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 $88 c m$ $88 cm$ to $28 c m$ $28 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-02-26T17:08:42

The change in centripetal force is $= 0.239 N$ $=0.239N$

The centripetal force is

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

$m = m a s s = 5 k g$ $m=mass=5kg$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=5*0.14^2/0.88=0.1114N$

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

$DeltaF=0.35-0.1114=0.239N$

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 88 cm88cm88 cm to 28 cm28cm28 cm, by how much must the centripetal force applied by the tracks change?