A model train with a mass of $2 k g$ $2 kg$ 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 $12 c m$ $12 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-08-12T07:05:39

The change in centripetal force is $= 0.03 N$ $=0.03N$

The centripetal force is

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

The mass, $m = (2) k g$ $m=(2)kg$

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

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=2*0.03^2/0.04=0.045N$

$F_2=mv^2/r_2=2*0.03^2/0.12=0.015N$

$DeltaF=0.045-0.015=0.03N$

A model train with a mass of 2 kg2kg2 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 12 cm12cm12 cm, by how much must the centripetal force applied by the tracks change?