A model train with a mass of $1 k g$ $1 kg$ is moving along a track at $24 \frac{c m}{s}$ $24 (cm)/s$ . If the curvature of the track changes from a radius of $16 c m$ $16 cm$ to $150 c m$ $150 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-05-12T06:26:01

The change in centripetal force is $= 0.32 N$ $=0.32N$

The centripetal force is

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

mass, $m = 1 k g$ $m=1kg$

speed, $v = 0.24 m s^{- 1}$ $v=0.24ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=1*0.24^2/0.16=0.36N$

$F_2=mv^2/r_2=1*0.24^2/1.50=0.038N$

$DeltaF=0.36-0.04=0.32N$

A model train with a mass of 1 kg1kg1 kg is moving along a track at 24 (cm)/s24cms24 (cm)/s. If the curvature of the track changes from a radius of 16 cm16cm16 cm to 150 cm150cm150 cm, by how much must the centripetal force applied by the tracks change?