A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $14 \frac{c m}{s}$ $14 (cm)/s$ . If the curvature of the track changes from a radius of $84 c m$ $84 cm$ to $42 c m$ $42 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-04-06T11:38:46

The change in centripetal force is $= 0.094 N$ $=0.094N$

The centripetal force is

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

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

speed, $v = 0.14 m s^{- 1}$ $v=0.14ms^-1$

radius, $= r$ $=r$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=4*0.14^2/0.84=0.093N$

$F_2=4*0.14^2/0.42=0.187N$

$DeltaF=0.187-0.093=0.094N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 14 (cm)/s14cms14 (cm)/s. If the curvature of the track changes from a radius of 84 cm84cm84 cm to 42 cm42cm42 cm, by how much must the centripetal force applied by the tracks change?