A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $21 \frac{c m}{s}$ $21 (cm)/s$ . If the curvature of the track changes from a radius of $420 c m$ $420 cm$ to $140 c m$ $140 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-07-20T08:20:06

The change in certipetal force is $= 0.084 N$ $=0.084N$

The centripetal force is

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

mass, $m = 4 k g$ $m=4kg$

speed, $v = 0.21 m s^{- 1}$ $v=0.21ms^-1$

radius, $= (r) m$ $=(r) m$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=4*0.21^2/4.2=0.042N$

$F_2=mv^2/r_2=4*0.21^2/1.4=0.126N$

$DeltaF=0.126-0.042=0.084N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 21 (cm)/s21cms21 (cm)/s. If the curvature of the track changes from a radius of 420 cm420cm420 cm to 140 cm140cm140 cm, by how much must the centripetal force applied by the tracks change?