2N$

$F_2=mv^2/r_2=5*0.04^2/0.24=0.033N$

$DeltaF=0.2-0.033=0.167N$

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 24 cm24cm24 cm, by how much must the centripetal force applied by the tracks change?