A model train with a mass of $4 k g$ $4 kg$ is moving along a track at $6 \frac{c m}{s}$ $6 (cm)/s$ . If the curvature of the track changes from a radius of $32 c m$ $32 cm$ to $15 c m$ $15 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-12-07T10:16:05

The change in centripetal force is $= 0.051 N$ $=0.051N$

The centripetal force is

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

The mass is $m = 4 k g$ $m=4kg$

The velocity is $v = 0.06 m s^{- 1}$ $v=0.06ms^-1$

The radii are

$r_{1} = 0.32 m$ $r_1=0.32m$

and

$r_{2} = 0.15 m$ $r_2=0.15m$

The variation in centripetal force is

$DeltaF=F_2-F_1$

The centripetal forces are

$F_1=4*0.06^2/0.32=0.045N$

$F_2=4*0.06^2/0.15=0.096N$

$DeltaF=0.096-0.045=0.051N$

A model train with a mass of 4 kg4kg4 kg is moving along a track at 6 (cm)/s6cms6 (cm)/s. If the curvature of the track changes from a radius of 32 cm32cm32 cm to 15 cm15cm15 cm, by how much must the centripetal force applied by the tracks change?