A model train with a mass of $3 k g$ $3 kg$ 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 $21 c m$ $21 cm$ to $18 c m$ $18 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-08-13T16:08:44

The centripetal force changes by $= 0.002 N$ $=0.002N$

The centripetal force is

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

The mass, $m = (3) k g$ $m=(3)kg$

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

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=3*0.03^2/0.21=0.013N$

$F_2=mv^2/r_2=3*0.03^2/0.18=0.015N$

$DeltaF=0.015-0.013=0.002N$

A model train with a mass of 3 kg3kg3 kg is moving along a track at 3 (cm)/s3cms3 (cm)/s. If the curvature of the track changes from a radius of 21 cm21cm21 cm to 18 cm18cm18 cm, by how much must the centripetal force applied by the tracks change?