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 $21 c m$ $21 cm$ , by how much must the centripetal force applied by the tracks change?

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Answer 1 · 2017-07-20T05:52:34

The change in centripetal force is $= 0.02357 N$ $=0.02357N$

The centripetal force is

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

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

speed, $v = 0.06 m s^{- 1}$ $v=0.06ms^-1$

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

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mv^2/r_1=4*0.06^2/0.32=0.045N$

$F_2=mv^2/r_2=4*0.06^2/0.21=0.06857N$

$DeltaF=0.06857-0.045=0.02357N$

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 21 cm21cm21 cm, by how much must the centripetal force applied by the tracks change?