A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $3 \frac{c m}{s}$ $3 (cm)/s$ . If the curvature of the track changes from a radius of $4 c m$ $4 cm$ to $27 c m$ $27 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-04-16T14:46:38

The centripetal force changes by $= 0.077 N$ $=0.077N$

The centripetal force is

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

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

speed, $v = 0.03 m s^{- 1}$ $v=0.03ms^-1$

radius, $= r$ $=r$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=4*0.03^2/0.04=0.09N$

$F_2=4*0.03^2/0.27=0.013N$

$DeltaF=0.09-0.013=0.077N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 3 (cm)/s3cms3 (cm)/s. If the curvature of the track changes from a radius of 4 cm4cm4 cm to 27 cm27cm27 cm, by how much must the centripetal force applied by the tracks change?