A model train with a mass of $9 k g$ $9 kg$ is moving along a track at $15 \frac{c m}{s}$ $15 (cm)/s$ . If the curvature of the track changes from a radius of $46 c m$ $46 cm$ to $35 c m$ $35 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-02-22T10:46:18

The variation in centripetal force is $= 0.14 N$ $=0.14N$

The centripetal force is

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

$m = m a s s = 9 k g$ $m=mass=9kg$

$v = 0.15 m s^{- 1}$ $v=0.15ms^-1$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=9*0.15^2/0.46=0.44N$

$F_2=9*0.15^5/0.35=0.58N$

$DeltaF=0.58-0.44=0.14N$

A model train with a mass of 9 kg9kg9 kg is moving along a track at 15 (cm)/s15cms15 (cm)/s. If the curvature of the track changes from a radius of 46 cm46cm46 cm to 35 cm35cm35 cm, by how much must the centripetal force applied by the tracks change?