A model train with a mass of $5 k g$ $5 kg$ is moving along a track at $14 \frac{c m}{s}$ $14 (cm)/s$ . If the curvature of the track changes from a radius of $84 c m$ $84 cm$ to $28 c m$ $28 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-04-06T12:59:39

The centripetal force changes by $= 0.23 N$ $=0.23N$

The centripetal force is

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

mass, $m = 5 k g$ $m=5kg$

speed, $v = 0.14 m s^{- 1}$ $v=0.14ms^-1$

radius, $= r$ $=r$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=5*0.14^2/0.84=0.12N$

$F_2=5*0.14^2/0.28=0.35N$

$DeltaF=0.35-0.12=0.23N$

A model train with a mass of 5 kg5kg5 kg is moving along a track at 14 (cm)/s14cms14 (cm)/s. If the curvature of the track changes from a radius of 84 cm84cm84 cm to 28 cm28cm28 cm, by how much must the centripetal force applied by the tracks change?